The algebraic fractions are type of fractions where algebraic formulas - identities are in nominator and denominator. All the rules that are valid for the set of real numbers are valid for algebraic fractions. The algebraic fractions can be canceled (shorten) only when nominator and denominator are written in the form of the product and have a common factor.
- Example 1 Shorten the fraction \(\frac{9-4a^2}{2a^2+3a}\)
Solution$$\frac{9-4a^2}{2a^2+3a} = \frac{(3-2a)(3+2a)}{a(2a+3)} = \frac{3-2a}{a}$$ - Example 2 Shorten the fraction \(\frac{(3a-6)^2}{a^2-2a}\)
Solution$$ \frac{(3a-6)^2}{a^2-2a} = \frac{9(a-2)^2}{a(a-2)} = \frac{9(a-2)}{a}$$ - Example 3 Shorten the fraction \(\frac{(2x-y)^2 - 2x+y}{4x^2-y^2-2x-y}\)
Solution$$ \frac{(2x-y)^2 - 2x+y}{4x^2-y^2-2x-y} = \frac{(2x-y)^2-(2x-y)}{(2x-y)(2y+y)-(2x+y)} = \frac{(2x-y)\left((2x-y)-1\right)}{(2x+y)\left((2x-y)-1\right)} = \frac{2x-y}{2x+y}$$ - Example 4 Shorten the fraction \(\frac{a^2-1}{1-4a^2}\cdot\left(\frac{3a}{a+1}-1\right)\)
Solution$$ \frac{a^2-1}{1-4a^2}\cdot\left(\frac{3a}{a+1}-1\right) = \frac{(a-1)(a+1)}{(1-2a)(1+2a)}\cdot \frac{3a-a-1}{a+1} = \frac{a-1}{(1-2a)(1+2a)} \cdot (2a-1) = $$ $$-\frac{a-1}{(1-2a)(1+2a)} \cdot (1-2a) = \frac{1-a}{1-2a}$$ - Example 5 Simplify the expression \(\frac{m^2 + 2mp + p^2}{m^2-p^2}:\frac{2m+2p}{m^2-mp}\)
Solution $$ \frac{m^2 + 2mp + p^2}{m^2-p^2}:\frac{2m+2p}{m^2-mp} = \frac{(m+p)^2}{(m-p)(m+p)}\cdot\frac{m(m-p)}{2(m+p)} = \frac{m}{2}.$$ - Example 6 Shorten the fraction \(\)
Solution $$ \frac{2a}{a+\frac{1}{a+\frac{1-a^2}{a}}} = \frac{2a}{a+\frac{1}{\frac{a^2 + 1-a^2}{a}}} = \frac{2a}{a+\frac{1}{\frac{1}{a}}} = \frac{2a}{a+a} = 1$$ - Example 7 Shorten the fraction \(\frac{9a^2-4}{6a+4}\)
Solution$$\frac{9a^2-4}{6a+4} = \frac{(3a-2)(3a+2)}{2(3a+2)} = \frac{3a-2}{2} $$ - Example 8 Shorten the fraction \(\frac{1}{a} - \frac{1}{b}\)
Solution$$ \frac{1}{a} - \frac{1}{b} = \frac{b-a}{ab} $$ - Example 9 Shorten the fraction \(\frac{x-3}{x^2 - 9}\)
Solution$$ \frac{x-3}{x^2 - 9} = \frac{x-3}{(x-3)(x+3)} = \frac{1}{x+3}$$ - Example 10 Shorten the fraction \(\frac{x^2 -10x+25}{x^2-25}\)
Solution$$ \frac{x^2 -10x+25}{x^2-25} = \frac{(x-5)^2}{(x-5)(x+5)} = \frac{x-5}{x-5}$$ - Example 11 Shorten the fraction \(\left(\frac{x-3}{x+3} - \frac{x+3}{x-3}\right):\frac{x}{x^2-9}\)
Solution$$ \left(\frac{x-3}{x+3} - \frac{x+3}{x-3}\right):\frac{x}{x^2-9} = \left(\frac{(x-3)^2 - (x+3)^2}{(x+3)(x-3)}\right)\cdot \frac{(x-3)(x+3)}{x} = \frac{x^2-6x+9-x^2-6x-9}{x} = -\frac{12x}{x} = -12$$ - Example 12 Shorten the fraction \(\frac{1}{a-3} - \frac{6}{a^2-9}\)
Solution $$ \frac{1}{a-3} - \frac{6}{a^2-9} = \frac{1}{(a-3)} - \frac{6}{(a-3)(a+3)} = \frac{a+3-6}{(a-3)(a+3)} = \frac{a-3}{(a-3)(a+3)} = \frac{1}{a+3}$$ - Example 13 Shorten the fraction \(\frac{1}{ab} - \frac{1}{ac} + \frac{1}{bc}\)
Solution$$ \frac{1}{ab} - \frac{1}{ac} + \frac{1}{bc} = \frac{c - b + a}{abc} = \frac{a-b+c}{abc}$$ - Example 14 Shorten the fraction \(\left(6-3a + \frac{18a^2}{6+3a}\right):\frac{9a^4-144}{6a^3+48}\)
Solution$$\left(6-3a + \frac{18a^2}{6+3a}\right):\frac{9a^4-144}{6a^3+48} = \frac{36-9a^2+18a^2}{6+3a}\cdot \frac{6(a+2)(a^2-2a+4)}{9(a^2-4)(a^2+4)} = \frac{9 (a^2+4)}{3(a+2)}\cdot \frac{6(a+2)(a-2)^2}{9(a^2-4)(a^2+4)}$$ $$ = \frac{2(a-2)^2}{(a^2-4)} = \frac{2(a^2-4a+4)}{(a^2-4)}$$ - Example 15 Shorten the fraction \(\frac{a^{-3}+a^{-2}}{a^{-2}-1}:\frac{1}{a^2}\)
Solution$$\frac{a^{-3}+a^{-2}}{a^{-2}-1}:\frac{1}{a^2} = \frac{\frac{1}{a^3} + \frac{1}{a^2}}{\frac{1}{a^2}-1}\cdot a^2 =\frac{\frac{a+1}{a^3}}{\frac{1-a^2}{a^2}}\cdot a^2 = \frac{\frac{a+1}{a^3}}{\frac{(1-a)(1+a)}{a^2}}\cdot a^2 $$ $$ = \frac{a^2}{a(1-a)} = \frac{1}{1-a}$$ - Example 16 Shorten the expression \(3-\frac{1+2a}{a}\)
Solution$$ 3-\frac{1+2a}{a} = \frac{3a-1-2a}{a} = \frac{a-1}{a}$$ - Example 17 Shorten the fraction \(\frac{1-x^{-3}y^{-3}}{x^{-2}y^{-2}+x^{-1}y^{-1}+1}\)
Solution$$ \frac{1-x^{-3}y^{-3}}{x^{-2}y^{-2}+x^{-1}y^{-1}+1} = \frac{\frac{x^3y^3-1}{x^3y^3}}{\frac{1}{x^2y^2} + \frac{1}{xy} + 1} = \frac{\frac{(xy-1)(x^2y^2 + xy + 1)}{x^3y^3}}{\frac{1+xy+x^2y^2}{x^2y^2}} = \frac{xy-1}{xy}$$ - Example 18 Shorten the fraction \(\frac{1}{(a-b)(a-c)} + \frac{1}{(b-c)(b-a)} + \frac{1}{(c-a)(c-d)}\)
Solution $$ \frac{1}{(a-b)(a-c)} + \frac{1}{(b-c)(b-a)} + \frac{1}{(c-a)(c-d)} = \frac{1}{(a-b)(a-c)} - \frac{1}{(a-b)(b-c)} + \frac{1}{(a-c)(b-c)} = \frac{b-c - (a-c) + a-b}{(a-b)(b-c)(a-c)} = 0$$ - Example 19 Shorten the fraction \(\frac{(xy+1)^2-(x+y)^2}{(x^2-1)(y^2-1)}\)
Solution$$\frac{(xy+1)^2-(x+y)^2}{(x^2-1)(y^2-1)} = \frac{x^2y^2 + 2xy + 1 - x^2 - 2xy - y^2}{x^2y^2-x^2 -y^2 +1} = \frac{x^2y^2-x^2 - y^2 + 1}{x^2y^2-x^2 - y^2 + 1} = 1$$ - Example 20 Shorten the fraction \(\frac{-(-a)^5\cdot(-a^{-2})^2 + 1}{(-a^{-3})^{-1}\cdot(-(-a)^4)\cdot a^{-5} -1}\)
Solution $$ \frac{-(-a)^5\cdot(-a^{-2})^2 + 1}{(-a^{-3})^{-1}\cdot(-(-a)^4)\cdot a^{-5} -1} = \frac{a^5a^{-4} +1}{a^3a^4a^{-5} - 1} = \frac{a+1}{a^2-1} = \frac{a+1}{(a-1)(a+1)} = \frac{1}{a-1}$$ - Example 21 Shorten the expression \(\frac{x+1}{x^2+x+1}:\frac{1}{x^3-1}\) and determine the value of the expression if the value of variable \(x\) is \(\sqrt{2}\)
Solution$$\frac{x+1}{x^2+x+1}:\frac{1}{x^3-1} = \frac{x+1}{x^2+x+1} \cdot (x-1)(x^2+x+1) = x^2-1$$ For \(x=\sqrt{2}\) $$ x^2-1 = 2-1 = 1$$ - Example 22 If \(a=\sqrt{3}\) and \(b = \sqrt{6}\) Then the expression \(\left[\left(\frac{8a^3-b^3}{4a^2+2ab+b^2}\right)^2 - \frac{16a^4-b^4}{4a^2-b^2}\right]^2\) is equal to?
Solution The frist and the second part of the expression must be simplified. The simplification of the first expression $$ \left(\frac{8a^3-b^3}{4a^2+2ab+b^2}\right)^2 = \left(\frac{(2a-b)(4a^2+2ab+b^2)}{4a^2+ 2ab+b^2}\right)^2 = \left(2a-b\right)^2 = 4a^2-4ab+b^2$$ The simplification of the second expression $$ \frac{16a^4-b^4}{4a^2-b^2} = \frac{(4a^2-b^2)(4a^2+b^2)}{4a^2-b^2} = 4a^2+b^2$$ Combining previous two expressions together the following simplified version of the expression is obtained. $$ \left[4a^2-4ab+b^2 - 4a^2 - b^2\right]^2 = \left(-4ab\right)^2 = 16a^2b^2$$ For \(a = \sqrt{3}\) , and \(b=\sqrt{6}\) the solution of the expression is $$ 16a^2b^2 = 16\cdot 3 \cdot 6 = 288.$$ - Example 23 Simplify the expression \(\left(\frac{1}{a}-\frac{1}{b}\right)\cdot\frac{a^2 + ab}{a^2-b^2}\)
Solution$$ \left(\frac{1}{a}-\frac{1}{b}\right)\cdot\frac{a^2 + ab}{a^2-b^2} = -\frac{a-b}{ab}\cdot \frac{a(a+b)}{(a-b)(a+b)} = -\frac{1}{b}. $$ - Example 24Simplify the expression \(\frac{\left(ab^{-3}-a^{-3}b\right)^{-1}\cdot \left(a^{-2} + b^{-2}\right)}{\left(b^{-2}-a^{-2}\right)^{-1}}\)
Solution$$ \frac{\left(ab^{-3}-a^{-3}b\right)^{-1}\cdot \left(a^{-2} + b^{-2}\right)}{\left(b^{-2}-a^{-2}\right)^{-1}} = \frac{\left(a^{-2} + b^{-2}\right)\left(b^{-2}-a^{-2}\right)}{\left(ab^{-3}-a^{-3}b\right)} = \frac{\left(\frac{a^2 + b^2}{a^2b^2}\right)\left(\frac{a^2 -b^2}{a^2b^2}\right)}{\frac{a^4 - b^4}{a^3b^3}} = $$ $$ \frac{\left(\frac{a^2 + b^2}{a^2b^2}\right)\left(\frac{a^2 -b^2}{a^2b^2}\right)}{\frac{a^4 - b^4}{a^3b^3}} = \frac{\frac{a^4-b^4}{a^4b^4}}{\frac{a^4-b^4}{a^3b^3}} = \frac{1}{ab}$$ - Example 25 - Simplify the expression \(\frac{a^{-2}-a^{-1}b^{-1}+b^{-2}}{a^{-3}+b^{-3}}:\left(\frac{a+b}{ab}\right)^{-1}\)
Solution$$\frac{a^{-2}-a^{-1}b^{-1}+b^{-2}}{a^{-3}+b^{-3}}\cdot \left(\frac{a+b}{ab}\right) = \frac{\frac{1}{a^2} - \frac{1}{ab} + \frac{1}{b^2}}{\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{1}{a^2}-\frac{1}{ab} + \frac{1}{b^2}\right)} \cdot \left(\frac{a+b}{ab}\right) = \frac{\frac{1}{1}}{\frac{a+b}{ab}}\cdot\left(\frac{a+b}{ab}\right) = $$ $$ = \frac{ab}{a+b} \cdot \frac{a+b}{ab} = 1$$ - Example 26 Simplify the expression \(\frac{2a^3 - 6a^2 +5a - 15}{9-a^2}\)
Solution$$\frac{2a^3 - 6a^2 +5a - 15}{9-a^2} = \frac{2a^2(a-3) + 5(a-3)}{(3-a)(3+a)} = \frac{(2a^2+5)(a-3)}{(a-3)(a+3)} = \frac{2a^2+5}{a+3}$$ - Example 27 Simplify the expression \(\frac{a^4+1-2a^2}{1-a-a^2 +a^3}\)
Solution$$\frac{a^4+1-2a^2}{1-a-a^2 +a^3} = \frac{(a^2-1)^2}{a^2(a-1)-(a-1)} = \frac{(a^2-1)^2}{(a-1)(a^2-1)} = \frac{a^2-1}{a-1} $$ $$\frac{(a-1)(a+1)}{a-1} = a+1 $$ - Example 28 Simplify the expression \(\left(5^n - \frac{4\cdot 35^n}{5^n+7^n} + 7^n\right):\left(\frac{5^n}{5^n+7^n} - \frac{7^n}{7^n-5^n}-\frac{2\cdot 35^n}{25^n-49^n}\right)\)
Solution Simplify the first part of the expression$$\left(5^n - \frac{4\cdot 35^n}{5^n+7^n} + 7^n\right) = \frac{5^n(5^n + 7^n)-4\cdot 35^n + 7^n(5^n + 7^n)}{5^n + 7^n} = \frac{(5^n+7^n)^2 - 4\cdot 35^n}{5^n + 7^n} = $$ $$\frac{5^{2n} + 2\cdot 35^n + 7^{2n} - 4\cdot 35^n}{5^n + 7^n} = \frac{(5^n-7^n)^2}{5^n + 7^n}$$ The next step is to simplify the second expression. $$ \left(\frac{5^n}{5^n+7^n} - \frac{7^n}{7^n-5^n}-\frac{2\cdot 35^n}{25^n-49^n}\right) = \frac{5^n}{5^n+7^n} + \frac{7^n}{5^n-7^n} -\frac{2\cdot 35^n}{\left(5^n-7^n\right)\left(5^n+7^n\right)} = $$ $$\frac{5^n\left(5^n-7^n\right) + 7^n\left(5^n+7^n\right) - 2\cdot 35^n}{\left(5^n-7^n\right)\left(5^n+7^n\right)} = \frac{5^{2n}-35^n+35^n+7^{2n}-2\cdot 35^n}{\left(5^n-7^n\right)\left(5^n+7^n\right)}$$ $$ \frac{(5^n-7^n)^2}{\left(5^n-7^n\right)\left(5^n+7^n\right)} = \frac{(5^n-7^n)}{\left(5^n+7^n\right)}$$ The final solution can be written in the following form $$ \frac{(5^n-7^n)^2}{5^n + 7^n}:\frac{(5^n-7^n)}{\left(5^n+7^n\right)} = \frac{(5^n-7^n)^2}{5^n + 7^n} \cdot \frac{5^n+7^n}{5^n-7^n} = 5^n-7^n$$ - Example 29 Simplify the expression \(\frac{4x-2}{2x^2 + 5x-3}\)
Solution $$ \frac{4x-2}{2x^2 + 5x-3} = \frac{2(2x-1)}{2x^2 + 6x - x - 3} = \frac{2(2x-1)}{x(2x-1)+3(2x-1)} = \frac{2(2x-1)}{(2x-1)(x-3)} = \frac{2}{x-3}$$ - Example 30 Simplify the expression \(\frac{x^2+7x}{4x^2-6x}:\frac{x^2+3x-28}{2x^2-32}\)
Solution The first step is to simplify the first algebraic fraction. $$ \frac{x^2+7x}{4x^2-6x} = \frac{x(x+7)}{2x(2x-3)} = \frac{x+7}{2(2x-3)}$$ Then the second algebraic fraction is simplified. $$ \frac{x^2+3x-28}{2x^2-32} = \frac{x^2 + 7x - 4x -28}{2(x^2-16)} = \frac{(x-4)(x+7)}{2(x-4)(x+4)} = \frac{x+7}{2(x+4)}$$ Putting together the simplified version of first and second algebraic fractions the following solution is obtained. $$ \frac{x+7}{2(2x-3)} : \frac{x+7}{2(x+4)} = \frac{x+7}{2(2x-3)}\cdot \frac{2(x+4)}{x+7} = \frac{x+4}{2x-3}$$ - Example 31 Simplify the expression \(\left[\left(\frac{a^2-16}{a^2+7a+12}\cdot \frac{a^2-4a-21}{a^2-4a}\right)\cdot \frac{1}{a-7}\right]^{-1}\)
Solution Since the entire expression is on the power of -1 the expression can be wrriten as $$\begin{eqnarray}\left(\frac{a^2+7a+12}{a^2-16}\cdot \frac{a^2-4a}{a^2-4a-21}\right)\cdot (a-7) &=& \frac{a^2+4a+3a+12}{(a-4)(a+4)}\cdot \frac{a(a-4)}{a^2+3a-7a-12} \cdot (a-7) = \\\nonumber \left(\frac{(a+3)(a+4)}{(a-4)(a+4)}\cdot \frac{a(a-4)}{(a+3)(a-7)}\right)\cdot (a-7) &=& a\end{eqnarray}$$ - Example 32 Simplify the expression \(\left(\frac{x-y}{x^2+xy}-\frac{x}{xy+y^2}\right):\left(\frac{y^2}{x^3-xy^2}+\frac{1}{x+y}\right)\)
Solution The first part of the expression in parenthesis can be simplifed as $$ \left(\frac{x-y}{x^2+xy}-\frac{x}{xy+y^2}\right) = \frac{x-y}{x(x+y)} - \frac{x}{y(x+y)} = \frac{xy-y^2-x^2}{xy(x+y)}$$ The second part of the expression (in parenthesis) can be simplifed as $$ \frac{y^2}{x^3-xy^2}+\frac{1}{x+y} = \frac{y^2}{x(x^2-y^2)}+\frac{1}{x+y} = \frac{y^2+x(x-y)}{x(x-y)(x+y)} = \frac{y^2+x^2-xy}{x(x-y)(x+y)}$$ Putting it all together $$ \frac{xy-y^2-x^2}{xy(x+y)}:\frac{y^2+x^2-xy}{x(x-y)(x+y)} = \frac{xy-y^2-x^2}{xy(x+y)} \cdot \frac{x(x-y)(x+y)}{y^2+x^2-xy} = - \frac{x-y}{y}$$ If \(x=\frac{1}{3}\) and \(y=\frac{1}{2}\) then the solution is $$ - \frac{x-y}{y} = -\frac{ \frac{1}{3}-\frac{1}{2}}{\frac{1}{2}} = -\frac{\frac{2-3}{6}}{\frac{1}{2}} = \frac{\frac{1}{6}}{\frac{1}{2}} = \frac{1}{3}$$ - Example 33 Simplify the expression \(\frac{4}{x+2}-\frac{3}{x-2}+\frac{12}{x^2-4}\)
Solution $$\frac{4}{x+2}-\frac{3}{x-2}+\frac{12}{x^2-4} = \frac{4(x-2)-3(x+2)+12}{(x-2)(x+2)} = $$ $$ \frac{4x-8-3x-6+12}{(x-2)(x+2)} = \frac{x-2}{(x-2)(x+2)} = \frac{1}{x+2}$$ - Example 34 Simplify the expression \(\frac{(1-3x)^2-2x+6x^2}{25x^3-x}\)
Solution $$ \frac{(1-3x)^2-2x+6x^2}{25x^3-x} = \frac{1-6x+9x^2-2x+6x^2}{x(25x^2-1)} = \frac{15x^2-8x+1}{x(5x-1)(5x+1)} = $$ $$ \frac{15x^2-5x-3x+1}{x(5x-1)(5x+1)} = \frac{3x(5x-1)-(5x-1)}{x(5x-1)(5x+1)} = \frac{(3x-1)(5x-1)}{x(5x-1)(5x+1)} = \frac{3x-1}{x(5x+1)}$$ - Example 35 Simplify the expression \(\frac{4x^2-20x+25}{4x-10}\)
Solution $$\frac{4x^2-20x+25}{4x-10} = \frac{(2x-5)^2}{2(2x-5)} = \frac{2x-5}{2} $$ - Example 36 Simplify the expression \(\frac{9x^2-6x+1}{9x^2-1}\)
Solution $$ \frac{9x^2-6x+1}{9x^2-1} = \frac{(3x-1)^2}{(3x-1)(3x+1)} = \frac{3x-1}{3x+1}$$ - Example 37 Simplify the expression \(\frac{a^5-8a^3+16a}{2a^3-4a^2-8a+16}\)
Solution $$\frac{a^5-8a^3+16a}{2a^3-4a^2-8a+16} = \frac{a(a^4-8a^2+16)}{2a(a^2-4)-4(a^2-4)} = \frac{a(a^2-4)^2}{(2a-4)(a^2-4)} = \frac{a(a-2)(a+2)}{2(a-2)} = \frac{a(a+2)}{2}$$ - Example 38 Simplify the expression \(\frac{(x-3)^2}{9-x^2}\)
Solution $$ \frac{(x-3)^2}{9-x^2} = -\frac{(x-3)^2}{(x-3)(x+3)} = \frac{3-x}{3+x}$$ - Example 39 Simplify the expression \(\frac{a^3-a^2+2a-2}{a^3+3a^2+2a+6}\)
Solution $$ \frac{a^3-a^2+2a-2}{a^3+3a^2+2a+6} = \frac{a(a^2+2) - (a^2+2)}{a(a^2+2)+3(a^2+2)} = \frac{(a-1)(a^2+2)}{(a+3)(a^2+2)} = \frac{a-1}{a+3}$$ - Example 40 Simplify the expression \(\left(4-\frac{4+b^2}{b}\right):\left(\frac{1}{2}-\frac{1}{b}\right)\)
Solution $$ \left(4-\frac{4+b^2}{b}\right):\left(\frac{1}{2}-\frac{1}{b}\right) = \frac{4b-4-b^2}{b}:\left(\frac{b-2}{2b}\right) = \frac{-(b-2)^2}{b}\cdot \frac{2b}{b-2} = -2(b-2) = 4-2b$$ - Example 41 Simplify the expression \(\left(2a-\frac{4a-1}{2a}\right)\cdot\frac{4a^2}{1-4a^2}:\left(\frac{1}{a}-\frac{4}{1+2a}\right)\)
Solution $$ \left(2a-\frac{4a-1}{2a}\right)\cdot\frac{4a^2}{1-4a^2}:\left(\frac{1}{a}-\frac{4}{1+2a}\right) = \frac{4a^2-4a+1}{2a}\cdot \frac{4a^2}{(1-2a)(1+2a)}:\frac{1+2a-4a}{a(1+2a)} = \frac{(1-2a)^2 4a^2}{2a(1-2a)(1+2a)}\cdot \frac{a(1+2a)}{1-2a} = 2a^2$$ - Example 42 Simplify the expression \(\frac{2c}{c-\frac{1}{c-\frac{1+c^2}{c}}}\)
Solution $$ \frac{2c}{c-\frac{1}{c-\frac{1+c^2}{c}}} = \frac{2c}{c-\frac{1}{\frac{c^2-1-c^2}{c}}} = \frac{2c}{c+\frac{1}{\frac{1}{c}}} = \frac{2c}{2c} = 1$$ - Example 43 Simplify the expression \(\left(d+\frac{9}{d-6}\right)\left(\frac{12}{d^2-3d}-\frac{d}{9-6d+d^2}\right)\)
Solution $$\left(d+\frac{9}{d-6}\right)\left(\frac{12}{d^2-3d}-\frac{d}{9-6d+d^2}\right) = \frac{d^2-6d+9}{d-6}\left(\frac{12}{d(d-3)}-\frac{d}{(d-3)^2}\right) = \frac{(d-3)^2}{d-6}\frac{12(d-3)-d^2}{d(d-3)^2} = \frac{1}{d-6}\frac{12d-36-d^2}{d} = \frac{-(d-6)^2}{d(d-6)} = -\frac{d-6}{d} =\frac{6-d}{d}$$ - Example 44 Simplify the expression \(\left(\frac{2}{x^2-4}+\frac{1}{2x-x^2}\right):\frac{1}{x^2+4x+4}\)
Solution $$ \left(\frac{2}{x^2-4}+\frac{1}{2x-x^2}\right):\frac{1}{x^2+4x+4} = \frac{2x-x-2}{x(x-2)(x+2)}\cdot (x+2)^2 = \frac{(x-2)(x+2)}{x(x-2)} = \frac{x+2}{x}$$ - Example 45 Simplify the expression \(\left(\frac{a-3}{a^2+3a}+\frac{a}{9-a^2}\right)\cdot \frac{a^2-3a}{2a-3}-\frac{a}{a+3}\)
Solution $$ \left(\frac{a-3}{a^2+3a}+\frac{a}{9-a^2}\right)\cdot \frac{a^2-3a}{2a-3}-\frac{a}{a+3} = -\frac{3(2a-3)}{a(a-3)(a+3)}\cdot \frac{a(a-3)}{2a-3}-\frac{a}{a+3} = -\frac{3}{a+3} - \frac{a}{a+3} =\frac{-(a+3)}{a+3} = -1 $$ - Example 46 Simplify the expression \(\frac{f^2}{f+5} + \frac{25f}{5-f}\cdot\left(\frac{f-5}{f^2-25}+ \frac{5}{f^2-5f}\right)\)
Solution $$ \frac{f^2}{f+5} + \frac{25f}{5-f}\cdot\left(\frac{f-5}{f^2-25}+ \frac{5}{f^2-5f}\right) = \frac{f^2}{f+5} + \frac{25f}{5-f}\cdot \frac{f^2-15f + 5f+25}{f(f-5)(f+5)} = $$ $$ \frac{f^2}{f+5} + \frac{25f}{5-f}\cdot \frac{(f-5)^2}{f(f-5)(f+5)} = \frac{f^2}{f+5} - \frac{25}{f+5} = \frac{f^2-25}{f+5} = \frac{(f-5)(f+5)}{f+5} = f-5$$ - Example 47 Simplify the expression \(\frac{2d}{d+s} + \frac{(d-s)^2}{s}\cdot\left(\frac{s}{(s-d)^2} + \frac{s}{s^2-d^2}\right)\)
Solution $$ \frac{2d}{d+s} + \frac{(d-s)^2}{s}\cdot\left(\frac{s}{(s-d)^2} + \frac{s}{s^2-d^2}\right) =\frac{2d}{d+s} + \frac{(d-s)^2}{s}\cdot\frac{s^2+sd+s^2-sd}{(s-d)^2(s+d)} = \frac{2d}{d+s} + \frac{(s-d)^2}{s}\cdot \frac{2s^2}{(s-d)^2(s+d)} = \frac{2d}{d+s} + \frac{2s}{s+d} = \frac{2(s+d)}{s+d} = 2$$ - Example 48 Simplify the expression \(\left(1+\frac{1}{y}\right)\left(1-\frac{1}{y}\right)\left(1-\frac{1}{y-1}\right)\left(1-\frac{1}{y+1}\right)\)
Solution $$ \left(1+\frac{1}{y}\right)\left(1-\frac{1}{y}\right)\left(1-\frac{1}{y-1}\right)\left(1-\frac{1}{y+1}\right) = \frac{y+1}{y}\frac{y-1}{y}\frac{y-1-1}{y-1} \frac{y+1-1}{y+1} = \frac{y-2}{y}$$
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