Simplify:
Q25. $\dfrac{a^{2} - 4b^{2}}{a^{2} - 9b^{2}} \times \dfrac{a - 3b}{a + 2b}$
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$\Rightarrow \dfrac{a^{2} - (2b)^{2}}{a^{2} - (3b)^{2}} \times \dfrac{a - 3b}{a + 2b}$
$\Rightarrow \dfrac{(a - 2b)^{2}}{(a - 3b)^{2}} \times \dfrac{a - 3b}{a + 2b}$
$\Rightarrow \dfrac{(a + 2b)(a - 2b)}{(a + 3b)(a - 3b)} \times \dfrac{a - 3b}{a + 2b}$
$\therefore \dfrac{a - 2b}{a + 3b}$ Ans
$\Rightarrow \dfrac{(a - 2b)^{2}}{(a - 3b)^{2}} \times \dfrac{a - 3b}{a + 2b}$
$\Rightarrow \dfrac{(a + 2b)(a - 2b)}{(a + 3b)(a - 3b)} \times \dfrac{a - 3b}{a + 2b}$
$\therefore \dfrac{a - 2b}{a + 3b}$ Ans
Q26. $\dfrac{a^{2} + 3a}{a^{2} + 4a + 3} \times \dfrac{a^{2} - 2a - 3}{a^{2} - 9}$
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$\Rightarrow \dfrac{a(a + 3)}{a^{2} + 3a + a + 3} \times \dfrac{a^{2} - 3a + a - 3}{a^{2} - 3^{2}}$
$\Rightarrow \dfrac{a(a + 3)}{a(a + 3) + 1(a + 3)} \times \dfrac{a(a - 3) + 1(a - 3)}{(a - 3)^{2}}$
$\Rightarrow \dfrac{a(a + 3)}{(a + 3)(a + 1)} \times \dfrac{(a - 3)(a + 1)}{(a + 3)(a - 3)}$
$\therefore \dfrac{a(a - 3)}{(a + 3)(a - 3)} = \dfrac{a}{a + 3}$ Ans
$\Rightarrow \dfrac{a(a + 3)}{a(a + 3) + 1(a + 3)} \times \dfrac{a(a - 3) + 1(a - 3)}{(a - 3)^{2}}$
$\Rightarrow \dfrac{a(a + 3)}{(a + 3)(a + 1)} \times \dfrac{(a - 3)(a + 1)}{(a + 3)(a - 3)}$
$\therefore \dfrac{a(a - 3)}{(a + 3)(a - 3)} = \dfrac{a}{a + 3}$ Ans
Q27. $\dfrac{a^{2} - 5a}{3a - 4b} \div \dfrac{a^{2} - 25}{9a^{2} - 16b^{2}}$
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$\Rightarrow \dfrac{a(a - 5)}{3a - 4b} \div \dfrac{a^{2} - 5^{2}}{(3a)^{2} - (4b)^{2}}$
$\Rightarrow \dfrac{a(a - 5)}{3a - 4b} \div \dfrac{(a - 5)^{2}}{(3a - 4b)^{2}}$
$\Rightarrow \dfrac{a(a - 5)}{3a - 4b} \div \dfrac{(a + 5)(a - 5)}{(3a + 4b)(3a - 4b)}$
$\Rightarrow \dfrac{a(a - 5)}{3a - 4b} \times \dfrac{(3a + 4b)(3a - 4b)}{(a + 5)(a - 5)}$
$\therefore \dfrac{a(3a + 4b)}{a + 5}$ Ans
$\Rightarrow \dfrac{a(a - 5)}{3a - 4b} \div \dfrac{(a - 5)^{2}}{(3a - 4b)^{2}}$
$\Rightarrow \dfrac{a(a - 5)}{3a - 4b} \div \dfrac{(a + 5)(a - 5)}{(3a + 4b)(3a - 4b)}$
$\Rightarrow \dfrac{a(a - 5)}{3a - 4b} \times \dfrac{(3a + 4b)(3a - 4b)}{(a + 5)(a - 5)}$
$\therefore \dfrac{a(3a + 4b)}{a + 5}$ Ans
Q28. $\dfrac{2a^{2} - 3a - 2}{a^{2} - a - 6} \div \dfrac{3a^{2} - 5a - 2}{3a^{2} - 8a - 3}$
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$\Rightarrow \dfrac{2a^{2} - 4a + a - 2}{a^{2} - 3a + 2a - 6} \div \dfrac{3a^{2} - 6a + a - 2}{3a^{2} - 9a + a - 3}$
$\Rightarrow \dfrac{2a(a - 2) + 1(a - 2)}{a(a - 3) + 2(a - 3)} \div \dfrac{3a(a - 2) + 1(a - 2)}{3a(a - 3) + 1(a - 3)}$
$\Rightarrow \dfrac{(a - 2)(2a + 1)}{(a - 3)(a + 2)} \div \dfrac{(a - 2)(3a + 1)}{(a - 3)(3a + 1)}$
$\Rightarrow \dfrac{(a - 2)(2a + 1)}{(a - 3)(a + 2)} \times \dfrac{(a - 3)(3a + 1)}{(a - 2)(3a + 1)}$
$\therefore \dfrac{(2a + 1)(3a + 1)}{(a + 2)(3a + 1)} = \dfrac{2a + 1}{a + 2}$ Ans
$\Rightarrow \dfrac{2a(a - 2) + 1(a - 2)}{a(a - 3) + 2(a - 3)} \div \dfrac{3a(a - 2) + 1(a - 2)}{3a(a - 3) + 1(a - 3)}$
$\Rightarrow \dfrac{(a - 2)(2a + 1)}{(a - 3)(a + 2)} \div \dfrac{(a - 2)(3a + 1)}{(a - 3)(3a + 1)}$
$\Rightarrow \dfrac{(a - 2)(2a + 1)}{(a - 3)(a + 2)} \times \dfrac{(a - 3)(3a + 1)}{(a - 2)(3a + 1)}$
$\therefore \dfrac{(2a + 1)(3a + 1)}{(a + 2)(3a + 1)} = \dfrac{2a + 1}{a + 2}$ Ans
Q29. $\dfrac{3x - 4}{4x} - \dfrac{2x - 3}{20x}$
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L.C.M. of dinominators = $20x$
$\Rightarrow \dfrac{5(3x - 4) - 2x - 3}{20x}$
$\Rightarrow \dfrac{15x - 20 - 2x - 3}{20x}$
$\therefore \dfrac{13x - 17}{20x}$ Ans
$\Rightarrow \dfrac{5(3x - 4) - 2x - 3}{20x}$
$\Rightarrow \dfrac{15x - 20 - 2x - 3}{20x}$
$\therefore \dfrac{13x - 17}{20x}$ Ans
Q30. $\dfrac{5}{a - 3} + \dfrac{7}{2a - 6}$
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L.C.M. of dinominators = $(a - 3)(2a - 6)$
$\Rightarrow \dfrac{5(2a - 6) + 7(a - 3)}{(a - 3)(2a - 6)}$
$\Rightarrow \dfrac{10a - 30 + 7a - 21}{(a - 3)(2a - 6)}$
$\Rightarrow \dfrac{17a - 51}{(a - 3)(2a - 6)}$
$\therefore \dfrac{17(a - 3)}{(a - 3)(2a - 6)} = \dfrac{17}{2a - 6}$ Ans
$\Rightarrow \dfrac{5(2a - 6) + 7(a - 3)}{(a - 3)(2a - 6)}$
$\Rightarrow \dfrac{10a - 30 + 7a - 21}{(a - 3)(2a - 6)}$
$\Rightarrow \dfrac{17a - 51}{(a - 3)(2a - 6)}$
$\therefore \dfrac{17(a - 3)}{(a - 3)(2a - 6)} = \dfrac{17}{2a - 6}$ Ans
Q31. $\dfrac{5}{a + b} - \dfrac{4}{a - b} + \dfrac{8a}{a^{2} - b^{2}}$
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L.C.M. of dinominators = $a^{2} - b^{2}$
$\Rightarrow \dfrac{5(a - b) - 4(a + b) + 8a}{a^{2} - b^{2}}$
$\Rightarrow \dfrac{5a - 5b - 4a - 4b + 8a}{a^{2} - b^{2}}$
$\Rightarrow \dfrac{9a - 9b}{a^{2} - b^{2}}$
$\therefore \dfrac{9(a - b)}{(a + b)(a - b)} = \dfrac{9}{a + b}$ Ans
$\Rightarrow \dfrac{5(a - b) - 4(a + b) + 8a}{a^{2} - b^{2}}$
$\Rightarrow \dfrac{5a - 5b - 4a - 4b + 8a}{a^{2} - b^{2}}$
$\Rightarrow \dfrac{9a - 9b}{a^{2} - b^{2}}$
$\therefore \dfrac{9(a - b)}{(a + b)(a - b)} = \dfrac{9}{a + b}$ Ans
Q32. $\dfrac{x^{2} - xy}{x^{2}y} - \dfrac{y + z}{yz} - \dfrac{2z^{2} - xz}{z^{2}x}$
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L.C.M. of dinominators = $x^{2}yz^{2}$
$\Rightarrow \dfrac{z^{2}(x^{2} - xy) - x^{2}z(y + z) - xy(2z^{2} - xz)}{x^{2}yz^{2}}$
$\Rightarrow \dfrac{x^{2}z^{2} - xyz^{2} - x^{2}yz - x^{2}z^{2} - 2xyz^{2} + x^{2}yz}{x^{2}yz^{2}}$
$\therefore \dfrac{-3xyz^{2}}{x^{2}yz^{2}} = -\dfrac{3}{x}$ Ans
$\Rightarrow \dfrac{z^{2}(x^{2} - xy) - x^{2}z(y + z) - xy(2z^{2} - xz)}{x^{2}yz^{2}}$
$\Rightarrow \dfrac{x^{2}z^{2} - xyz^{2} - x^{2}yz - x^{2}z^{2} - 2xyz^{2} + x^{2}yz}{x^{2}yz^{2}}$
$\therefore \dfrac{-3xyz^{2}}{x^{2}yz^{2}} = -\dfrac{3}{x}$ Ans
Q33. $\dfrac{a}{a - 1} - \dfrac{a^{2}}{a^{2} - 1}$
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L.C.M. of dinominators = $(a - 1)(a^{2} - 1)$
$\Rightarrow \dfrac{a(a^{2} - 1) - a^{2}(a - 1)}{(a - 1)(a^{2} - 1)}$
$\Rightarrow \dfrac{a^{3} - a - a^{3} + a^{2}}{(a - 1)(a^{2} - 1)}$
$\Rightarrow \dfrac{- a + a^{2}}{(a - 1)(a^{2} - 1)} = \dfrac{a^{2} - a}{(a - 1)(a^{2} - 1)}$
$\Rightarrow \dfrac{a^{2} - a}{(a - 1)(a^{2} - 1)} = \dfrac{a(a - 1)}{(a - 1)(a^{2} - 1)}$
$\therefore \dfrac{a(a - 1)}{(a - 1)(a^{2} - 1)} = \dfrac{a}{a^{2} - 1}$ Ans
$\Rightarrow \dfrac{a(a^{2} - 1) - a^{2}(a - 1)}{(a - 1)(a^{2} - 1)}$
$\Rightarrow \dfrac{a^{3} - a - a^{3} + a^{2}}{(a - 1)(a^{2} - 1)}$
$\Rightarrow \dfrac{- a + a^{2}}{(a - 1)(a^{2} - 1)} = \dfrac{a^{2} - a}{(a - 1)(a^{2} - 1)}$
$\Rightarrow \dfrac{a^{2} - a}{(a - 1)(a^{2} - 1)} = \dfrac{a(a - 1)}{(a - 1)(a^{2} - 1)}$
$\therefore \dfrac{a(a - 1)}{(a - 1)(a^{2} - 1)} = \dfrac{a}{a^{2} - 1}$ Ans
Q34. $\dfrac{1}{2x^{2} - x - 3} - \dfrac{1}{2x^{2} + x - 1}$
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First denominator: $2x^{2} - x - 3$
$\Rightarrow 2x^{2} - x - 3 = 2x^{2} - 3x + 2x - 3$
$\Rightarrow 2x^{2} - 3x + 2x - 3 = x(2x - 3) + 1(2x - 3)$
$\Rightarrow x(2x - 3) + 1(2x - 3) = (2x - 3)(x + 1)$
Second denominator: $2x^{2} + x - 1$
$\Rightarrow 2x^{2} + x - 1 = 2x^{2} + 2x - x - 1$
$\Rightarrow 2x^{2} + 2x - x - 1 = 2x(x + 1) - 1(x + 1)$
$\Rightarrow 2x(x + 1) - 1(x + 1) = (x + 1)(2x - 1)$
Now, L.C.M. of dinominators = $(2x - 3)(x + 1)(2x - 1)$
$\Rightarrow \dfrac{(2x - 1) - (2x - 3)}{(2x - 3)(x + 1)(2x - 1)}$
$\Rightarrow \dfrac{2x - 1 - 2x + 3}{(2x - 3)(x + 1)(2x - 1)}$
$\therefore \dfrac{2}{(2x - 3)(x + 1)(2x - 1)}$ Ans
$\Rightarrow 2x^{2} - x - 3 = 2x^{2} - 3x + 2x - 3$
$\Rightarrow 2x^{2} - 3x + 2x - 3 = x(2x - 3) + 1(2x - 3)$
$\Rightarrow x(2x - 3) + 1(2x - 3) = (2x - 3)(x + 1)$
Second denominator: $2x^{2} + x - 1$
$\Rightarrow 2x^{2} + x - 1 = 2x^{2} + 2x - x - 1$
$\Rightarrow 2x^{2} + 2x - x - 1 = 2x(x + 1) - 1(x + 1)$
$\Rightarrow 2x(x + 1) - 1(x + 1) = (x + 1)(2x - 1)$
Now, L.C.M. of dinominators = $(2x - 3)(x + 1)(2x - 1)$
$\Rightarrow \dfrac{(2x - 1) - (2x - 3)}{(2x - 3)(x + 1)(2x - 1)}$
$\Rightarrow \dfrac{2x - 1 - 2x + 3}{(2x - 3)(x + 1)(2x - 1)}$
$\therefore \dfrac{2}{(2x - 3)(x + 1)(2x - 1)}$ Ans
Q35. $\dfrac{5}{4x^{2} + 3x - 1} - \dfrac{1}{3x^{2} + 4x + 1}$
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First denominator: $4x^{2} + 3x - 1$
$\Rightarrow 4x^{2} + 3x - 1 = 4x^{2} + 4x - x - 1$
$\Rightarrow 4x^{2} + 4x - x - 1 = 4x(x + 1) - 1(x + 1)$
$\Rightarrow 4x(x + 1) - 1(x + 1) = (x + 1)(4x - 1)$
Second denominator: $3x^{2} + 4x + 1$
$\Rightarrow 3x^{2} + 4x + 1 = 3x^{2} + 3x + x + 1$
$\Rightarrow 3x^{2} + 3x + x + 1 = 3x(x + 1) + 1(x + 1)$
$\Rightarrow 3x(x + 1) + 1(x + 1) = (x + 1)(3x + 1)$
Now, L.C.M. of dinominators = $(x + 1)(4x - 1)(3x + 1)$
$\Rightarrow \dfrac{5(3x + 1) - (4x - 1)}{(x + 1)(4x - 1)(3x + 1)}$
$\Rightarrow \dfrac{15x + 5 - 4x + 1}{(x + 1)(4x - 1)(3x + 1)}$
$\therefore \dfrac{11x + 6}{(x + 1)(4x - 1)(3x + 1)}$ Ans
$\Rightarrow 4x^{2} + 3x - 1 = 4x^{2} + 4x - x - 1$
$\Rightarrow 4x^{2} + 4x - x - 1 = 4x(x + 1) - 1(x + 1)$
$\Rightarrow 4x(x + 1) - 1(x + 1) = (x + 1)(4x - 1)$
Second denominator: $3x^{2} + 4x + 1$
$\Rightarrow 3x^{2} + 4x + 1 = 3x^{2} + 3x + x + 1$
$\Rightarrow 3x^{2} + 3x + x + 1 = 3x(x + 1) + 1(x + 1)$
$\Rightarrow 3x(x + 1) + 1(x + 1) = (x + 1)(3x + 1)$
Now, L.C.M. of dinominators = $(x + 1)(4x - 1)(3x + 1)$
$\Rightarrow \dfrac{5(3x + 1) - (4x - 1)}{(x + 1)(4x - 1)(3x + 1)}$
$\Rightarrow \dfrac{15x + 5 - 4x + 1}{(x + 1)(4x - 1)(3x + 1)}$
$\therefore \dfrac{11x + 6}{(x + 1)(4x - 1)(3x + 1)}$ Ans