Question

Please help me to solve the problem.

${\mathbb{\underline{\huge{\pink{\fbox{Explanation\:Required}}}}}}$

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Std 10, C.B. S. E. Assignment.​

Answer 1

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Solve the Equation :-

$\bf\dfrac{5}{2 - x} +\dfrac{x - 5}{x + 2} + \dfrac{3x + 8}{ {x}^{2} - 4} = 0$

$\large\star{\underline{\tt{\red{Answer}}}}\star$

$\purple\longmapsto\tt\dfrac{5}{2 - x} +\dfrac{x - 5}{x + 2} + \dfrac{3x + 8}{ {x}^{2} - 4} = 0\\\\\bf \: taking \: ( - 1) \: common \: from \: first \: part \: denominator \: we \: get\\\\ \purple\longmapsto\tt\dfrac{ - 5}{x - 2} +\dfrac{x - 5}{x + 2} + \dfrac{3x + 8}{ {x}^{2} - 4} = 0 \\ \\ \bf \: now \: using \: {x}^{2} - {y}^{2} = (x + y)(x - y) \\ \\ \purple\longmapsto\tt\dfrac{- 5}{x - 2} +\dfrac{x - 5}{x + 2} + \dfrac{3x + 8}{(x + 2)(x - 2)} = 0 \\ \\ \bf taking \: \red{LCM} \: now: \\ \\ \purple\longmapsto\tt \: \frac{ - 5(x + 2) + (x - 5)(x - 2) +(3x +8 ) }{(x^{2} - 4)} = 0 \\ \\ \bf \: cross - multiply \: we \: get \\\\\purple\longmapsto\tt \:- 5(x + 2) + (x - 5)(x - 2) +(3x +8 ) = 0 \\ \\ \purple\longmapsto\tt \: - 5x - 10 + {x}^{2} - 2x - 5x + 10 + 3x + 8 = 0 \\ \\ \purple\longmapsto\tt \: {x}^{2} - 9x + 8 = 0 \\ \\ \bf \: splitting \: the \: middle \: term \: now \\ \\ \purple\longmapsto\tt \: {x}^{2} - x - 8x+8 = 0 \\ \\ \purple\longmapsto\tt \: x(x - 1) - 8(x - 1) = 0 \\ \\ \purple\longmapsto\tt \: (x - 1)(x - 8) = 0 \\ \\ \bf \: putting \: both \: equal \: to \: zero \: now \: we \: get \\ \\\purple\longmapsto\: \Large\red{\boxed{\tt\blue{x}\purple{=} \green{1}\orange,\pink{8}}}$

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Answer 2

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$\huge\tt{TO~SOLVE:}$

• (5/2-x) + (x-5/x+2) + ( 3x +8/x² -4 ) = 0

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$\huge\tt{SOLUTION:}$

↪(5/2-x) + (x-5/x+2) + ( 3x +8/x² -4 ) = 0

↪(-5/2-x)+ (x-5/x+2)+ (3x +8/x²-4 ) = 0(taking -1 as common)

↪{-5/(x+2) + (x - 5)(x - 2)+(3x + 8) / (x²- 4)} = 0 (Taking L.C.M)

↪-5(x+2) + (x-5) (x-2) + (3x+8) = 0 (cross multiplying)

↪-5x - 10 + x² - 2x - 5x + 10 + 3x + 8 = 0

↪x² - 9x + 8 = 0

↪x²- x - 8x + 8 = 0 (splitting the terms)

↪(x - 1) (x - 8) = 0

↪x = 1,8 (putting both equals to zero)

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${\fbox{\fbox{\huge\tt\purple{X=1,8}}}}$

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