Evaluate - 1-x - 1x - 2 - 1-x - 2x - 3 - 1-x - 3x - 4- - 1-6

⇒[ 1 ( x 1 ) ( x 2 ) #160; + 1 ( x 2 ) ( x 3 ) #160; + 1 ( x 3 ) ( x 4 ) #160; #160; ] = 1 6 ⇒ 1 x 2 1 x 1 + 1 x 3 1 x 2 + 1 ...

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Standard X

Mathematics

Multiplication of Rational Expressions

Evaluate [ ...

Question

Evaluate $[\frac{1}{(x - 1) (x - 2)} + \frac{1}{(x - 2) (x - 3)} + \frac{1}{(x - 3) (x - 4)}] = \frac{1}{6}$

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Solution

$\Rightarrow [\frac{1}{(x - 1) (x - 2)} + \frac{1}{(x - 2) (x - 3)} + \frac{1}{(x - 3) (x - 4)}] = \frac{1}{6}$
$\Rightarrow \frac{1}{x - 2} - \frac{1}{x - 1} + \frac{1}{x - 3} - \frac{1}{x - 2} + \frac{1}{x - 4} - \frac{1}{x - 3} = \frac{1}{6}$
$\Rightarrow \frac{1}{x - 4} - \frac{1}{x - 1} = \frac{1}{6}$
$\Rightarrow \frac{3}{(x - 4) (x - 1)} = \frac{1}{6}$
$\Rightarrow 18 = x^{2} - 5 x + 4$
$\Rightarrow x^{2} - 5 x - 14 = 0$
$\Rightarrow x^{2} - 7 x + 2 x - 14 = 0$
$\Rightarrow x (x - 7) + 2 (x - 7) = 0$
$\Rightarrow (x + 2) (x - 7) = 0$
$x = 7$ or $- 2$
(Answer)

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Similar questions

\frac{1}{(x - 1) (x - 2)} + \frac{1}{(x - 2) (x - 3)} + \frac{1}{(x - 3) (x - 4)} = \frac{1}{6}

Q. Solve for

x : \frac{1}{(x - 1) (x - 2)} + \frac{1}{(x - 2) (x - 3)} + \frac{1}{(x - 3) (x - 4)} = \frac{1}{6}

Q. Solve for x, 1÷(x-1)(x-2)+1÷(x-2)(x-3)+1÷(x-3)(x-4)=1÷6

Solve the following quadratic equation by factorization.
$\frac{1}{(x - 1) (x - 2)} + \frac{1}{(x - 2) (x - 3)} + \frac{1}{(x - 3) (x - 4)} = \frac{1}{6}$

Q. Evaluate:

\frac{x - 4}{3} - \frac{2 x + 1}{6} = \frac{5 x + 1}{2}

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Evaluate [1(x−1)(x−2)+1(x−2)(x−3)+1(x−3)(x−4)]=16

⇒[1(x−1)(x−2)+1(x−2)(x−3)+1(x−3)(x−4)]=16⇒1x−2−1x−1+1x−3−1x−2+1x−4−1x−3=16⇒1x−4−1x−1=16⇒3(x−4)(x−1)=16⇒18=x2−5x+4⇒x2−5x−14=0⇒x2−7x+2x−14=0⇒x(x−7)+2(x−7)=0⇒(x+2)(x−7)=0x=7 or −2 (Answer)

Evaluate $[\frac{1}{(x - 1) (x - 2)} + \frac{1}{(x - 2) (x - 3)} + \frac{1}{(x - 3) (x - 4)}] = \frac{1}{6}$