Solve for x- 1-x-1x-2-1-x-2x-3-2-3- y-1-2-3

We have, 1/(x 1)(x 2)+1/(x 2)(x 3)=2/3,x≠ 1,2,3. 3(x 3)+3(x 1)=2(x 1)(x 2)(x 3) 3x 9+3x 3=2(x 1)(x 2)(x 3) 6x 12=2(x 1)(x 2)(x 3) 6(x 2)=2(x 1)(x 2)(x 3) ...

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Byju's Answer

Standard X

Mathematics

Sum of N Terms of an AP

Solve for x: ...

Question

Solve for x : $\frac{1}{(x - 1) (x - 2)} + \frac{1}{(x - 2) (x - 3)}$

= $\frac{2}{3}, x \neq 1, 2, 3$

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Solution

We have,

$\frac{1}{(x - 1) (x - 2)} + \frac{1}{(x - 2) (x - 3)} = \frac{2}{3}, x \neq 1, 2, 3$ .

$3 (x - 3) + 3 (x - 1) = 2 (x - 1) (x - 2) (x - 3)$

$3 x - 9 + 3 x - 3 = 2 (x - 1) (x - 2) (x - 3)$

$6 x - 12 = 2 (x - 1) (x - 2) (x - 3)$

$6 (x - 2) = 2 (x - 1) (x - 2) (x - 3)$

$3 = (x - 1) (x - 3)$

$3 = x^{2} - 3 x - x + 3$

$x^{2} - 4 x = 0$

$x (x - 4) = 0$

$∴ x = 0$ or 4

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Solve for x :1(x−1)(x−2)+1(x−2)(x−3) = 23,x≠1,2,3

We have, 1(x−1)(x−2)+1(x−2)(x−3)=23,x≠1,2,3. 3(x−3)+3(x−1)=2(x−1)(x−2)(x−3) 3x−9+3x−3=2(x−1)(x−2)(x−3) 6x−12=2(x−1)(x−2)(x−3) 6(x−2)=2(x−1)(x−2)(x−3) 3=(x−1)(x−3) 3=x2−3x−x+3 x2−4x=0 x(x−4)=0 ∴ x=0 or 4

Solve for x : $\frac{1}{(x - 1) (x - 2)} + \frac{1}{(x - 2) (x - 3)}$

= $\frac{2}{3}, x \neq 1, 2, 3$