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

1/(x 1)(x 2)+1/(x 3)(x 2)=2/3⇒(x 3)+(x 1)/(x 1)(x 2)(x 3)=2/3⇒x 3+x 1/(x 1)(x 2)(x 3)=2/3⇒2x 4/(x 1)(x 2)(x 3)=2/3⇒2(x 2)/(x 1)(x 2)(x 3)=2/3⇒2/(x 1)(x 3)=2/3⇒2 ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard VII

Mathematics

Solving an Equation

Solve for :1/...

Question

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

Open in App

Solution

$\frac{1}{(x - 1) (x - 2)} + \frac{1}{(x - 3) (x - 2)} = \frac{2}{3}$
$\Rightarrow \frac{(x - 3) + (x - 1)}{(x - 1) (x - 2) (x - 3)} = \frac{2}{3}$
$\Rightarrow \frac{x - 3 + x - 1}{(x - 1) (x - 2) (x - 3)} = \frac{2}{3}$
$\Rightarrow \frac{2 x - 4}{(x - 1) (x - 2) (x - 3)} = \frac{2}{3}$
$\Rightarrow \frac{2 (x - 2)}{(x - 1) (x - 2) (x - 3)} = \frac{2}{3}$
$\Rightarrow \frac{2}{(x - 1) (x - 3)} = \frac{2}{3}$
$\Rightarrow \frac{2}{x^{2} - 3 x - x + 3} = \frac{2}{3}$
$\Rightarrow \frac{1}{x^{2} - 4 x + 3} = \frac{1}{3}$
$x^{2} - 4 x + 3 = 3 \Rightarrow x^{2} - 4 x + 3 - 3 = 0$
$\Rightarrow x^{2} - 4 x = 0 \Rightarrow x (x - 4) = 0$
$\Rightarrow x = 0 o r x - 4 = 0 \Rightarrow x = 4 ∴ x = 0 o r x = 4$

Suggest Corrections

Similar questions

Q. Let

f (x) = \frac{(x - 3) (x + 2) (x + 6)}{(x + 1) (x - 5)}

Find intervals, where

f (x)

is positive or negative.

If $f = \frac{x}{1 + x^{2}} + \frac{1}{3} {(\frac{x}{1 + x^{2}})}^{3} + \frac{1}{5} {(\frac{x}{1 + x^{2}})}^{5} + . . .$ to $\infty$ and $g = x - \frac{2}{3} x^{3} + \frac{1}{5} x^{5} + \frac{1}{7} x^{7} - \frac{2}{9} x^{9} + . . .$ , then $f = d \times g$ . Find 4d.

Q. Let

\to a, \to b

and

\to c

be mutually perpendicular vectors of the same magnitude.If a vector

\to x

satisfies the equation

\to a \times [(\to x - \to b) \times \to a] + \to b \times [(\to x - \to c) \times \to b] + \to c \times [(\to x - \to a) \times \to c] = 0,

then

\to x

is given by

Let $[k]$ denotes the greatest integer less than or equal to $k .$ Then the number of positive integral solutions of the equation $[\frac{x}{[π^{2}]}] = ⎡ ⎢ ⎢ ⎣ \frac{x}{[11 \frac{1}{2}]} ⎤ ⎥ ⎥ ⎦$ is

Determine whether the following numbers are in proportion or not $:$

$\frac{1}{3}, \frac{1}{4}, \frac{1}{6}, \frac{1}{7}$

Join BYJU'S Learning Program

Solve for :1x-1x-2+1x-3x-2=23,x≠1,2,3

1x-1x-2+1x-3x-2=23⇒x-3+x-1x-1x-2x-3=23⇒x-3+x-1x-1x-2x-3=23⇒2x-4x-1x-2x-3=23⇒2x-2x-1x-2x-3=23⇒2x-1x-3=23⇒2x2-3x-x+3=23⇒1x2-4x+3=13x2-4x+3=3⇒x2-4x+3-3=0⇒x2-4x=0⇒xx-4=0⇒x=0orx-4=0⇒x=4∴x=0orx=4

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