The correct options are A 6/xWe have 1/x 2+2/x 1=6/x ⇒(x 1)+2(x 2)/(x 2)(x 1)=6/x ⇒3x 5/x^2 3x+2=6/x ⇒ 3x^2 5x=6x^2 18x+12 ⇒ 3x^2 13x+12=0 ⇒ 3x^2 9x 4x +1 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Subtraction of Rational Expressions

1/x-2+2/x-1=6...

Question

$\frac{1}{x - 2} + \frac{2}{x - 1} = \frac{6}{x}$

\frac{6}{x}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{4}{x}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{7 x}{11}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{x}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{6}{x}$
We have

$\frac{1}{x - 2} + \frac{2}{x - 1} = \frac{6}{x}$

$\Rightarrow \frac{(x - 1) + 2 (x - 2)}{(x - 2) (x - 1)} = \frac{6}{x}$

$\Rightarrow \frac{3 x - 5}{x^{2} - 3 x + 2} = \frac{6}{x}$

$\Rightarrow 3 x^{2} - 5 x = 6 x^{2} - 18 x + 12$

$\Rightarrow 3 x^{2} - 13 x + 12 = 0$

$\Rightarrow 3 x^{2} - 9 x - 4 x + 12 = 0$

$\Rightarrow 3 x (x - 3) - 4 (x - 3) = 0$

$\Rightarrow (x - 3) (3 x - 4) = 0$

$\Rightarrow x - 3 = 0 o r 3 x - 4 = 0$

$\Rightarrow x = 3 o r x = \frac{4}{3}$

Suggest Corrections

Similar questions

$(1 - 2 x)^{4} + 4 x^{2} (1 - 2 x)^{3} + 6 x^{4} (1 - 2 x)^{2} + 4 x^{6} (1 - 2 x) + x^{8} =$

Q. Multiply

(2 x - \frac{1}{x}) (3 x + \frac{2}{x}) .

Q. Evaluate

[x + \sqrt{(x^{2} - 1)}]^{6} + [x - \sqrt{(x^{2} - 1)}]^{6}

Q. Equation of the circle inscribed in

| x + 1 | + | y - 6 | = 2

x^{2} + \frac{1}{x^{2}} = 6

Find
(i)

x^{3} - \frac{1}{x^{3}}

(i)

x^{6} + \frac{1}{x^{6}}

Join BYJU'S Learning Program

1x−2+2x−1=6x

The correct option is A 6xWe have 1x−2+2x−1=6x ⇒(x−1)+2(x−2)(x−2)(x−1)=6x ⇒3x−5x2−3x+2=6x ⇒3x2−5x=6x2−18x+12 ⇒3x2−13x+12=0 ⇒3x2−9x−4x+12=0 ⇒3x(x−3)−4(x−3)=0 ⇒(x−3)(3x−4)=0 ⇒x−3=0 or 3x−4=0 ⇒x=3 or x=43

$\frac{1}{x - 2} + \frac{2}{x - 1} = \frac{6}{x}$