Question

2.giai phuong trinh sau:

a.\(\sqrt{\frac{2x-3}{x-1}}=2\)

b.\(\frac{\sqrt{2x-3}}{\sqrt{x-1}}=2\)

Answer 1

Đkiện: x <1 hoặc x \(\ge\frac{3}{2}\)

\(\sqrt{\frac{2x-3}{x-1}}=2\) (1)

(1) => \(\frac{2x-3}{x-1}=4\)

=> 2x - 3 = 4x - 4

<=> 2x - 4x = -4 + 3

<=> -2x = -1

<=> x = \(\frac{1}{2}\)( TMĐK)

Vậy x = \(\frac{1}{2}\)

b, Đkiện: x \(\ge\frac{3}{2}\)

(1) => \(\sqrt{2x-3}=2\sqrt{x-1}\)

=>2x - 3 = 4(x - 1)

<=> 2x -3 = 4x -4

<=> -2x = -1

<=> x = \(\frac{1}{2}\)(ko TMĐK)

Vậy pt vô nghiệm

Answer 2

b. \(x>0;x\ne1\)

\(\Rightarrow\sqrt{\frac{2x-3}{x-1}}=2\Rightarrow\frac{2x-3}{x-1}=4\Rightarrow2x-3=4x-4\Rightarrow2x=1\Rightarrow x=\frac{1}{2}\)