Question

giúp mình vs ạ !

Answer 1

a) \(\dfrac{\sqrt{2x-3}}{\sqrt{x-1}}=2\left(x\ge\dfrac{3}{2}\right)\Rightarrow\sqrt{2x-3}=2\sqrt{x-1}\)

\(\Rightarrow2x-3=4x-4\Rightarrow2x=1\Rightarrow x=\dfrac{1}{2}\) (loại)

b) \(\sqrt{\dfrac{x-3}{2x+1}}=2\left(\dfrac{x-3}{2x+1}\ge0,x\ne-\dfrac{1}{2}\right)\)

\(\Rightarrow\dfrac{x-3}{2x+1}=4\Rightarrow x-3=8x+4\Rightarrow7x=-7\Rightarrow x=-1\)

Thế vào \(\Rightarrow\) chọn

c) \(\dfrac{10x-3}{\sqrt{2x+1}}=\sqrt{2x+1}\)

Vì \(VP>0\Rightarrow VT>0\) mà \(\sqrt{2x+1}>0\Rightarrow10x-3>0\Rightarrow x>\dfrac{3}{10}\)

\(\Rightarrow10x-3=2x+1\Rightarrow8x=4\Rightarrow x=\dfrac{1}{2}\)

d) \(\sqrt{4x^2-9}=2\sqrt{2x-3}\left(x\ge\dfrac{3}{2}\right)\Rightarrow\sqrt{2x-3}.\sqrt{2x+3}=2\sqrt{2x-3}\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{2x-3}=0\\\sqrt{2x+3}=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{2}\left(l\right)\end{matrix}\right.\)