Question

Giải phương trình : \(\sqrt{4x+1}-\sqrt{3x-2}\) = \(\dfrac{x+3}{5}\)

Answer 1

ĐKXĐ: x\(\ge\dfrac{2}{3}\)

\(\sqrt{4x+1}-\sqrt{3x-2}=\dfrac{x+3}{5}\)

<=> \(\left(\sqrt{4x+1}-3\right)-\left(\sqrt{3x-2}-2\right)=\dfrac{x+3}{5}-1\)

<=> \(\dfrac{4\left(x-2\right)}{\sqrt{4x+1}+3}-\dfrac{3\left(x-2\right)}{\sqrt{3x-2}+2}=\dfrac{x-2}{5}\)

<=> \(\left(x-2\right)\left(\dfrac{4}{\sqrt{4x+1}+3}-\dfrac{3}{\sqrt{3x-2}+2}-\dfrac{1}{5}\right)=0\)

<=> x-2=0 ( Vì \(\left(\dfrac{4}{\sqrt{4x+1}+3}-\dfrac{3}{\sqrt{3x-2}+2}-\dfrac{1}{5}\right)\)<0 với mọi x\(\ge\dfrac{2}{3}\))

<=> x=2

Vậy S={2}