Question

A=\(\dfrac{\sqrt{x}}{\sqrt{x}+3}\) và B=\(\dfrac{8\sqrt{x}-3}{14}\)

Tìm x sao cho A=2B

Answer 1

\(\dfrac{\sqrt{x}}{\sqrt{x}+3}=2\left(\dfrac{8\sqrt{x}-3}{14}\right)\left(x\ge0\right)\)

<=> \(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{8\sqrt{x}-3}{7}=0\)

<=> \(\dfrac{7\sqrt{x}}{7\left(\sqrt{x}+3\right)}-\dfrac{8x+21\sqrt{x}-9}{7\left(\sqrt{x}+3\right)}=0\)

<=>\(7\sqrt{x}-8x+21\sqrt{x}-9=0\)

<=>\(8x-28\sqrt{x}+9=0\)     *

Sau đó tính đenta 

\(\Delta=496>0\)

=> pt * có 2 nghiệm phân biệt 

<=> \(\left[{}\begin{matrix}\sqrt{x1}=\dfrac{7-\sqrt{31}}{4}\\\sqrt{x2}=\dfrac{7+\sqrt{31}}{4}\end{matrix}\right.\)  => \(\left[{}\begin{matrix}x1=\dfrac{40-7\sqrt{31}}{8}\\x2=\dfrac{40+7\sqrt{31}}{8}\end{matrix}\right.\)   \(\left(tm\right)\)

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