Question

[Julian Edward]

giải pt

a) \(2\sqrt{\frac{x}{x-1}}-\sqrt{\frac{x-1}{x}}=\frac{5x-2}{x}\)

b) \(3\sqrt{\frac{2x}{x-1}}+4\sqrt{\frac{x-1}{2x}}=\frac{5x-3}{2x}+9\)

c) \(\sqrt{\frac{x}{3-2x}}+5\sqrt{\frac{3-2x}{x}}=\frac{12-9x}{x}+6\)

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

e) \(\sqrt{\frac{x}{x-1}}+\sqrt{\frac{x-1}{x}}=\frac{3}{\sqrt{2}}\)

f) \(\sqrt{x-\frac{1}{x}}=\frac{1}{\sqrt{x}}-\sqrt{x}\)

Answer 1

[Nguyễn Việt Lâm]

a/ ĐKXĐ: ...

\(\Leftrightarrow2\sqrt{\frac{x}{x-1}}-\sqrt{\frac{x-1}{x}}=\frac{2\left(x-1\right)}{x}+3\)

Đặt \(\sqrt{\frac{x-1}{x}}=a>0\)

\(\frac{2}{a}-a=2a^2+3\Leftrightarrow2a^3+a^2+3a-2=0\)

\(\Leftrightarrow\left(2a-1\right)\left(a^2+a+2\right)=0\Leftrightarrow a=\frac{1}{2}\)

\(\Rightarrow\sqrt{\frac{x-1}{x}}=\frac{1}{2}\Leftrightarrow4\left(x-1\right)=x\)

b/ ĐKXĐ: ...

\(\Leftrightarrow3\sqrt{\frac{2x}{x-1}}+4\sqrt{\frac{x-1}{2x}}=\frac{3\left(x-1\right)}{2x}+10\)

Đặt \(\sqrt{\frac{x-1}{2x}}=a>0\)

\(\frac{3}{a}+4a=3a^2+10\Leftrightarrow3a^3-4a^2+10a-3=0\)

\(\Leftrightarrow\left(3a-1\right)\left(a^2-a+3\right)=0\Leftrightarrow a=\frac{1}{3}\)

\(\Leftrightarrow\sqrt{\frac{x-1}{2x}}=\frac{1}{3}\Leftrightarrow9\left(x-1\right)=2x\)

Answer 2

[Nguyễn Việt Lâm]

c/ ĐKXĐ: ...

\(\Leftrightarrow\sqrt{\frac{x}{3-2x}}+5\sqrt{\frac{3-2x}{x}}=\frac{4\left(3-2x\right)}{x}+5\)

Đặt \(\sqrt{\frac{3-2x}{x}}=a>0\)

\(\frac{1}{a}+5a=4a^2+5\Leftrightarrow4a^3-5a^2+5a-1=0\)

\(\Leftrightarrow\left(4a-1\right)\left(a^2-a+1\right)=0\Leftrightarrow a=\frac{1}{4}\)

\(\Leftrightarrow\sqrt{\frac{3-2x}{x}}=\frac{1}{4}\Leftrightarrow16\left(3-2x\right)=x\)

d/ ĐKXĐ: ...

Đặt \(\sqrt{\frac{x-1}{x}}=a>0\)

\(a^2-2a=3\Leftrightarrow a^2-2a-3=0\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=3\end{matrix}\right.\)

\(\Leftrightarrow\sqrt{\frac{x-1}{x}}=3\Leftrightarrow x-1=9x\)