Question

[Nguyen Thi Bich Huong]

Giải phương trình vô tỉ:

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

b) \(\sqrt{2x+3}+\sqrt{x+1}=3x+2\sqrt{2x^2+5x+3}-2\)

c) \(\sqrt{7x+7}+\sqrt{7x-6}+2\sqrt{49x^2+7x-42}=181-4x\)

d) \(\frac{\sqrt{x+4}+\sqrt{x-4}}{2}=x+\sqrt{x^2-16}-6\)

e) \(5\sqrt{x}+\frac{5}{2\sqrt{x}}=2x+\frac{1}{2x}+4\)

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

Answer 1

[Nguyễn Việt Lâm]

a/ Giải rồi

b/ ĐKXĐ: \(x\ge-1\)

Đặt \(\sqrt{2x+3}+\sqrt{x+1}=t>0\)

\(\Rightarrow t^2=3x+4+2\sqrt{2x^2+5x+3}\) (1)

Pt trở thành:

\(t=t^2-6\Leftrightarrow t^2-t-6=0\Rightarrow\left[{}\begin{matrix}t=3\\t=-2\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2x+3}+\sqrt{x+1}=3\)

\(\Leftrightarrow3x+4+2\sqrt{2x^2+5x+3}=9\)

\(\Leftrightarrow2\sqrt{2x^2+5x+3}=5-3x\left(x\le\frac{5}{3}\right)\)

\(\Leftrightarrow4\left(2x^2+5x+3\right)=\left(5-3x\right)^2\)

\(\Leftrightarrow...\)

Answer 2

[Nguyễn Việt Lâm]

e/ ĐKXD: \(x>0\)

\(5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)

Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=t\ge\sqrt{2}\)

\(\Rightarrow t^2=x+\frac{1}{4x}+1\)

Pt trở thành:

\(5t=2\left(t^2-1\right)+4\)

\(\Leftrightarrow2t^2-5t+2=0\Rightarrow\left[{}\begin{matrix}t=2\\t=\frac{1}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=2\)

\(\Leftrightarrow2x-4\sqrt{x}+1=0\)

\(\Rightarrow\sqrt{x}=\frac{2\pm\sqrt{2}}{2}\)

\(\Rightarrow x=\frac{3\pm2\sqrt{2}}{2}\)