Question

[Nguyễn Thế Mãnh]

Giải các phương trình sau:

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

b) \(x^2+2x+3\sqrt{x^2+2x+2}-6=0\)

c) \(\sqrt{\left(x+1\right)\left(2-x\right)}-1+2x=2x^2\)

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

Answer 1

[Nguyễn Việt Lâm]

a/ ĐKXĐ: \(x\ge2\)

\(\Leftrightarrow2x-3+2\sqrt{x^2-3x+2}=x+1\)

\(\Leftrightarrow2\sqrt{x^2-3x+2}=4-x\) (\(x\le4\))

\(\Leftrightarrow4\left(x^2-3x+2\right)=x^2-8x+16\)

\(\Leftrightarrow3x^2-4x-8=0\Rightarrow\left[{}\begin{matrix}x=\frac{2+\sqrt{7}}{3}\\x=\frac{2-\sqrt{7}}{3}\left(l\right)\end{matrix}\right.\)

b/ Đặt \(x^2+2x+2=a>0\)

\(a^2+3a-8=0\Rightarrow\left[{}\begin{matrix}a=\frac{-3+\sqrt{41}}{2}\\a=\frac{-3-\sqrt{41}}{2}\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow x^2+2x+2-\frac{-3+\sqrt{41}}{2}=0\)

Bạn tự giải nốt, nghiệm quá xấu, chắc bạn ghi sai đề

Answer 2

[Nguyễn Việt Lâm]

c/ ĐKXĐ: \(-1\le x\le2\)

\(\Leftrightarrow2\left(-x^2+x+2\right)+\sqrt{-x^2+x+2}-5=0\)

Đặt \(\sqrt{-x^2+x+2}=a\ge0\)

\(2a^2+a-5=0\Rightarrow\left[{}\begin{matrix}a=\frac{-1+\sqrt{41}}{2}\\a=\frac{-1-\sqrt{41}}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow-x^2+x+2-\frac{-1+\sqrt{41}}{2}=0\)

??? Lại 1 nghiệm khủng khiếp nữa???

d/ ĐKXĐ: \(\left[{}\begin{matrix}x>0\\x< -1\end{matrix}\right.\)

Đặt \(\sqrt{\frac{2x}{x+1}}=a>0\)

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

\(\Rightarrow\sqrt{\frac{2x}{x+1}}=1\Rightarrow2x=x+1\Rightarrow x=1\)