Question

Giải các phương trình sau:

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

\(\sqrt{x^2+3x}+2\sqrt{x+2}=2x+\sqrt{x+\frac{6}{x}+5}\)

\(\left(x+2\right)\left(x+4\right)+5\left(x+2\right)\sqrt{\frac{x+4}{x+2}}=6\)

Answer 1

a.

ĐKXĐ: \(x\ge1\)

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

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

\(\Leftrightarrow x^2-x\sqrt{x-1}-2\left(x-1\right)=0\)

Đặt \(\sqrt{x-1}=a\ge0\)

\(\Rightarrow x^2-ax-2a^2=0\)

\(\Leftrightarrow\left(x+a\right)\left(x-2a\right)=0\)

\(\Leftrightarrow x=2a\)

\(\Leftrightarrow x=2\sqrt{x-1}\)

\(\Leftrightarrow x^2=4\left(x-1\right)\)

\(\Leftrightarrow x=2\)

Answer 2

b.

ĐKXĐ: \(x>0\)

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

\(\Leftrightarrow\sqrt{x\left(x+3\right)}-2x+2\sqrt{x+2}-\sqrt{\frac{\left(x+2\right)\left(x+3\right)}{x}}=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x+3}-2\sqrt{x}\right)+\sqrt{\frac{x+2}{x}}\left(2\sqrt{x}-\sqrt{x+3}\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-\sqrt{\frac{x+2}{x}}\right)\left(\sqrt{x+3}-2\sqrt{x}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\sqrt{\frac{x+2}{x}}\\\sqrt{x+3}=2\sqrt{x}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{x+2}{x}\\x+3=4x\end{matrix}\right.\)

\(\Leftrightarrow...\)

Answer 3

c.

ĐKXĐ: \(\left[{}\begin{matrix}x>-2\\x\le-4\end{matrix}\right.\)

-TH1: với \(x>-2\Rightarrow x+2=\sqrt{\left(x+2\right)^2}\)

Pt trở thành:

\(\left(x+2\right)\left(x+4\right)+5\sqrt{\frac{\left(x+4\right)\left(x+2\right)^2}{x+2}}=6\)

\(\Leftrightarrow\left(x+2\right)\left(x+4\right)+5\sqrt{\left(x+2\right)\left(x+4\right)}-6=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{\left(x+2\right)\left(x+4\right)}=1\\\sqrt{\left(x+2\right)\left(x+4\right)}=-6\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow x^2+6x+7=0\Rightarrow\left[{}\begin{matrix}x=-3+\sqrt{2}\\x=-3-\sqrt{2}< -2\left(l\right)\end{matrix}\right.\)

TH2: \(x\le-4\Rightarrow x+2=-\sqrt{\left(x+2\right)^2}\)

Pt trở thành:

\(\left(x+2\right)\left(x+4\right)-5\sqrt{\left(x+2\right)\left(x+4\right)}-6=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{\left(x+2\right)\left(x+4\right)}=-1< 0\left(l\right)\\\sqrt{\left(x+2\right)\left(x+4\right)}=6\end{matrix}\right.\)

\(\Leftrightarrow x^2+6x-28=0\Rightarrow\left[{}\begin{matrix}x=-3+\sqrt{37}>-4\left(l\right)\\x=-3-\sqrt{37}\end{matrix}\right.\)

Vậy ...