Question

giải phương trình:

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

Answer 1

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{3x+2}\\2\left(3x+2\right)+3\sqrt{3x+2}+2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{3x+2}\\2\left(\sqrt{3x+2}+\frac{3}{4}\right)^2+\frac{7}{6}=0\text{ }\left(\text{vô nghiệm}\right)\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x\ge0\\x^2=3x+2\end{cases}}\Leftrightarrow x=\frac{3+\sqrt{17}}{2}.\)

Answer 2

(-X² - 2X +2X²-X-2)-(X+2)(\(\sqrt{3X+2}\)- X)=0       ĐK tự đặt

NHÂN LƯỢNG LIÊN HỢP CHO \(\sqrt{3X+2}\)+ X    >0 

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

nhóm \(x^2-3x-2\\ \)

pt tương đương :\(\left(x^2-3x-3\right)\left(x+3\right)\)

\(\hept{\begin{cases}x=-3\left(loai\right)\\x^2-3x-3=0\end{cases}}\)

đến đây la xong