Tải Qanda về

Answer:

\(\sqrt{\left(x-3\right).\left(x^2-x-6\right)}=x^2-7x+12\)

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

Điều kiện: \(\hept{\begin{cases}\left(x-3\right)^2.\left(x+2\right)\ge0\\\left(x-3\right).\left(x-4\right)\ge0\end{cases}}\Leftrightarrow\orbr{\begin{cases}-2\le x\le3\\x\ge4\end{cases}}\)

PT \(\Leftrightarrow\left(x-3\right)^2.\left(x+2\right)=\left(x-3\right)^2.\left(x-4\right)^2\)

\(\Leftrightarrow\left(x-3\right)^2.\left(x^2-9x+14\right)=0\)

Trường hợp 1: \(x=3\)

Trường hợp 2: \(x=7\)

Trường hợp 3: \(x=2\) (TMĐK)

Answer:

b) \(2\sqrt{x+3}=9x^2-x-4\)

ĐK: x\(x\ge-3\) phương trình tương đương:

Ta có: \(2\sqrt{x+3}=9x^2-x-4\)

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

\(\Leftrightarrow x+3+2\sqrt{x+3}+1=9x^2\)

\(\Leftrightarrow\left(1+\sqrt{3+x}\right)^2=9x^2\)

\(\left(1+\sqrt{3+x}\right)^2=9x^2\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x+3}+1=3x\\\sqrt{x+3}+1=-3x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{-5-\sqrt{97}}{18}\end{cases}}\)

1.111111e+103