Đk: x>=-3
\(pt\Leftrightarrow4\left(x+3\right)=81x^4-18x^3-71x^2+8x+16-4x-12\)
\(\Leftrightarrow81x^4-18x^3-71x^2+4x+4=0\)
\(\Leftrightarrow81x^3\left(x-1\right)+63x^2\left(x-1\right)-8x\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(81x^3+63x^2-8x-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(81x^3+18x^2+45x^2+10x-18x-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[9x^2\left(9x+2\right)+5x\left(9x+2\right)-2\left(9x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(9x+2\right)\left(9x^2+5x-2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(9x+2\right)\left[9\left(x+\frac{5}{18}\right)^2-\frac{97}{36}\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-2}{9}\\x=\frac{-5+\sqrt{97}}{18}\\x=\frac{-5-\sqrt{97}}{18}\end{matrix}\right.\)(tmđk)
Thay vì cách làm dài bình phương 2 vế, ta có cách ngắn hơn như sau: ĐK: \(x\ge-3;9x^2-x-4\ge0\)
Phương trình tương đương:
\(9x^2=x+3+2\sqrt{x+3}+1=\left(\sqrt{x+3}+1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=\sqrt{x+3}+1\\3x=-\left(\sqrt{x+3}+1\right)\end{matrix}\right.\). Đặt \(\sqrt{x+3}=a\ge0\)
\(\Rightarrow\left[{}\begin{matrix}3a^2-a-10=0\\3a^2+a-8=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{-5}{3}\\a=...\\a=...\end{matrix}\right.\)
Từ đó suy ra x
