\(2-x^2=\sqrt{2-x}\) ĐKXĐ: x ≤ 2
pt <=> \(x^4-4x^2+4=2-x\)
\(\Leftrightarrow x^4-4x^2+x+2=0\)
\(\Leftrightarrow x^2\left(x^2-4\right)+\left(x+2\right)=0\)
\(\Leftrightarrow x^2\left(x-2\right)\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^3-2x^2+1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^3-x^2-x^2+1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left[x^2\left(x-1\right)-\left(x^2-1\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right)\left[x^2\left(x-1\right)-\left(x-1\right)\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)\left(x^2-x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-1=0\\x^2-x-1=0\end{matrix}\right.\)
+) x + 2 = 0 <=> x = -2
+) x - 1 = 0 <=> x = 1
+) \(x^2-x-1=0\)
\(\Leftrightarrow x^2-2\cdot\dfrac{1}{2}x+\dfrac{1}{4}-\dfrac{5}{4}=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{5}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{\sqrt{5}}{2}\\x-\dfrac{1}{2}=-\dfrac{\sqrt{5}}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{5}+1}{2}\\x=\dfrac{1-\sqrt{5}}{2}\end{matrix}\right.\)
Thử lại thấy x = -2 không thỏa mãn
Vậy pt có 3 nghiệm: \(\left[{}\begin{matrix}x_1=1\\x_2=\dfrac{\sqrt{5}+1}{2}\\x_3=\dfrac{1-\sqrt{5}}{2}\end{matrix}\right.\)
