\(5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}-2x^2-4=0\) (Đk \(x\ge-1\))
\(\Leftrightarrow5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}-2\left(x^2-x+1\right)-2\left(x+1\right)=0\)
Đặt \(\left\{{}\begin{matrix}a=\sqrt{x+1}\\b=\sqrt{x^2-x+1}\end{matrix}\right.\)
\(pt\Leftrightarrow-2a^2+5ab-2b^2=0\)
\(\Leftrightarrow-2a^2+4ab+ab-2b^2=0\)
\(\Leftrightarrow-2a\left(a-2b\right)+b\left(a-2b\right)=0\)
\(\Leftrightarrow\left(a-2b\right)\left(b-2a\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2b\\b=2a\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=2\sqrt{x^2-x+1}\\\sqrt{x^2-x+1}=2\sqrt{x+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=4x^2-4x+4\\x^2-x+1=4x+4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x^2-5x+3=0\\x^2-5x-3=0\end{matrix}\right.\)
tự ra kết quả giùm mình nha
cách 2:
\(2\left(x^2+2\right)=5\sqrt{x^3+1}\)
\(VP=5\sqrt[]{x^3+1}=5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)
\(VT=2\left(x^2-x+1\right)+2\left(x+1\right)\)
Pt: \(2\left(x^2-x+1\right)+2\left(x+1\right)=5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)
chia \(x^2-x+1\ge0\) ta được:
\(2+2.\frac{x+1}{x^2-x+1}=5\sqrt{\frac{x+1}{x^2-x+1}}\)
Đặt \(\sqrt{\frac{x+1}{x^2-x+1}}=t\left(t\ge0\right)\)
Pt: \(2+2t^2=5t\)
tới đây bạn tự giải ẩn t rồi thay vào giải nốt ẩn x nha
