2) \(Đk:x\ge-1\)
\(3\left(x^2+2x+2\right)=10\sqrt{x^3+2x^2+2x+1}\)
\(\Leftrightarrow3\left(x^2+2x+2\right)=10\sqrt{\left(x+1\right)\left(x^2+x+1\right)}\)
- Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\left(a\ge0\right)\\\sqrt{x^2+x+1}=b\left(b\ge\dfrac{3}{4}\right)\end{matrix}\right.\)
- Khi đó phương trình trở thành:
\(3\left(a^2+b^2\right)=10ab\)
\(\Leftrightarrow3a^2-10ab+3b^2=0\)
\(\Leftrightarrow\left(a-3b\right)\left(3a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=3b\\3a=b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=3\sqrt{x^2+x+1}\\3\sqrt{x+1}=\sqrt{x^2+x+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=9\left(x^2+x+1\right)\\9\left(x+1\right)=x^2+x+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}9x^2+8x+8=0\left(1\right)\\x^2-8x-8=0\left(2\right)\end{matrix}\right.\)
- Giải pt (1): \(\Delta=8^2-4.9.8=-224< 0\)
\(\Rightarrow Pt\left(1\right)\) vô nghiệm.
- Giải pt (2): \(\Delta=\left(-8\right)^2-4.1.\left(-8\right)=96>0\)
\(\Rightarrow Pt\left(2\right)\) có hai nghiệm phân biệt:
\(x_1=\dfrac{8+\sqrt{96}}{2}=4+2\sqrt{6};x_2=\dfrac{8-\sqrt{96}}{2}=4-2\sqrt{6}\)
- Vậy phương trình đã cho có tập nghiệm \(S=\left\{4+2\sqrt{6};4-2\sqrt{6}\right\}\)
