9: ĐKXĐ: \(\begin{cases}7x^2-17x+7\ge0\\ x\ge0\end{cases}\Rightarrow\begin{cases}x^2-\frac{17}{7}x+1\ge0\\ x\ge0\end{cases}\)
=>\(\begin{cases}x^2-2\cdot x\cdot\frac{17}{14}+\frac{289}{196}-\frac{93}{196}\ge0\\ x\ge0\end{cases}\Rightarrow\begin{cases}\left(x-\frac{17}{14}\right)^2\ge\frac{93}{196}\\ x>=0\end{cases}\)
=>\(\begin{cases}x\ge0\\ \left[\begin{array}{l}x-\frac{17}{14}\ge\frac{\sqrt{93}}{14}\\ x-\frac{17}{14}\le\frac{-\sqrt{93}}{14}\end{array}\right.\end{cases}\Rightarrow\left[\begin{array}{l}x\ge\frac{\sqrt{93}+17}{14}\\ 0\le x\le\frac{-\sqrt{93}+17}{14}\end{array}\right.\)
\(x-1+\sqrt{x}=\sqrt{7x^2-17x+7}\)
=>\(x-1+\sqrt{x}=\sqrt{7x^2-14x+7-3x}\)
=>\(\left(x-1\right)+\sqrt{x}=\sqrt{7\left(x-1\right)^2-3x}\)
Đặt \(x-1=a;\sqrt{x}=b\) (Điều kiện: b>0; a>=1)
Ta sẽ có hệ: \(\begin{cases}a=b^2-1\\ a+b=\sqrt{7a^2-3b^2}\end{cases}\Rightarrow\begin{cases}a=b^2-1\\ 7a^2-3b^2=a^2+2ab+b^2\end{cases}\)
=>\(\begin{cases}a=b^2-1\\ 6a^2-2ab-4b^2=0\end{cases}\Rightarrow\begin{cases}a=b^2-1\\ 3a^2-ab-2b^2=0\end{cases}\)
=>\(\begin{cases}a=b^2-1\\ 3a^2-3ab+2ab-2b^2=0\end{cases}\Rightarrow\begin{cases}a=b^2-1\\ 3a\left(a-b\right)+2b\left(a-b\right)=0\end{cases}\)
=>\(\begin{cases}\left(a-b\right)\left(3a+2b\right)=0\\ a=b^2-1\end{cases}\)
TH1: a-b=0
=>a=b
\(b^2-1=a\)
=>\(b^2-1=b\)
=>\(b^2-b-1=0\)
=>\(b^2-b+\frac14-\frac54=0\)
=>\(\left(b-\frac12\right)^2=\frac54\)
=>\(\left[\begin{array}{l}b-\frac12=\frac{\sqrt5}{2}\\ b-\frac12=-\frac{\sqrt5}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}b=\frac{\sqrt5+1}{2}\left(nhận\right)\\ b=\frac{-\sqrt5+1}{2}\left(loại\right)\end{array}\right.\)
=>\(b=\frac{\sqrt5+1}{2}\)
=>\(x=b^2=\left(\frac{\sqrt5+1}{2}\right)^2=\left(\frac{6+2\sqrt5}{4}\right)=\frac{3+\sqrt5}{2}\) (nhận)
TH2: 3a+2b=0
=>3a=-2b
=>\(a=-\frac23b\)
\(b^2-1=a\)
=>\(b^2-1=-\frac23b\)
=>\(b^2+\frac23b-1=0\)
=>\(b^2+\frac23b+\frac19-\frac{10}{9}=0\)
=>\(\left(b+\frac13\right)^2=\frac{10}{9}\)
mà \(b+\frac13\ge\frac13\left(b\ge0\right)\)
nên \(b+\frac13=\frac{\sqrt{10}}{3}\)
=>\(b=\frac{\sqrt{10}-1}{3}\) (nhận)
=>\(x=\frac{\left(\sqrt{10}-1\right)^2}{3^2}=\frac{11-2\sqrt{10}}{9}\) (nhận)
