\(已知x=\sqrt{1+\sqrt{x+1}},则x^2+7x-6的整数部分是多少？\)
\[
\begin{align*}
解：x=\sqrt{1+\sqrt{x+1}},且x \geq 0,两边平方\\
x^2-1 = \sqrt{x+1},现次平方\\
\left(x^2-1\right)^2 = x+1 \\
(x+1)(x-1)^2=1 ,展开并移项\\
(x+1)(x^2-2x+1)-1=0\\
x^3-2x^2+x+x^2-2x+1-1=0 \\
x^3-x^2-x=0 \\
x(x^2-x-1)=0 \\
因为 x \geq 0,所以x^2-x-1 = 0,变形x^2 = x+1\\
x = \frac{1\pm\sqrt{5}}{2}\\
因为x \geq 0 所以x = \frac{1+\sqrt{5}}{2}\\\\\\
x^2+7x-6 \\
&=x+1+7x-6 \\
&=8x-5\\
&=8\times \frac{1+\sqrt{5}}{2} -5 \\
&=4+4\sqrt{5}-5 \\
&=4\sqrt{5}-1 \\
&=7.944 \\
\\所以原式的整数部分为7
\end{align*}
\]