Dạng 4:
a: ĐKXĐ: x>=0
\(2\sqrt{x}-9=0\)
=>\(2\sqrt{x}=9\)
=>\(\sqrt{x}=\frac92\)
=>\(x=\left(\frac92\right)^2=\frac{81}{4}\) (nhận)
c: \(\sqrt{x^2+6x+9}=3x-1\)
=>\(\sqrt{\left(x+3\right)^2}=3x-1\)
=>|x+3|=3x-1
=>\(\begin{cases}3x-1\ge0\\ \left(3x-1\right)^2=\left(x+3\right)^2\end{cases}\Rightarrow\begin{cases}3x\ge1\\ \left(3x-1-x-3\right)\left(3x-1+x+3\right)=0\end{cases}\)
=>\(\begin{cases}x\ge\frac13\\ \left(2x-4\right)\left(4x+2\right)=0\end{cases}\Rightarrow x\in\left\lbrace2\right\rbrace\)
Dạng 3:
a: \(x^2-7=\left(x-\sqrt7\right)\left(x+\sqrt7\right)\)
b: \(x^2-9=x^2-3^2=\left(x-3\right)\left(x+3\right)\)
c: \(x^2-2005=\left(x-\sqrt{2005}\right)\left(x+\sqrt{2005}\right)\)
d: \(x^2-2\sqrt2x+2=x^2-2\cdot x\cdot\sqrt2+\left(\sqrt2\right)^2=\left(x-\sqrt2\right)^2\)
e: \(x^2-\sqrt5=\left(x-\sqrt[4]{5}\right)\left(x+\sqrt[4]{5}\right)\)
f: \(x^2-2\sqrt{12}x+12=x^2-2\cdot x\cdot2\sqrt3+\left(2\sqrt3\right)^2=\left(x-2\sqrt3\right)^2\)
