Bài 2:
a: ĐKXĐ: x>=-3
Ta có: \(\sqrt{9x+27}+6=\frac32\cdot\sqrt{16x+48}\)
=>\(3\sqrt{x+3}+6=\frac32\cdot4\sqrt{x+3}=6\sqrt{x+3}\)
=>\(3\sqrt{x+3}=6\)
=>\(\sqrt{x+3}=2\)
=>x+3=4
=>x=4-3
=>x=1
b: ĐKXĐ: x∈R
\(\sqrt{x^2+2x+2}=x-1\)
=>\(\begin{cases}x-1\ge0\\ x^2+2x+2=\left(x-1\right)^2\end{cases}\)
=>\(\begin{cases}x\ge1\\ x^2+2x+2=x^2-2x+1\end{cases}\Rightarrow\begin{cases}x\ge1\\ 2x+2x=1-2\end{cases}\)
=>\(\begin{cases}x\ge1\\ 4x=-1\end{cases}\Rightarrow\begin{cases}x\ge1\\ x=-\frac14\end{cases}\)
=>x∈∅
c: ĐKXĐ: x∈R
\(\sqrt{4x^2-12x+9}-7=0\)
=>\(\sqrt{\left(2x-3\right)^2}-7=0\)
=>|2x-3|=7
=>\(\left[\begin{array}{l}2x-3=7\\ 2x-3=-7\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=10\\ 2x=-4\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=-2\end{array}\right.\)
Bài 1:
a: \(\sqrt{18}-\frac12\cdot\sqrt{72}+4\cdot\sqrt{\frac12}+\sqrt{72}\)
\(=3\sqrt2-\frac12\cdot6\sqrt2+2\sqrt2+6\sqrt2\)
\(=3\sqrt2-3\sqrt2+8\sqrt2=8\sqrt2\)
b: \(\frac{2}{3+2\sqrt2}-\frac{2}{3-2\sqrt2}\)
\(=\frac{2\left(3-2\sqrt2\right)-2\left(3+2\sqrt2\right)}{\left(3+2\sqrt2\right)\left(3-2\sqrt2\right)}\)
\(=\frac{6-4\sqrt2-6-4\sqrt2}{9-8}=-8\sqrt2\)
c: \(\frac{2\sqrt3-\sqrt6}{\sqrt2-1}+4\cdot\sqrt{\frac32}+\frac{9}{3+\sqrt6}\)
\(=\frac{\sqrt6\left(\sqrt2-1\right)}{\sqrt2-1}+4\cdot\frac{\sqrt3}{\sqrt2}+\frac{9\left(3-\sqrt6\right)}{\left(3-\sqrt6\right)\left(3+\sqrt6\right)}\)
\(=\sqrt6+2\sqrt6+3\left(3-\sqrt6\right)=3\sqrt6+9-3\sqrt6=9\)
