Bài 1:
1: Thay x=16 vào A, ta được:
\(A=\frac{16+3}{\sqrt{16}+3}=\frac{19}{4+3}=\frac{19}{7}\)
2: \(B=\left(\frac{x+3\sqrt{x}-2}{x-9}-\frac{1}{\sqrt{x}+3}\right)\cdot\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{x+3\sqrt{x}-2-\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}=\frac{\left.\left(\sqrt{x}+1\right)^2\right.}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+1}{\sqrt{x}+3}\)
Bài 3:
a: ĐKXĐ: x>=2
Ta có: \(\sqrt{x^2-4}-\sqrt{x-2}=0\)
=>\(\sqrt{x-2}\left(\sqrt{x+2}-1\right)=0\)
TH1: \(\sqrt{x-2}=0\)
=>x-2=0
=>x=2(nhận)
TH2: \(\sqrt{x+2}-1=0\)
=>x+2=1
=>x=-1(loại)
b: ĐKXĐ: x>=3
\(\sqrt{x^2-9}-\sqrt{4x-12}=0\)
=>\(\sqrt{\left(x-3\right)\left(x+3\right)}-2\cdot\sqrt{\left(x-3\right)}=0\)
=>\(\sqrt{x-3}\left(\sqrt{x+3}-2\right)=0\)
TH1: \(\sqrt{x-3}=0\)
=>x-3=0
=>x=3(nhận)
Th2: \(\sqrt{x+3}-2=0\)
=>\(\sqrt{x+3}=2\)
=>x+3=4
=>x=1(loại)
Bài 2:
a: \(\sqrt{\left(2x-1\right)^2}=3\)
=>|2x-1|=3
=>\(\left[\begin{array}{l}2x-1=3\\ 2x-1=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=4\\ 2x=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2\\ x=-1\end{array}\right.\)
b: \(\sqrt{x^2-2x+1}=12\)
=>\(\sqrt{\left(x-1\right)^2}=12\)
=>|x-1|=12
=>\(\left[\begin{array}{l}x-1=12\\ x-1=-12\end{array}\right.\Rightarrow\left[\begin{array}{l}x=13\\ x=-11\end{array}\right.\)
