Bài 52:
a: ĐKXĐ: 2x-1>=0
=>x>=1/2
\(\sqrt{2x-1}=\sqrt{2}-1\)
=>\(2x-1=\left(\sqrt{2}-1\right)^2=3-2\sqrt{2}\)
=>\(2x=3-2\sqrt{2}+1=4-2\sqrt{2}\)
=>\(x=2-\sqrt{2}\left(nhận\right)\)
b: ĐKXĐ: 3x+11>=0
=>3x>=-11
=>x>=-11/3
\(\sqrt{3x+11}=3+\sqrt{2}\)
=>\(3x+11=\left(3+\sqrt{2}\right)^2=11+6\sqrt{2}\)
=>\(3x=6\sqrt{2}\)
=>\(x=2\sqrt{2}\left(nhận\right)\)
c:
ĐKXĐ: x>=-5
\(\sqrt{x+5}=\sqrt{3}-2\)
mà \(\sqrt{3}-2=\sqrt{3}-\sqrt{4}< 0\)
nên \(x\in\varnothing\)
d: ĐKXĐ: x>=-38
\(\sqrt{x+38}=3+\sqrt{5}\)
=>\(x+38=\left(3+\sqrt{5}\right)^2=14+6\sqrt{5}\)
=>\(x=14+6\sqrt{5}-38=6\sqrt{5}-24\left(nhận\right)\)
Bài 1:
a: \(\dfrac{1}{3+\sqrt{2}}+\dfrac{1}{3-\sqrt{2}}\)
\(=\dfrac{3-\sqrt{2}+3+\sqrt{2}}{9-2}\)
\(=\dfrac{6}{7}\)
b: \(\dfrac{2}{3\sqrt{2}-4}-\dfrac{2}{3\sqrt{2}+4}\)
\(=\dfrac{2\left(3\sqrt{2}+4\right)-2\left(3\sqrt{2}-4\right)}{\left(3\sqrt{2}-4\right)\left(3\sqrt{2}+4\right)}\)
\(=\dfrac{6\sqrt{2}+8-6\sqrt{2}+8}{18-16}=\dfrac{16}{2}=8\)
c: \(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\dfrac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}\)
\(=\dfrac{\left(\sqrt{5}-\sqrt{3}\right)^2+\left(\sqrt{5}+\sqrt{3}\right)^2}{\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)}\)
\(=\dfrac{8-2\sqrt{15}+8+2\sqrt{15}}{2}=\dfrac{16}{2}=8\)
d: \(\dfrac{3}{2\sqrt{2}-3\sqrt{3}}-\dfrac{3}{2\sqrt{2}+3\sqrt{3}}\)
\(=\dfrac{3\left(2\sqrt{2}+3\sqrt{3}\right)-3\left(2\sqrt{2}-3\sqrt{3}\right)}{8-27}\)
\(=\dfrac{6\sqrt{2}+9\sqrt{3}-6\sqrt{2}+9\sqrt{3}}{-19}=-\dfrac{18\sqrt{3}}{19}\)
