Question

bài 1: giải pt

a. \(\sqrt{x-1}+\sqrt{2x-1}=5\)

b. \(x+\sqrt{2x-1}-2=0\)

bài 2: tính

A=\(\sqrt{7-4\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)

B=\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

C=\(\left(\frac{1}{3-\sqrt{5}}-\frac{1}{3+\sqrt{5}}\right)\).\(\frac{\sqrt{5}-1}{5-\sqrt{5}}\)

Answer 1

Bài 1:

a/ ĐKXĐ: \(x\ge1\)

\(\Leftrightarrow\sqrt{x-1}-2+\sqrt{2x-1}-3=0\)

\(\Leftrightarrow\frac{x-5}{\sqrt{x-1}+2}+\frac{2\left(x-5\right)}{\sqrt{2x-1}+3}=0\)

\(\Leftrightarrow\left(x-5\right)\left(\frac{1}{\sqrt{x-1}+2}+\frac{2}{\sqrt{2x-1}+3}\right)=0\)

\(\Rightarrow x=5\)

b/ĐKXĐ:...

\(x-1+\sqrt{2x-1}-1=0\)

\(\Leftrightarrow x-1+\frac{2\left(x-1\right)}{\sqrt{2x-1}+1}=0\)

\(\Leftrightarrow\left(x-1\right)\left(1+\frac{2}{\sqrt{2x-1}+1}\right)=0\)

\(\Rightarrow x=1\)

Answer 2

Bài 2:

\(A=\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)

\(=\left|2-\sqrt{3}\right|-\left|2+\sqrt{3}\right|\)

\(=2-\sqrt{3}-2-\sqrt{3}=-2\sqrt{3}\)

\(B=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}\)

\(=\left(3-\sqrt{6}\right)+\left(2\sqrt{6}-3\right)\)

\(=\sqrt{6}\)

\(C=\left(\frac{3+\sqrt{5}-3+\sqrt{5}}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right).\frac{\left(\sqrt{5}-1\right)}{\sqrt{5}\left(\sqrt{5}-1\right)}\)

\(=\frac{2\sqrt{5}}{4}.\frac{1}{\sqrt{5}}=\frac{1}{2}\)