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\(A=\frac{1}{\sqrt{3}+1}+\frac{1}{\sqrt{3}-1}\)
\(=\frac{\sqrt{3}-1}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}+\frac{\sqrt{3}+1}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=\frac{\sqrt{3}-1+\sqrt{3}+1}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\sqrt{3}}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\sqrt{3}}{3-1}\)
\(=\frac{2\sqrt{3}}{2}\)
\(=\sqrt{3}\)
\(B=\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\)
\(=\frac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}\left(\sqrt{5}-1\right)}+\frac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}\left(\sqrt{5}+1\right)}\)
\(=\frac{\left(\sqrt{5}+1\right)}{\left(\sqrt{5}-1\right)}+\frac{\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)}\)
\(=\frac{\left(\sqrt{5}+1\right)^2}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}+\frac{\left(\sqrt{5}-1\right)^2}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\)
\(=\frac{5+2\sqrt{5}+1+5-2\sqrt{5}+1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)
\(=\frac{12}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)
\(=\frac{12}{5-1}\)
\(=\frac{12}{4}\)
\(=3\)
a) Ta có: \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
\(=\left(-\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
\(=-2+2\sqrt{5}-\sqrt{5}\)
\(=-2+\sqrt{5}\)
b) \(\left(\frac{1}{2}\sqrt{\frac{1}{2}}-\frac{3}{2}\sqrt{2}+\frac{4}{5}\sqrt{200}\right)\div\frac{1}{8}\)
\(=\left(\frac{\sqrt{2}}{4}-\frac{3\sqrt{2}}{2}+8\sqrt{2}\right)\cdot8\)
\(=\frac{27\sqrt{2}}{4}\cdot8\)
\(=54\sqrt{2}\)
tu lam di cau nao kho thi hoi hoi vay ko ai tra loi cho dau
cau e)
\(A=\sqrt{2+\sqrt{3}}.\sqrt{2-\sqrt{3}}\)(suy ra A>=0)
\(A^2=\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)\)
\(A^2=1\)
A=1
(bai toan co nhieu cach)
cau m)
\(=\frac{\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}}\)
\(=\frac{\sqrt{3}+\sqrt{2}+\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{2}}\)
\(=1\)
cau G)
\(=\frac{5\sqrt{7}}{\sqrt{35}}-\frac{7\sqrt{5}}{\sqrt{35}}+\frac{2\sqrt{70}}{\sqrt{35}}\)
\(=\frac{5}{\sqrt{5}}-\frac{7}{\sqrt{7}}+2\sqrt{2}\)
\(=\sqrt{5}-\sqrt{7}+2\sqrt{2}\)
Bài 1:
a) \(5\sqrt{\frac{1}{5}}+\frac{1}{3}\sqrt{45}+\frac{5-\sqrt{5}}{\sqrt{5}}\)
\(=\sqrt{5}+\sqrt{5}+\sqrt{5}-1\)
\(=3\sqrt{5}-1\)
b) \(\sqrt{48}-6\sqrt{\frac{1}{3}}+\frac{\sqrt{3}-3}{\sqrt{3}}\)
\(=4\sqrt{3}-2\sqrt{3}+1-\sqrt{3}\)
\(=\sqrt{3}+1\)
c) \(\left(\frac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\frac{5}{\sqrt{5}}\right)\div\left(\frac{1}{\sqrt{5}-\sqrt{2}}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\div\frac{\sqrt{5}+\sqrt{2}}{5-2}\)
\(=-\left(\sqrt{2}+\sqrt{5}\right)\cdot\frac{3}{\sqrt{5}+\sqrt{2}}\)
\(=-3\)
Bài 2:
đk: \(x\ge1\)
Ta có: \(\sqrt{4x+4}-\sqrt{9x-9}-8\sqrt{\frac{x+1}{16}}=5\)
\(\Leftrightarrow2\sqrt{x+1}-3\sqrt{x-1}-2\sqrt{x+1}=5\)
\(\Leftrightarrow-3\sqrt{x-1}=5\)
\(\Rightarrow\sqrt{x-1}=-\frac{5}{3}\) (vô lý)
=> PT vô nghiệm