\(\frac{\sqrt{8-2\sqrt{12}}}{\sqrt{3}-1}\) - \(\sqrt{8}\)
= \(\frac{\sqrt{6-2\sqrt{12}+2}}{\sqrt{3}-1}\) - \(\sqrt{8}\)
= \(\frac{\sqrt{\left(\sqrt{6}-\sqrt{2}\right)^2}}{\sqrt{3}-1}-\sqrt{8}\)
= \(\frac{\sqrt{6}-\sqrt{2}}{\sqrt{3}-1}-\sqrt{8}\)
= \(\frac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\sqrt{8}\)
= \(\sqrt{2}-\sqrt{8}\)
Rút gọn biểu thức:
\(a,\frac{3\sqrt{8}-2\sqrt{12}+\sqrt{20}}{3\sqrt{18}-2\sqrt{27}+\sqrt{45}}\)
\(b,\frac{1+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{1-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
Rút gọn biểu thức sau :
A=\(\frac{\sqrt{8-2\sqrt{12}}}{\sqrt{3}-1}\)-\(\sqrt{8}\)
Rút gọn biểu thức :
a) \(A=\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\)
b) \(B=\frac{2+\sqrt{6}+\sqrt{10}+\sqrt{2}+\sqrt{3}+\sqrt{5}}{\sqrt{8}+\sqrt{12}+\sqrt{20}}.\frac{\sqrt{2}-1}{3}\)
1,Rút gọn:
a, \(\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+2}\)
b,\(\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-\frac{1}{\sqrt{4}-\sqrt{5}}+\frac{1}{\sqrt{5}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{9}}\)
Rút gọn \(A=\frac{\sqrt{\sqrt[4]{8}+\sqrt{\sqrt{2}-1}}-\sqrt{\sqrt[4]{8}-\sqrt{\sqrt{2}-1}}}{\sqrt{\sqrt[4]{8}-\sqrt{\sqrt{2}+1}}}\)
Rút gọn biểu thức:
\(a,\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(b,\frac{2+\sqrt{3}}{1-\sqrt{4-2\sqrt{3}}}+\frac{2-\sqrt{3}}{1+\sqrt{4+2\sqrt{3}}}\)
P=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}+\dfrac{8\sqrt{x}+8}{x+2\sqrt{x}}-\dfrac{\sqrt{x}+2}{\sqrt{x}}\right):\left(\dfrac{x+\sqrt{x}+3}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}}\right)\)
a. rút gọn P
b. chứng minh rằng với mọi giá trị x ta luôn có P\(\le1\)
1) Rút gọn biểu thức
P=\(\left(\dfrac{x+2\sqrt{x}+4}{x\sqrt{x}-8}+\dfrac{x+2\sqrt{x}+4}{x-1}\right):\left(3+\dfrac{1}{\sqrt{x}-2}+\dfrac{2}{\sqrt{x}+1}\right)\)
Rút gọn A = \(\frac{4}{3+\sqrt{5}+\sqrt{2+2\sqrt{5}}}\)
Rút gọn B = \(\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}\)
