Ta có: \(\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)\)
\(=\left(1+\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2\)
\(=1+2\sqrt{3}+3-2\)
\(=2+2\sqrt{3}\)
Ta có: \(\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)\)
\(=\left(1+\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2\)
\(=1+2\sqrt{3}+3-2\)
\(=2+2\sqrt{3}\)
D = \(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\)
A = \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
C = \(\dfrac{2\sqrt{4-\sqrt{5+\sqrt{21+\sqrt{80}}}}}{\sqrt{10}-\sqrt{2}}\)
F = \(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
B = \(\dfrac{1}{\sqrt{1}+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...+\dfrac{1}{\sqrt{n-1}+\sqrt{n}}\)
E = \(\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-...-\dfrac{1}{\sqrt{24}-\sqrt{25}}\)
Tính:
\(A=\sqrt{20}-2\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
\(B=4\sqrt{\left(\sqrt{3}-1\right)^2}+2\sqrt{12}+4\sqrt{\dfrac{1}{2}}\)
\(C=\left(3+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)\left(3-\dfrac{3+\sqrt{3}}{1+\sqrt{3}}\right)\)
\(D=\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}\)
Tính:
\(A=\dfrac{\sqrt{10}-\sqrt{2}}{\sqrt{5}-1}-\dfrac{2-\sqrt{2}}{\sqrt{2}-1}\)
\(B=\left(1+\dfrac{2+\sqrt{2}}{1+\sqrt{2}}\right)\cdot\left(1-\dfrac{2-\sqrt{2}}{1-\sqrt{2}}\right)\)
\(C=\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{2}-\sqrt{3}}\)
\(D=\dfrac{3\sqrt{2}-2\sqrt{3}}{\sqrt{3}-\sqrt{2}}:\dfrac{1}{\sqrt{6}}\)
\(E=\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{1+\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}\)
bài 1:
a) D = \(\sqrt{2-\sqrt{3}}+\sqrt{2+\sqrt{3}}\)
b) E = \(\sqrt[3]{\sqrt{5}-2}+\sqrt[3]{\sqrt{5}+2}\)
c) F =\(\sqrt[3]{182+\sqrt{33125}}+\sqrt[3]{182-\sqrt{33125}}\)
bài 2:
a) C = \(\frac{1}{\sqrt{2}+1}+\frac{1}{\sqrt{3}+\sqrt{2}}+\frac{1}{\sqrt{4}+\sqrt{3}}\)
b) D = \(\frac{3+2\sqrt{3}}{\sqrt{3}}+\frac{2+\sqrt{2}}{\sqrt{2}+1}-\frac{1}{2-\sqrt{3}}\)
c) E =\(\frac{3-x^2}{x+\sqrt{3}}\) với x\(\ne-\sqrt{3}\)
d) F = \(\frac{1}{1+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{2019}+\sqrt{2020}}\)
e) G = \(\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2+...}}}}\) (có vô hạn dấu căn)
* Rút gọn biểu thức
a. \(\sqrt{72}-5\sqrt{2}+3\sqrt{12}\)
b. \(6\sqrt{\dfrac{1}{2}}-\dfrac{2}{\sqrt{2}}-5\sqrt{2}\)
c. \(\dfrac{\sqrt{8}-2}{\sqrt{2}-1}+\dfrac{2}{\sqrt{3}-1}-\dfrac{3}{\sqrt{3}}\)
d. \(\sqrt[3]{64}+\sqrt[3]{27}-2\sqrt[3]{-8}\)
Tính:
\(A=\left(\sqrt{72}-3\sqrt{24}+5\sqrt{8}\right)\sqrt{2}+4\sqrt{27}\)
\(B=\dfrac{1}{\sqrt{2}-1}+\dfrac{14}{3+\sqrt{2}}\)
\(C=\dfrac{5+3\sqrt{5}}{\sqrt{5}}+\dfrac{3\sqrt{3}}{\sqrt{3}+1}-\left(\sqrt{5}+3\right)\)
\(D=\sqrt{\left(1-\sqrt{2}\right)^2}-3\sqrt{18}+4\sqrt{\dfrac{1}{2}}\)
Rút gọn biểu thức sau
\(a.\dfrac{\sqrt{5}-2}{5+2\sqrt{5}}-\dfrac{1}{2+\sqrt{5}}+\dfrac{1}{\sqrt{5}}\)
\(b.\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{2}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\)
\(c.\dfrac{2\sqrt{3}-4}{\sqrt{3}-1}+\dfrac{2\sqrt{2}-1}{\sqrt{2}-1}-\dfrac{1+\sqrt{6}}{\sqrt{2}+3}\)
a,\(\frac{2\sqrt{2}}{\sqrt{2\sqrt{2}+1}+1}-\frac{2\sqrt{2}}{\sqrt{2\sqrt{2}+1}-1}\)
b,\(\frac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}+\frac{1}{2-\sqrt{3}}\)
c,\(\frac{2}{\sqrt{3}-\sqrt{5}}+\frac{3-2\sqrt{3}}{\sqrt{3}-2}\)
d,\(\frac{-4}{\sqrt{7}-\sqrt{5}}+\frac{1}{\sqrt{3}-1}+\frac{4-2\sqrt{5}}{\sqrt{5}-2}\)
e,\(\frac{6}{\sqrt{5}-1}+\frac{7}{1-\sqrt{3}}-\frac{2}{\sqrt{3}-\sqrt{5}}\)
Rút gọn:
C=\(\dfrac{3+\sqrt{5}}{2\sqrt{5}+\sqrt[]{3+\sqrt{5}}}+\dfrac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
D=\(\dfrac{1+\sqrt{\dfrac{3}{2}}}{1+\sqrt{1+\dfrac{\sqrt{3}}{2}}}+\dfrac{1-\sqrt{\dfrac{3}{2}}}{1-\sqrt{1-\dfrac{\sqrt{3}}{2}}}\)
rút gọn
\(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{1+\sqrt{2}}-\dfrac{1}{2-\sqrt{3}}\)
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right).\left(\sqrt{5}-\sqrt{2}\right)\)
\(\dfrac{2}{\sqrt{5}+1}-\sqrt{\dfrac{2}{3-\sqrt{5}}}\)
\(\left(\dfrac{1}{2-\sqrt{5}}+\dfrac{2}{\sqrt{5}+\sqrt{3}}\right)/\dfrac{1}{\sqrt{21-12\sqrt{3}}}\)
\(\dfrac{2}{\sqrt{3}-1}-\sqrt{\dfrac{2}{6-3\sqrt{3}}}\)
\(\dfrac{\sqrt{2}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}\)