a) \(\sqrt{40}.\sqrt{12,1}.\sqrt{0,09}=\sqrt{40.12,1.0,09}=\sqrt{\frac{1089}{25}}=\frac{33}{5}\)
b) \(\sqrt{3,5}.\sqrt{2,5}.\sqrt{7}.\sqrt{\frac{1}{5}}=\sqrt{3,5.2,5.7.\frac{1}{5}}=\sqrt{\frac{49}{4}}=\frac{7}{2}\)
\(a,\sqrt{40}.\sqrt{12,1}.\sqrt{0,09}=\sqrt{40.12,1.0,09}=\sqrt{43,56}=6,6\)
\(b,\sqrt{3,5}.\sqrt{2,5}.\sqrt{7}.\sqrt{\frac{1}{5}}=\sqrt{3,5.2,5.7.\frac{1}{5}}=\sqrt{12,25}=3,5\)
tính:\(\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-\frac{1}{\sqrt{4}-\sqrt{5}}+\frac{1}{\sqrt{5}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{7}}+\frac{1}{\sqrt{7}-\sqrt{8}}-\frac{1}{\sqrt{8}-\sqrt{9}}\)
Tính :
a ) \(S=\frac{1}{3\sqrt{1}+3\sqrt{3}}+\frac{1}{3\sqrt{3}+3\sqrt{5}}+...+\)\(\frac{1}{3\sqrt{2017}+3\sqrt{2019}}\)
b ) \(S=\frac{7}{\sqrt{2.2}+\sqrt{2.3}}+\frac{7}{\sqrt{2.3}+\sqrt{2.4}}\)\(+...+\frac{7}{\sqrt{2.2018}+\sqrt{2.2019}}\)
Tính :
a ) \(S=\frac{1}{\sqrt{1}\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+.....+\)\(\frac{1}{\sqrt{2017}+\sqrt{2019}}\)
b ) \(S=\frac{1}{\sqrt{2}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{6}}+....+\frac{1}{\sqrt{100}+\sqrt{102}}\)
c ) \(S=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+.....+\frac{1}{\sqrt{100}+\sqrt{101}}\)
d ) \(S=\frac{1}{\sqrt{3}+\sqrt{6}}+\frac{1}{\sqrt{6}+\sqrt{9}}+\frac{1}{\sqrt{9}+\sqrt{12}}+....+\frac{1}{\sqrt{2016}+\sqrt{2019}}\)
Rút gọn:
a) \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+\sqrt{48}}}}\)
b) \(\dfrac{\sqrt{3}-\sqrt{5+\sqrt{24}}+\sqrt{\sqrt{72}+11}}{\sqrt{6+\sqrt{20}}+\sqrt{2}-\sqrt{7+\sqrt{40}}}\)
Bài 1: Tính giá trị của biểu thức:
1)\(H=\sqrt[3]{3+\sqrt{9+\frac{125}{7}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{7}}}\)
2)\(P=\frac{2}{\sqrt{4-3\sqrt[4]{5}+2\sqrt{5}-\sqrt[4]{125}}}\)
Bài 2: Tính giá trị biểu thức: \(Q=\sqrt[10]{\frac{19+6\sqrt{10}}{2}}.\sqrt[5]{3\sqrt{2}-2\sqrt{5}}\)
\(K=\frac{\sqrt{\sqrt[4]{8}+\sqrt{\sqrt{2}-1}}-\sqrt{\sqrt[4]{8}-\sqrt{\sqrt{2}-1}}}{\sqrt{\sqrt[4]{8}-\sqrt{\sqrt{2}+1}}}\)
\(A=\left(\sqrt{3}+1\right)^2+\frac{5}{4}\sqrt{48}-\frac{2}{\sqrt{3+1}}\)
\(B=\frac{4}{3-\sqrt{5}}-\frac{3}{\sqrt{5}+\sqrt{2}}-\frac{1}{\sqrt{2}-1}\)
\(C=\sqrt{4-2\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
Rút gọn
Giúp mình với mình cần gấp
rút gọn biểu thức \(A=\frac{\sqrt{20}+2}{\sqrt{3}-1}-\frac{\sqrt{112}+4}{\sqrt{5}+1}+\sqrt{5}\left(\sqrt{7}-\sqrt{3}\right)\)
Cho a,b,c>0 và\(a+b+c=\sqrt{a}+\sqrt{b}+\sqrt{c}=2\)Tính\(A=\frac{1+a}{\sqrt{a}+\sqrt{b}}+\frac{1+b}{\sqrt{b}+\sqrt{c}}+\frac{1+c}{\sqrt{c}+\sqrt{a}}\)
\(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{2}=\sqrt{\frac{2\left(4-\sqrt{7}\right)}{2}}-\sqrt{\frac{2\left(4+\sqrt{7}\right)}{2}}+\sqrt{2}\)
=\(\sqrt{\frac{8-2\sqrt{7}}{2}}-\sqrt{\frac{8+2\sqrt{7}}{2}}+\sqrt{2}\)
=\(\sqrt{\frac{\left(\sqrt{7}-1\right)^2}{2}}-\sqrt{\frac{\left(\sqrt{7}+1\right)^2}{2}}+\sqrt{2}\)
=\(\frac{\sqrt{7}-1}{\sqrt{2}}-\frac{\sqrt{7}+1}{\sqrt{2}}+\sqrt{2}\)
=\(\frac{-2}{\sqrt{2}}+\sqrt{2}\)
=\(-\sqrt{2}+\sqrt{2}\)
=0
