\(VT=\sqrt{6+2\sqrt{3}+2\sqrt{2}+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{3+2+1+2\sqrt{6}+2\sqrt{2}+2\sqrt{3}}-\sqrt{3-2\sqrt{3}.\sqrt{2}+2}=\sqrt{\left(\sqrt{3}+\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}+\sqrt{2}+1-\sqrt{3}+\sqrt{2}=1+2\sqrt{2}=VP\)
Vậy , đẳng thức được chứng minh .
Trục căn ở mẫu:
\(a)\frac{5}{\sqrt{10}}\\ b)\frac{-2}{1-\sqrt{5}}\\ c)\frac{4}{\sqrt{3}+\sqrt{2}}\\ d)\frac{1}{3-2\sqrt{2}}\\ e)\frac{6-\sqrt{6}}{1-\sqrt{6}}\\ g)\frac{3\sqrt{2}-2\sqrt{3}}{2\left(\sqrt{3}-\sqrt{2}\right)}\\ h)\frac{\sqrt{3}-3}{\sqrt{3}-1}\\ i)\frac{\sqrt{15}}{5\sqrt{3}+3\sqrt{5}}\)
Rút gọn
H=\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
F=\(\frac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)
G=\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
E=\(\frac{2\sqrt{3+\sqrt{5-13+\sqrt{48}}}}{\sqrt{6}+\sqrt{2}}\)
D=\(\sqrt{4+\sqrt{15}}+\sqrt{4-\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)
Z=\(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10-2\sqrt{5}}}\)
Thực hiện phép tính:
\(\frac{\sqrt{2+\sqrt{3}}}{2}:\left(\frac{\sqrt{2+\sqrt{3}}}{2}+\frac{2}{\sqrt{6}}+\frac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\right)\)
\(\frac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}\)
a. P=\(\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{6}-\sqrt{2}}+\sqrt{6+2\sqrt{5}-\sqrt{29-12\sqrt{5}}}\)
b.P= (\(\frac{2}{\sqrt{3}-1}-\frac{52}{3\sqrt{3}-1}+\frac{12}{3-\sqrt{3}}\)) ( 5+\(\sqrt{27}\))
c. P= (\(\frac{2+\sqrt{2}}{\sqrt{2}+1}+1\))(\(\frac{2-\sqrt{2}}{\sqrt{2}-1}-1\))
d. P=\(\sqrt{9+\sqrt{17}}-\sqrt{9-\sqrt{17}}-\sqrt{2}\)
đ. P=(2+\(\sqrt{4+\sqrt{6+2\sqrt{5}}}\) )(\(\sqrt{10}-\sqrt{2}\) )
e. P= \(\frac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
ê. P= \(\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}\)
g. G= \(\frac{2\sqrt{3-\sqrt{3+\sqrt{13+\sqrt{48}}}}}{\sqrt{6}-\sqrt{2}}\)
h. H=\(\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}-\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}\)
i. I= \(\frac{2+\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{2-\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}\)
a. P= (\(3+\sqrt{2}+\sqrt{6}\))(\(\sqrt{6-3\sqrt{3}}\))
b. A=(\(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\)): (\(\sqrt{6}+11\))
c. B= \(\frac{\sqrt{8-2\sqrt{12}}}{\sqrt{3}-1}\)-\(\sqrt{8}\)
d. C= \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}\)
đ. D=\(\frac{1}{\sqrt{2}-\sqrt{3}}\sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}\)
e. E= \(\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
ê. G= \(\sqrt{4+5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
g. H=\(\frac{2\sqrt{4+\sqrt{5+21+\sqrt{80}}}}{\sqrt{10}-\sqrt{2}}\)
i. I=\(\sqrt{\frac{4-\sqrt{7}}{4+\sqrt{7}}}+\sqrt{\frac{4+\sqrt{7}}{4-\sqrt{7}}}\)
k. K=\(\frac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
Chứng minh các đẳng thức:
a) \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)=1
b)\(\dfrac{\left(5+2\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{\sqrt{2}+\sqrt{3}}\)-1 =0
c) \(\sqrt{26+15\sqrt{3}}+\sqrt{26-15\sqrt{3}}-5\sqrt{\dfrac{3}{2}}=\dfrac{\sqrt{6}}{2}\)
Rút gọn các biểu thức :
\(a,\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
\(b,\frac{\sqrt{3}+\sqrt{11+6\sqrt{2}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}\)
Rút gọn các biểu thức sau:
a) \(\dfrac{3-2\sqrt{2}}{1-\sqrt{2}}\) b)\(\dfrac{5\sqrt{6}-15}{6-2\sqrt{6}}\)
c) \(\sqrt{\left(3-2\sqrt{2}\right)\left(4-2\sqrt{3}\right)}\)
d) \(\sqrt{\left(6+2\sqrt{5}\right)^3}-\sqrt{\left(6-2\sqrt{5}\right)^3}\)
1. Rút gọn \(A=\frac{\sqrt{14+6\sqrt{5}}-\sqrt{14-6\sqrt{5}}}{\sqrt{\left(\sqrt{5}+1\right)\cdot\sqrt{6-2\sqrt{5}}}}\)
2.Tính a) \(B=\left(\sqrt[3]{2}+1\right)^3\cdot\left(\sqrt[3]{2}-1\right)^3\)
b)Tìm C=\(a^3b-ab^3\) với \(a=\frac{6}{2\sqrt[3]{2}-2+\sqrt[3]{4}}\); \(b=\frac{2}{2\sqrt[3]{2}+2+\sqrt[3]{4}}\)
3. Giải \(\left|x^2-x+1\right|-\left|x-2\right|=6\)
