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\(A=\sqrt{19-3\sqrt{40}}-\sqrt{19+3\sqrt{40}}=\sqrt{19-2\sqrt{90}}-\sqrt{19+2\sqrt{90}}=\sqrt{10-2.\sqrt{10}.3+9}-\sqrt{10+2.\sqrt{10}.3+9}=\sqrt{\left(\sqrt{10}-3\right)^2}-\sqrt{\left(\sqrt{10}+3\right)^2}=\sqrt{10}-3-\sqrt{10}-3=-6\)\(B=\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}=\sqrt{18-2.\sqrt{18}.\sqrt{3}+3}+\sqrt{6+2.\sqrt{3}.\sqrt{6}+3}-\sqrt{24+12\sqrt{3}}=\sqrt{\left(\sqrt{18}-\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{6}+\sqrt{\sqrt{3}}\right)^2}-\sqrt{\left(\sqrt{18}+\sqrt{6}\right)^2}=\sqrt{18}-\sqrt{3}+\sqrt{6}+\sqrt{3}-\sqrt{18}-\sqrt{6}=0\)
\(C=\sqrt{6+2\sqrt{2\sqrt{3-\sqrt{4+2\sqrt{3}}}}}\)
\(C=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(C=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\) \(=\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\) \(=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
\(D=\sqrt{\frac{8+2\sqrt{15}}{2}}-\sqrt{\frac{14-6\sqrt{5}}{2}}\) \(=\sqrt{\frac{\left(\sqrt{5}+\sqrt{3}\right)^2}{2}}-\sqrt{\frac{\left(3-\sqrt{5}\right)^2}{2}}\)
\(=\frac{\sqrt{5}+\sqrt{3}-3+\sqrt{5}}{\sqrt{2}}=\frac{2\sqrt{10}+\sqrt{6}-3\sqrt{2}}{2}\)
\(E=\sqrt{\frac{4+2\sqrt{3}}{2}}+\sqrt{\frac{4-2\sqrt{3}}{2}}\) \(=\sqrt{\frac{\left(\sqrt{3}+1\right)^2}{2}}+\sqrt{\frac{\left(\sqrt{3}-1\right)^2}{2}}\)
\(=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
\(F=\sqrt{\frac{24-6\sqrt{7}}{2}}-\sqrt{\frac{24+6\sqrt{7}}{2}}\) \(=\sqrt{\frac{21-2\sqrt{21\cdot3}+3}{2}}-\sqrt{\frac{21+2\sqrt{21\cdot3}+3}{2}}\)
\(=\sqrt{\frac{\left(\sqrt{21}-\sqrt{3}\right)^2}{2}}-\sqrt{\frac{\left(\sqrt{21}+\sqrt{3}\right)^2}{2}}\)
\(=\frac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}=\frac{-2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
\(G=\left(3+\sqrt{3}\right)\cdot\sqrt{12-6\sqrt{3}}\) \(=\left(3+\sqrt{3}\right)\cdot\sqrt{\left(3-\sqrt{3}\right)^2}\)
\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)=9-3=6\)
\(H=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(3-\sqrt{5}\right)^2}\) \(=\sqrt{5}-2-3-\sqrt{5}=-5\)
\(I=\sqrt{\left(2\sqrt{2}-1\right)^2}-\sqrt{\left(2\sqrt{3}-1\right)^2}\)
\(=2\sqrt{2}-1-2\sqrt{3}+1=2\sqrt{2}-2\sqrt{3}\)
a, \(\frac{1}{\left(\sqrt{3}+\sqrt{2}\right)^2}\) +\(\frac{1}{\left(\sqrt{3}-\sqrt{2}\right)^2}\) =\(\frac{\left(\sqrt{3}+\sqrt{2}\right)^2+\left(\sqrt{3}-\sqrt{2}\right)^2}{\left(\sqrt{3}+\sqrt{2}\right)^2\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\frac{10}{1}=10\)
mấy câu còn lại bạn tự làm nốt nhé mk ban rồi
\(1)\) Ta có :
\(\left(\sqrt{3\sqrt{2}}\right)^4=\left[\left(\sqrt{3\sqrt{2}}\right)^2\right]^2=\left(3\sqrt{2}\right)^2=9.2=18\)
\(\left(\sqrt{2\sqrt{3}}\right)^4=\left[\left(\sqrt{2\sqrt{3}}\right)^2\right]^2=\left(2\sqrt{3}\right)^2=4.3=12\)
Vì \(18>12\) nên \(\left(\sqrt{3\sqrt{2}}\right)^4>\left(\sqrt{2\sqrt{3}}\right)^4\)
\(\Rightarrow\)\(\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)
Vậy \(\sqrt{3\sqrt{2}}>\sqrt{2\sqrt{3}}\)
Chúc bạn học tốt ~
a: \(=2\cdot3+\sqrt{15}-2\sqrt{15}=6-\sqrt{15}\)
b: \(=5\sqrt{10}+2\cdot5-5\sqrt{10}=10\)
c: \(=2\sqrt{7}\cdot\sqrt{7}-\sqrt{12}\cdot\sqrt{7}-\sqrt{7}\cdot\sqrt{7}+2\sqrt{21}=2\cdot7-7=7\)
d: \(=\left(2\sqrt{11}-3\sqrt{2}\right)\cdot\sqrt{11}+3\sqrt{22}=2\cdot11=22\)
Phân tích số dưới dấu căn ra thừa số nguyên tố hoặc đổi thành phân số.
3\(\sqrt{ }\)512 = 3\(\sqrt{ }\)29 = 3\(\sqrt{ }\)(23)3= 23 = 8
3\(\sqrt{ }\)-729 = – 3\(\sqrt{ }\)729 = – 3\(\sqrt{ }\)36=- 3\(\sqrt{ }\)(32)3 = – (32)= -9
3\(\sqrt{ }\)-216 = -3/5
3\(\sqrt{ }\)-0,008 = -1/5