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a)
\(A=\sqrt{26+15\sqrt{3}}=\sqrt{\frac{52+30\sqrt{3}}{2}}=\sqrt{\frac{27+25+2\sqrt{27.25}}{2}}\)
\(=\sqrt{\frac{(\sqrt{27}+\sqrt{25})^2}{2}}=\frac{\sqrt{27}+\sqrt{25}}{\sqrt{2}}=\frac{3\sqrt{3}+5}{\sqrt{2}}=\frac{3\sqrt{6}+5\sqrt{2}}{2}\)
b)
\(B\sqrt{2}=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2\)
\(=\sqrt{7+1+2\sqrt{7}}-\sqrt{7+1-2\sqrt{7}}-2\)
\(=\sqrt{(\sqrt{7}+1)^2}-\sqrt{(\sqrt{7}-1)^2}-2=\sqrt{7}+1-(\sqrt{7}-1)-2=0\)
\(\Rightarrow B=0\)
c)
\(C=\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}=\sqrt{3+5-2\sqrt{3.5}}-\sqrt{3+5+2\sqrt{3.5}}\)
\(=\sqrt{(\sqrt{5}-\sqrt{3})^2}-\sqrt{(\sqrt{5}+\sqrt{3})^2}=(\sqrt{5}-\sqrt{3})-(\sqrt{5}+\sqrt{3})=-2\sqrt{3}\)
d)
\(D=(\sqrt{6}-2)(5+2\sqrt{6})\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{2}(\sqrt{3}-\sqrt{2})(2+3+2\sqrt{2.3})\sqrt{2+3-2\sqrt{2.3}}\)
\(=\sqrt{2}(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})^2\sqrt{(\sqrt{3}-\sqrt{2})^2}\)
\(=\sqrt{2}(\sqrt{3}-\sqrt{2})^2(\sqrt{3}+\sqrt{2})^2=\sqrt{2}[(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})]^2\)
\(=\sqrt{2}.1^2=\sqrt{2}\)
e)
\(E=(\sqrt{10}-\sqrt{2})\sqrt{3+\sqrt{5}}=(\sqrt{5}-1).\sqrt{2}.\sqrt{3+\sqrt{5}}\)
\(=(\sqrt{5}-1)\sqrt{6+2\sqrt{5}}=(\sqrt{5}-1)\sqrt{5+1+2\sqrt{5.1}}\)
\(=(\sqrt{5}-1)\sqrt{(\sqrt{5}+1)^2}=(\sqrt{5}-1)(\sqrt{5}+1)=4\)
f)
\(F=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20+9-2\sqrt{20.9}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{(\sqrt{20}-3)^2}}}=\sqrt{\sqrt{5}-\sqrt{3-(\sqrt{20}-3)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{5+1-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{(\sqrt{5}-1)^2}}=\sqrt{\sqrt{5}-(\sqrt{5}-1)}=\sqrt{1}=1\)
Sửa đề
\(A=\left(2-\sqrt{3}\right)\sqrt[3]{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt[3]{26-15\sqrt{3}}\)
\(=\left(2-\sqrt{3}\right)\sqrt[3]{8+12\sqrt{3}+18+3\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt[3]{8-12\sqrt{3}+18-3\sqrt{3}}\)
\(=\left(2-\sqrt{3}\right)\sqrt[3]{\left(2+\sqrt{3}\right)^3}-\left(2+\sqrt{3}\right)\sqrt[3]{\left(2-\sqrt{3}\right)^3}\)
\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)-\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=0\)
a) Ta có: \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2\cdot\sqrt{20}\cdot3+9}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5-2\cdot\sqrt{5}\cdot1+1}}\)
\(=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}\)
\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}\)
\(=\sqrt{1}=1\)
b) Ta có: \(\sqrt{6+2\sqrt{5}-\sqrt{29-12\sqrt{5}}}\)
\(=\sqrt{6+2\sqrt{5}-\sqrt{20-2\cdot2\sqrt{5}\cdot3+9}}\)
\(=\sqrt{6+2\sqrt{5}-\sqrt{\left(2\sqrt{5}-3\right)^2}}\)
\(=\sqrt{6+2\sqrt{5}-\left(2\sqrt{5}-3\right)}\)
\(=\sqrt{6+3}=3\)
c) Sửa đề: \(\sqrt{2+\sqrt{5+\sqrt{13-\sqrt{48}}}}\)
Ta có: \(\sqrt{2+\sqrt{5+\sqrt{13-\sqrt{48}}}}\)
\(=\sqrt{2+\sqrt{5+\sqrt{12-2\cdot2\sqrt{3}\cdot1+1}}}\)
\(=\sqrt{2+\sqrt{5+\sqrt{\left(2\sqrt{3}-1\right)^2}}}\)
\(=\sqrt{2+\sqrt{5+2\sqrt{3}-1}}\)
\(=\sqrt{2+\sqrt{3+2\sqrt{3}\cdot1+1}}\)
\(=\sqrt{2+\sqrt{\left(\sqrt{3}+1\right)^2}}\)
\(=\sqrt{3+\sqrt{3}}\)
d) Ta có: \(\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}\)
\(=\dfrac{\left(6-2\sqrt{5}\right)\sqrt{6+2\sqrt{5}}+\left(6+2\sqrt{5}\right)\sqrt{6-2\sqrt{5}}}{2\sqrt{2}}\)
\(=\dfrac{\left(\sqrt{5}-1\right)^2\cdot\left(\sqrt{5}+1\right)+\left(\sqrt{5}+1\right)^2\cdot\left(\sqrt{5}-1\right)}{2\sqrt{2}}\)
\(=\dfrac{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)\left(\sqrt{5}-1+\sqrt{5}+1\right)}{2\sqrt{2}}\)
\(=\dfrac{4\cdot2\sqrt{5}}{2\sqrt{2}}\)
\(=\dfrac{8\sqrt{5}}{2\sqrt{2}}=\dfrac{4\sqrt{5}}{\sqrt{2}}=2\sqrt{10}\)
a: \(\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)
\(=4-\sqrt{15}+\sqrt{15}=4\)
b: \(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=2+\sqrt{3}-2+\sqrt{3}\)
\(=2\sqrt{3}\)
c: \(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(=\sqrt{\left(2\sqrt{5}+3\right)^2}-\sqrt{\left(2\sqrt{5}-3\right)^2}\)
\(=2\sqrt{5}+3-2\sqrt{5}+3=6\)
9: \(A=\dfrac{\sqrt{8+2\sqrt{15}}-\sqrt{14-6\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}-3+\sqrt{5}}{\sqrt{2}}=\dfrac{2\sqrt{10}+\sqrt{6}-3\sqrt{2}}{2}\)
10: \(A=\dfrac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
11: \(A=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{24+6\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}=-\dfrac{2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
12: \(B=\left(3+\sqrt{3}\right)\sqrt{12-6\sqrt{3}}\)
\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)\)
=9-3=6
13: \(A=\sqrt{5}-2-\left(3-\sqrt{5}\right)\)
\(=\sqrt{5}-2-3+\sqrt{5}=2\sqrt{5}-5\)
a) Ta có: \(M=\dfrac{2}{\sqrt{7}-\sqrt{6}}-\sqrt{28}+\sqrt{54}\)
\(=\dfrac{2\left(\sqrt{7}+\sqrt{6}\right)}{\left(\sqrt{7}-\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right)}-2\sqrt{7}+3\sqrt{6}\)
\(=2\sqrt{7}+2\sqrt{6}-2\sqrt{7}+3\sqrt{6}\)
\(=5\sqrt{6}\)
b) Ta có: \(N=\left(2-\sqrt{3}\right)\left(\sqrt{26+15\sqrt{3}}\right)-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{52+30\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{52-30\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{27+2\cdot3\sqrt{3}\cdot5+25}-\left(2+\sqrt{3}\right)\sqrt{27-2\cdot3\sqrt{3}\cdot5+25}}{\sqrt{2}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{\left(3\sqrt{3}+5\right)^2}-\left(2+\sqrt{3}\right)\sqrt{\left(3\sqrt{3}-5\right)^2}}{\sqrt{2}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\left(3\sqrt{3}+5\right)-\left(2+\sqrt{3}\right)\left(3\sqrt{3}-5\right)}{\sqrt{2}}\)
\(=\dfrac{6\sqrt{3}+10-9-5\sqrt{3}-\left(6\sqrt{3}-10+9-5\sqrt{3}\right)}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}+1-\sqrt{3}+1}{\sqrt{2}}\)
\(=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
a: Sửa đề: \(A=\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)
\(=4-\sqrt{15}+\sqrt{15}=4\)
b: \(A=2-\sqrt{3}+\sqrt{3}-1=1\)
c: \(C=3\sqrt{5}-2-3\sqrt{5}-2=-4\)
d: Sửa đề: \(D=\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(=2\sqrt{5}+3-2\sqrt{5}+3\)
=6
a) \(A=\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}\)
\(A=\left|4-\sqrt{15}\right|+\sqrt{15}\)
\(A=4-\sqrt{15}+\sqrt{15}\)
\(A=4\)
b) \(B=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1-\sqrt{3}\right)}\)
\(B=\left|2-\sqrt{3}\right|+\left|1-\sqrt{3}\right|\)
\(B=2-\sqrt{3}-1+\sqrt{3}\)
\(B=1\)
c) \(C=\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}\)
\(C=\sqrt{\left(3\sqrt{5}\right)^2-2\cdot3\sqrt{15}\cdot2+2^2}-\sqrt{\left(3\sqrt{5}\right)^2+2\cdot3\sqrt{5}\cdot2+2^2}\)
\(C=\sqrt{\left(3\sqrt{5}-2\right)^2}-\sqrt{\left(3\sqrt{5}+2\right)^2}\)
\(C=\left|3\sqrt{5}-2\right|-\left|3\sqrt{5}+2\right|\)
\(C=3\sqrt{5}-2-3\sqrt{5}-2\)
\(C=-4\)
d) \(D=\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)
\(D=\sqrt{\left(2\sqrt{5}\right)^2+2\cdot2\sqrt{5}\cdot3+3^2}-\sqrt{\left(2\sqrt{5}\right)^2-2\cdot2\sqrt{5}\cdot3+3^3}\)
\(D=\sqrt{\left(2\sqrt{5}+3\right)^2}-\sqrt{\left(2\sqrt{5}-3\right)^2}\)
\(D=\left|2\sqrt{5}+3\right|-\left|2\sqrt{5}-3\right|\)
\(D=2\sqrt{5}+3-2\sqrt{5}+3\)
\(D=6\)
\(\sqrt{242}.\sqrt{26}.\sqrt{130}.\sqrt{0,9}-\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)\)
\(=\sqrt{121}.\sqrt{2}.\sqrt{2}.\sqrt{13}.\sqrt{13}.\sqrt{10}.\sqrt{0,9}-\left(2-1\right)\)
\(=11.2.13.\sqrt{9}-1=286.3-1=857\)
\(\frac{3-\sqrt{6}}{\sqrt{12}-\sqrt{8}}-\frac{\sqrt{15}-\sqrt{5}}{2\sqrt{12}-4}+\frac{\sqrt{17-4\sqrt{15}}}{4}\)
\(=\frac{\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{2\left(\sqrt{3}-\sqrt{2}\right)}-\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{4\left(\sqrt{3}-1\right)}+\frac{\sqrt{\left(2\sqrt{3}-\sqrt{5}\right)^2}}{4}\)
\(=\frac{\sqrt{3}}{2}-\frac{\sqrt{5}}{4}+\frac{2\sqrt{3}-\sqrt{5}}{4}\)
\(=\sqrt{3}-\frac{\sqrt{5}}{4}\)