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\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=|2+\sqrt{3}|-|2-\sqrt{3}|\)
\(=2+\sqrt{3}-2+\sqrt{3}\)
\(=2\sqrt{3}\)
\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=|3+\sqrt{2}|-|3-\sqrt{2}|\)
\(=3+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}\)
\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(=\sqrt{\left(3+2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}\)
\(=|3+2\sqrt{2}|+|3-2\sqrt{2}|\)
\(=3+2\sqrt{2}+3-2\sqrt{2}\)
\(=6\)
\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=|2+\sqrt{5}|-|2-\sqrt{5}|\)
\(=2+\sqrt{5}-\sqrt{5}+2\)
\(=4\)
\(E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{\left(1-\sqrt{5}\right)^2}\)
\(=|1+\sqrt{5}|-|1-\sqrt{5}|\)
\(=1+\sqrt{5}-\sqrt{5}+1\)
\(=2\)
\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(A=\sqrt{3}+2+2-\sqrt{3}\)
A = 2 + 2
A = 4
\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(B=\sqrt{2}+3+3-\sqrt{2}\)
B = 3 + 3
B = 6
\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(C=3+2\sqrt{2}+3-2\sqrt{2}\)
C = 3 + 3
C = 6
\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(D=\sqrt{5}+2-\sqrt{5}+2\)
D = 2 + 2
D = 4
\(E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(E=\sqrt{5}+1-\sqrt{5}+1\)
E = 1 + 1
E = 2
a) Ta có: \(\sqrt{11-2\sqrt{10}}\)
\(=\sqrt{10-2\cdot\sqrt{10}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{10}-1\right)^2}\)
\(=\left|\sqrt{10}-1\right|=\sqrt{10}-1\)
b) Ta có: \(\sqrt{9-2\sqrt{14}}\)
\(=\sqrt{7-2\cdot\sqrt{7}\cdot\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{7}-\sqrt{2}\right|\)
\(=\sqrt{7}-\sqrt{2}\)
c) Ta có: \(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{3+2\cdot\sqrt{3}\cdot1+1}+\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\left|\sqrt{3}+1\right|+\left|\sqrt{3}-1\right|\)
\(=\sqrt{3}+1+\sqrt{3}-1\)
\(=2\sqrt{3}\)
d) Ta có: \(\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}-\sqrt{5+2\cdot\sqrt{5}\cdot2+4}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(=\left|\sqrt{5}-2\right|-\left|\sqrt{5}+2\right|\)
\(=\sqrt{5}-2-\left(\sqrt{5}+2\right)\)
\(=\sqrt{5}-2-\sqrt{5}-2\)
\(=-4\)
e) Ta có: \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)
\(=\frac{\sqrt{2}\cdot\left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\right)}{\sqrt{2}}\)
\(=\frac{\sqrt{2}\cdot\left(\sqrt{4-\sqrt{7}}\right)-\sqrt{2}\cdot\left(\sqrt{4+\sqrt{7}}\right)}{\sqrt{2}}\)
\(=\frac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}}{\sqrt{2}}\)
\(=\frac{\sqrt{7-2\cdot\sqrt{7}\cdot1+1}-\sqrt{7+2\cdot\sqrt{7}\cdot1+1}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{7}-1\right|-\left|\sqrt{7}+1\right|}{\sqrt{2}}\)
\(=\frac{\sqrt{7}-1-\left(\sqrt{7}+1\right)}{\sqrt{2}}\)
\(=\frac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}\)
\(=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)
g) Ta có: \(\sqrt{3}+\sqrt{11+6\sqrt{2}}+\sqrt{5+2\sqrt{6}}\)
\(=\sqrt{3}+\sqrt{9+2\cdot3\cdot\sqrt{2}+2}+\sqrt{2+2\cdot\sqrt{2}\cdot\sqrt{3}+3}\)
\(=\sqrt{3}+\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}\)
\(=\sqrt{3}+\left|3+\sqrt{2}\right|+\left|\sqrt{2}+\sqrt{3}\right|\)
\(=\sqrt{3}+3+\sqrt{2}+\sqrt{2}+\sqrt{3}\)
\(=3+2\sqrt{3}+2\sqrt{2}\)
h) Ta có: \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\cdot\sqrt{3+2\cdot\sqrt{3}\cdot2+4}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\cdot\sqrt{\left(\sqrt{3}+2\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{48-10\cdot\left(\sqrt{3}+2\right)}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{48-10\sqrt{3}-20}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{25-2\cdot5\cdot\sqrt{3}+3}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\sqrt{\left(5-\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+5\cdot\left(5-\sqrt{3}\right)}\)
\(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)
\(=\sqrt{25}=5\)
k) Ta có: \(\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}\)
\(=\sqrt{49-2\cdot7\cdot\sqrt{45}+45}-\sqrt{49+2\cdot7\cdot\sqrt{45}+45}\)
\(=\sqrt{\left(7-\sqrt{45}\right)^2}-\sqrt{\left(7+\sqrt{45}\right)^2}\)
\(=\left|7-\sqrt{45}\right|-\left|7+\sqrt{45}\right|\)
\(=7-\sqrt{45}-\left(7+\sqrt{45}\right)\)
\(=7-\sqrt{45}-7-\sqrt{45}\)
\(=-2\sqrt{45}=-6\sqrt{5}\)
i) Đặt \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)
\(\Leftrightarrow A^2=\left(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\right)^2\)
\(=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\cdot\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\cdot\left(4-\sqrt{10+2\sqrt{5}}\right)}\)
\(=8+2\cdot\sqrt{16-\left(10+2\sqrt{5}\right)}\)
\(=8+2\cdot\sqrt{6-2\sqrt{5}}\)
\(=8+2\cdot\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=8+2\cdot\left(\sqrt{5}-1\right)\)
\(=8+2\sqrt{5}-2\)
\(=6+2\sqrt{5}\)
\(=\left(\sqrt{5}+1\right)^2\)
\(\Leftrightarrow A=\sqrt{5}+1\)
\(A=\frac{\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-2\sqrt{6}\right)^2\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}{9\sqrt{3}-11\sqrt{2}}=\frac{\left(\sqrt{3}+\sqrt{2}\right)^2\left(\sqrt{3}-\sqrt{2}\right)\left(5-2\sqrt{6}\right)^2}{9\sqrt{3}-11\sqrt{2}}\)
\(=\left(\sqrt{3}+\sqrt{2}\right)\left(9\sqrt{3}+11\sqrt{3}\right)\left(5-2\sqrt{6}\right)^2\)
\(=\left(49+20\sqrt{6}\right)\left(5-2\sqrt{6}\right)^2=\left(5+2\sqrt{6}\right)^2\left(5-2\sqrt{6}\right)^2=1\)
\(A=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)
\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)
\(=\sqrt{4+5}=3\)
\(A=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{3}-1}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\left(\sqrt{3}-1\right)}\)
\(=\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=2\)
\(A=\left(2-\sqrt{3}\right)\sqrt{4+2.2.\sqrt{3}+3}=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=1\)
các câu còn lại làm tương tự nhé bạn !
Giải:
a) \(A=\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)
\(\Leftrightarrow A=\sqrt{5-4\sqrt{5}+4}-\sqrt{5+4\sqrt{5}+4}\)
\(\Leftrightarrow A=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(\Leftrightarrow A=\left|\sqrt{5}-2\right|-\left|\sqrt{5}+2\right|\)
\(\Leftrightarrow A=\sqrt{5}-2-\sqrt{5}-2\left(\sqrt{5}>\sqrt{4}=2\right)\)
\(\Leftrightarrow A=-4\)
Vậy ...
b) \(B=\left(11-4\sqrt{3}\right)\left(11+4\sqrt{3}\right)\)
\(\Leftrightarrow B=11^2-\left(4\sqrt{3}\right)^2\)
\(\Leftrightarrow B=121-48=73\)
Vậy ...
c) \(C=\sqrt{4-2\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)
\(\Leftrightarrow C=\sqrt{3-2\sqrt{3}+1}+\sqrt{3+4\sqrt{3}+4}\)
\(\Leftrightarrow C=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+4\right)^2}\)
\(\Leftrightarrow C=\left|\sqrt{3}-1\right|+\left|\sqrt{3}+4\right|\)
\(\Leftrightarrow C=\sqrt{3}-1+\sqrt{3}+4\)
\(\Leftrightarrow C=2\sqrt{3}+3\)
Vậy ...
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