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B = \(\sqrt{\sqrt{75-2.2\sqrt{2}.5\sqrt{3}+8}+\sqrt{50-2.2\sqrt{3}.5\sqrt{2}+12}}.\sqrt{3\sqrt{3}-3\sqrt{2}}\)
= \(\sqrt{\sqrt{\left(5\sqrt{3}-2\sqrt{2}\right)^2}+\sqrt{\left(5\sqrt{2}+2\sqrt{3}\right)^2}}.\sqrt{3\sqrt{3}-3\sqrt{2}}\)
= \(\sqrt{5\sqrt{3}-2\sqrt{2}+5\sqrt{2}-2\sqrt{3}}.\sqrt{3\sqrt{3}-3\sqrt{2}}\)
= \(\sqrt{3\sqrt{3}+3\sqrt{2}}.\sqrt{3\sqrt{3}-3\sqrt{2}}=\sqrt{\left(3\sqrt{3}+3\sqrt{2}\right)\left(3\sqrt{3}-3\sqrt{2}\right)}\)
= \(\sqrt{27-18}=\sqrt{9}=3\)
a. \(\sqrt{49-20\sqrt{6}}-\sqrt{106+20\sqrt{6}}=\sqrt{\left(5-2\sqrt{6}\right)^2}-\sqrt{\left(10+\sqrt{6}\right)^2}=5-2\sqrt{6}-10-\sqrt{6}=-5-3\sqrt{6}\)
b. \(\sqrt{83-20\sqrt{6}}+\sqrt{62-20\sqrt{6}}=\sqrt{\left(5\sqrt{3}-2\sqrt{2}\right)^2}+\sqrt{\left(5\sqrt{2}-2\sqrt{3}\right)^2}=5\sqrt{3}-2\sqrt{2}+5\sqrt{2}-2\sqrt{3}=3\sqrt{3}+3\sqrt{2}\)
c. \(\sqrt{302-20\sqrt{6}}+\sqrt{203-20\sqrt{6}}=\sqrt{\left(10\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(10\sqrt{2}-\sqrt{3}\right)^2}=10\sqrt{3}-\sqrt{2}+10\sqrt{2}-\sqrt{3}=9\sqrt{3}+9\sqrt{2}\)
d. \(\sqrt{601-20\sqrt{6}}-\sqrt{154-20\sqrt{6}}=\sqrt{\left(10\sqrt{6}-1\right)^2}-\sqrt{\left(5\sqrt{6}-2\right)^2}=10\sqrt{6}-1-5\sqrt{6}+2=1+5\sqrt{6}\)
- \(5-2\sqrt{6}=3-2\sqrt{2}\cdot\sqrt{3}+2=\left(\sqrt{3}-\sqrt{2}\right)^2\Rightarrow\sqrt{5-2\sqrt{6}}=\sqrt{3}-\sqrt{2}\)
- Tương tự \(5+2\sqrt{6}=\left(\sqrt{3}+\sqrt{2}\right)^2\)
- Tử số: \(TS=\left(\sqrt{3}+\sqrt{2}\right)^2\left(49-20\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)=\)
\(=\left(\sqrt{3}+\sqrt{2}\right)\left(49-20\sqrt{6}\right)\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)=\)
\(=49\sqrt{3}+49\sqrt{2}-20\cdot3\sqrt{2}-20\cdot2\sqrt{3}=9\sqrt{3}-11\sqrt{2}\)
- Vậy C = 1.
\(=\frac{21}{2}\left(\sqrt{4+2\sqrt{3}}+\sqrt{6-2\sqrt{5}}\right)^2-3\left(\sqrt{4-2\sqrt{3}}+\sqrt{6+2\sqrt{5}}\right)^2-15\sqrt{15}\)
\(=\frac{21}{2}\left(\sqrt{3}+1+\sqrt{5}-1\right)^2-3\left(\sqrt{3}-1+\sqrt{5}+1\right)^2-15\sqrt{15}\)
\(=\frac{15}{2}\left(\sqrt{3}+\sqrt{5}\right)^2-15\sqrt{15}\)
\(=\frac{15}{2}\left(8+2\sqrt{15}\right)-15\sqrt{15}\)
\(=60+15\sqrt{15}-15\sqrt{15}=60\)
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@@22@22@22@@222@@2@@2@@@2@2
\(A=\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\sqrt{2}+1+\sqrt{2}-1=2\sqrt{2}\)
\(B=\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\frac{\sqrt{3}.\sqrt{5}-\sqrt{2}.\sqrt{3}}{\sqrt{5}.\sqrt{7}-\sqrt{2}.\sqrt{7}}=\frac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\frac{\sqrt{3}}{\sqrt{7}}=\sqrt{\frac{3}{7}}\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3}-1}}\)
\(C=\sqrt{6+2\sqrt{2}.\sqrt{2-\sqrt{3}}}\)
\(C=\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(C=\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(C=\sqrt{6+2.\left(\sqrt{3}-1\right)}\)
\(C=\sqrt{6+2\sqrt{3}-2}\)
\(C=\sqrt{4+2\sqrt{3}}\)
\(C=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
1) Ta có: \(\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}\)
\(=\sqrt{2+2\sqrt{2}+1}+\sqrt{2-2\sqrt{2}+1}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(\sqrt{2}-1\right)^2}\)
\(=\sqrt{2}+1+\sqrt{2}-1\)
\(=2\sqrt{2}\approx2,82843\)
2) Ta có: \(B=\frac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\)
\(\Leftrightarrow B=\frac{\sqrt{5}.\sqrt{3}-\sqrt{2}.\sqrt{3}}{\sqrt{5}.\sqrt{7}-\sqrt{2}.\sqrt{7}}\)
\(\Leftrightarrow B=\frac{\sqrt{3}.\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}.\left(\sqrt{5}-\sqrt{2}\right)}\)
\(\Leftrightarrow B=\frac{\sqrt{3}}{\sqrt{7}}\approx0,65465\)
3) Ta có: \(C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
\(\Leftrightarrow C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3+2\sqrt{3}+1}}}\)
\(\Leftrightarrow C=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{8}.\sqrt{3-\sqrt{3}-1}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{2.8-2.2.\sqrt{3}.2}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{12-2.\sqrt{4.3}.2+1}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{12-2.\sqrt{12}.2+4}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{\left(\sqrt{12}-2\right)^2}}\)
\(\Leftrightarrow C=\sqrt{6+\sqrt{12}-2}\)
\(\Leftrightarrow C=\sqrt{3+2\sqrt{3}+1}\)
\(\Leftrightarrow C=\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(\Leftrightarrow C=\sqrt{3}+1\approx2,73205\)
Phân tích cái trong ngặc đầu thành: (5 căn 3 - 2 căn 2)^2
cái thứ 2 là ( 5 căn 2 - 2 căn 3)^2
sau đó phá đc 1 ngặc làm tiếp
fan vũ đúng ko