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B = \(\sqrt{\sqrt{75-2.2\sqrt{2}.5\sqrt{3}+8}+\sqrt{50-2.2\sqrt{3}.5\sqrt{2}+12}}.\sqrt{3\sqrt{3}-3\sqrt{2}}\)
= \(\sqrt{\sqrt{\left(5\sqrt{3}-2\sqrt{2}\right)^2}+\sqrt{\left(5\sqrt{2}+2\sqrt{3}\right)^2}}.\sqrt{3\sqrt{3}-3\sqrt{2}}\)
= \(\sqrt{5\sqrt{3}-2\sqrt{2}+5\sqrt{2}-2\sqrt{3}}.\sqrt{3\sqrt{3}-3\sqrt{2}}\)
= \(\sqrt{3\sqrt{3}+3\sqrt{2}}.\sqrt{3\sqrt{3}-3\sqrt{2}}=\sqrt{\left(3\sqrt{3}+3\sqrt{2}\right)\left(3\sqrt{3}-3\sqrt{2}\right)}\)
= \(\sqrt{27-18}=\sqrt{9}=3\)
Bài 20:
a) \(\sqrt{9-4\sqrt{5}}\cdot\sqrt{9+4\sqrt{5}}=\sqrt{81-80}=1\)
b) \(\left(2\sqrt{2}-6\right)\cdot\sqrt{11+6\sqrt{2}}=2\left(\sqrt{2}-3\right)\left(3+\sqrt{2}\right)\)
\(=2\left(2-9\right)=2\cdot\left(-7\right)=-14\)
c: \(\sqrt{2}\cdot\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}+1\right)\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
=2
d) \(\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)
\(=\sqrt{4-2\sqrt{3}}\cdot\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)\)
\(=\left(4-2\sqrt{3}\right)\left(2+\sqrt{3}\right)\)
\(=8+4\sqrt{3}-4\sqrt{3}-6\)
=2
- \(5-2\sqrt{6}=3-2\sqrt{2}\cdot\sqrt{3}+2=\left(\sqrt{3}-\sqrt{2}\right)^2\Rightarrow\sqrt{5-2\sqrt{6}}=\sqrt{3}-\sqrt{2}\)
- 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.
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. \(\sqrt{601-20\sqrt{6}}-\sqrt{154-20\sqrt{6}}=\sqrt{\left(10\sqrt{6}-1\right)^2}-\sqrt{\left(5\sqrt{6}-2\right)^2}=10\sqrt{6}-1-5\sqrt{6}+2=1+5\sqrt{6}\)
\(=\frac{21}{2}\left(\sqrt{4+2\sqrt{3}}+\sqrt{6-2\sqrt{5}}\right)^2-3\left(\sqrt{4-2\sqrt{3}}+\sqrt{6+2\sqrt{5}}\right)^2-15\sqrt{15}\)
\(=\frac{21}{2}\left(\sqrt{3}+1+\sqrt{5}-1\right)^2-3\left(\sqrt{3}-1+\sqrt{5}+1\right)^2-15\sqrt{15}\)
\(=\frac{15}{2}\left(\sqrt{3}+\sqrt{5}\right)^2-15\sqrt{15}\)
\(=\frac{15}{2}\left(8+2\sqrt{15}\right)-15\sqrt{15}\)
\(=60+15\sqrt{15}-15\sqrt{15}=60\)
Biểu thức trên = \(\frac{21.\left(\sqrt{4+2\sqrt{3}}+\sqrt{6-2\sqrt{5}}\right)^2}{2}\)\(-\frac{6.\left(\sqrt{4-2\sqrt{3}}+\sqrt{6+2\sqrt{5}}\right)^2}{2}\)\(-15\sqrt{15}\)
\(=\frac{21.\left(\sqrt{3+2\sqrt{3}+1}+\sqrt{5-2\sqrt{5}+1}\right)^2}{2}-\frac{6.\left(\sqrt{3-2\sqrt{3}+1}+\sqrt{5+2\sqrt{5}+1}\right)^2}{2}-15\sqrt{15}\)
\(=\frac{21.\left(\sqrt{3}+\sqrt{5}\right)^2}{2}-\frac{6.\left(\sqrt{3}+\sqrt{5}\right)^2}{2}-15\sqrt{15}\) (đoạn này làm tắt)
\(=\frac{15.\left(\sqrt{3}+\sqrt{5}\right)^2}{2}-15\sqrt{15}\)\(=\frac{15.\left(8+2\sqrt{15}\right)}{2}-15\sqrt{15}\)
\(=60+15\sqrt{15}-15\sqrt{15}=60\)
a: \(A=\dfrac{2\sqrt{2}\left(\sqrt{3}+1\right)}{3\cdot\sqrt{2+\sqrt{3}}}=\dfrac{4\left(\sqrt{3}+1\right)}{3\cdot\sqrt{4+2\sqrt{3}}}\)
\(=\dfrac{4\left(\sqrt{3}+1\right)}{3\left(\sqrt{3}+1\right)}=\dfrac{4}{3}\)
b: \(B=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\left|3\sqrt{5}-3\right|\)
\(=\sqrt{5}-\sqrt{3}-3\sqrt{5}+3=3-\sqrt{3}-2\sqrt{5}\)
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