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\(\frac{3\sqrt{10}+\sqrt{20}-3\sqrt{6}-\sqrt{12}}{\sqrt{5}-\sqrt{3}}\)
\(=\frac{3\sqrt{10}+2\sqrt{5}-3\sqrt{6}-2\sqrt{3}}{\sqrt{5}-\sqrt{3}}\)
\(=\frac{\left(3\sqrt{10}-3\sqrt{6}\right)+\left(2\sqrt{5}-2\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}\)
\(=\frac{3\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)+2\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}\)
\(=3\sqrt{2}+2\)
\(a)\)\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)
\(=\)\(\sqrt{6-6\sqrt{6}+9}+\sqrt{24-12\sqrt{6}+9}\)
\(=\)\(\sqrt{\left(\sqrt{6}+3\right)}+\sqrt{\left(\sqrt{24}+3\right)}\)
\(=\)\(\left|\sqrt{6}+3\right|+\left|\sqrt{24}+3\right|\)
\(=\)\(\sqrt{6}+3+\sqrt{24}+3\)
\(=\)\(\sqrt{6}\left(1+\sqrt{4}\right)+9\)
\(=\)\(3\sqrt{6}+9\)
Chúc bạn học tốt ~
\(b)\)\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)
\(=\)\(\left|2-\sqrt{3}\right|+\sqrt{3-2\sqrt{3}+1}\)
\(=\)\(2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}\) ( vì \(2=\sqrt{4}>\sqrt{3}\) )
\(=\)\(2-\sqrt{3}+\left|\sqrt{3}-1\right|\)
\(=\)\(2-\sqrt{3}+\sqrt{3}-1\) ( vì \(\sqrt{3}>\sqrt{1}=1\) )
\(=\)\(1\)
Chúc bạn học tốt ~
PS : mới lớp 8 sai thì thông cảm >.<
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
1. Sửa đề:
\(\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}+\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}=\frac{(\sqrt{2+\sqrt{3}})^2+(\sqrt{2-\sqrt{3}})^2}{\sqrt{(2+\sqrt{3})(2-\sqrt{3})}}\)
\(=\frac{2+\sqrt{3}+2-\sqrt{3}}{\sqrt{2^2-3}}=\frac{4}{1}=4\)
2.
\(\sqrt{\frac{2+\sqrt{3}}{2-\sqrt{3}}}-\sqrt{\frac{2-\sqrt{3}}{2+\sqrt{3}}}=\frac{(\sqrt{2+\sqrt{3}})^2-(\sqrt{2-\sqrt{3}})^2}{\sqrt{(2+\sqrt{3})(2-\sqrt{3})}}\)
\(=\frac{2+\sqrt{3}-(2-\sqrt{3})}{\sqrt{2^2-3}}=\frac{2\sqrt{3}}{1}=2\sqrt{3}\)
\(\sqrt{28-6\sqrt{3}}-\sqrt{12+6\sqrt{3}}\)
\(=\sqrt{\left(3\sqrt{3}-1\right)^2}-\sqrt{\left(3+\sqrt{3}\right)^2}\)
\(=3\sqrt{3}-1-3-\sqrt{3}=2\sqrt{3}-4\)
học tốt ~
\(a,=\left(2\sqrt{6}-4\sqrt{3}\right)\sqrt{6}+12\sqrt{2}=12-12\sqrt{2}+12\sqrt{2}=12\\ b,=\dfrac{6\left(3-\sqrt{3}\right)}{6}+\sqrt{3}=3-\sqrt{3}+\sqrt{3}=3\)
mình ghi nhầm pn ơi.. bài 2 là \(\left(3-\sqrt{2}\right)\cdot\sqrt{11+6\sqrt{6}}\)
Ta có :
A=\(\sqrt{12+6\sqrt{3}}+\sqrt{12-6\sqrt{3}}\)
=\(\sqrt{9+6\sqrt{3}+3}+\sqrt{9-6\sqrt{3+3}}\)
=\(\sqrt{3^2+2.3.\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{3^2-2.3\sqrt{3}+\left(\sqrt{3}\right)^2}\)
=\(\sqrt{\left(3+\sqrt{3}\right)^2}+\sqrt{\left(3-\sqrt{3}\right)^2}\)
=\(3+\sqrt{3}+3-\sqrt{3}=6\)
Vậy A =6