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a: Theo đề, ta có hệ phương trình:
\(\left\{{}\begin{matrix}-a+b=-20\\3a+b=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=7\\b=8-3a=8-3\cdot7=-13\end{matrix}\right.\)
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Ta có: \(\sqrt{\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}}\)
\(=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2\cdot\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)}\)
\(=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2-2\cdot\dfrac{x+y+z}{xyz}}\)
\(=\sqrt{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}\)
\(=\left|\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right|\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a) \(2\sqrt{18}-7\sqrt{32}-\sqrt{72}+3\sqrt{8}\)
\(=2\sqrt{9.2}-7\sqrt{16.2}-\sqrt{36.2}+3\sqrt{4.2}\)
\(=6\sqrt{2}-28\sqrt{2}-6\sqrt{2}+6\sqrt{2}\)
\(=-22\sqrt{2}\)
b) \(\sqrt{\left(1+2\sqrt{3}\right)^2-\sqrt{4+2\sqrt{3}}}\)
\(=1+2\sqrt{3}-\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=1+2\sqrt{3}-\sqrt{3}-1\)
\(=\sqrt{3}\)
c) \(\dfrac{5\sqrt{3}-3}{5-\sqrt{3}}-\dfrac{4}{\sqrt{3}+\sqrt{7}}-\dfrac{6}{\sqrt{3}}\)
\(=\dfrac{\sqrt{3}\left(5-\sqrt{3}\right)}{5-\sqrt{3}}-\dfrac{4\left(\sqrt{7}-\sqrt{3}\right)}{7-3}-2\sqrt{3}\)
\(=5-\sqrt{3}-\sqrt{7}+\sqrt{3}-2\sqrt{3}\)
\(=5-\sqrt{7}-2\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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Bài 1:
\(\sin\widehat{A}=\dfrac{BC}{BA}\)
\(\cos\widehat{A}=\dfrac{CA}{AB}\)
\(\tan\widehat{A}=\dfrac{BC}{CA}\)
\(\cot\widehat{A}=\dfrac{CA}{BC}\)
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b: =(m-1)^2-4(-m^2-2)
=m^2+2m+1+4m^2+8
=5m^2+2m+9
=5(m^2+2/5m+9/5)
=5(m^2+2*m*1/5+1/25+44/25)
=5(m+1/5)^2+44/5>=44/5>0 với mọi m
=>PT luôn có hai nghiệm pb