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1: \(\dfrac{3}{2}\sqrt{6}+2\sqrt{\dfrac{2}{3}}-4\sqrt{\dfrac{3}{2}}\)
\(=\dfrac{3}{2}\sqrt{6}+2\cdot\sqrt{\dfrac{6}{9}}-4\cdot\sqrt{\dfrac{6}{4}}\)
\(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{6}\)
\(=\dfrac{1}{6}\sqrt{6}\)
2: \(2\sqrt{48}+6\sqrt{\dfrac{1}{3}}-4\sqrt{12}\)
\(=2\cdot4\sqrt{3}+\dfrac{6}{\sqrt{3}}-4\cdot2\sqrt{3}\)
\(=8\sqrt{3}-8\sqrt{3}+2\sqrt{3}=2\sqrt{3}\)
8: \(\sqrt{4-\sqrt{7}}\cdot\sqrt{4+\sqrt{7}}\)
\(=\sqrt{\left(4-\sqrt{7}\right)\left(4+\sqrt{7}\right)}\)
\(=\sqrt{16-7}=\sqrt{9}=3\)
9: \(\sqrt{2}\cdot\sqrt{2+\sqrt{2}}\cdot\sqrt{2-\sqrt{2}}\)
\(=\sqrt{2}\sqrt{\left(2+\sqrt{2}\right)\left(2-\sqrt{2}\right)}\)
\(=\sqrt{2}\cdot\sqrt{4-2}\)
\(=\sqrt{2\cdot2}=2\)
Kẻ CD//AB thì CD//MN
Do đó \(\widehat{ACD}=\widehat{CAB}=41^0;\widehat{MCD}=\widehat{CMN}=54^0\) (so le trong)
Vậy \(\widehat{ACM}=\widehat{ACD}+\widehat{DCM}=41^0+54^0=95^0\)
Ta có: \(3x=4y\Rightarrow\dfrac{x}{4}=\dfrac{y}{3}\Rightarrow\dfrac{x}{20}=\dfrac{y}{15}\)
\(2y=5z\Rightarrow\dfrac{y}{5}=\dfrac{z}{2}\Rightarrow\dfrac{y}{15}=\dfrac{z}{6}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{20}=\dfrac{y}{15}=\dfrac{z}{6}=\dfrac{x+z}{20+6}=\dfrac{52}{26}=2\)
\(\Rightarrow\left\{{}\begin{matrix}x=20.2=40\\y=15.2=30\\z=6.2=12\end{matrix}\right.\)
.Đề dou ọ