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![](https://rs.olm.vn/images/avt/0.png?1311)
Giá trị của \(\sqrt{0,16}\) là
(A) 0,04 (B) 0,4
(C) 0,04 và -0,04 (D) 0,4 và -0,4
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(2\sqrt{31}=\sqrt{4.31}=\sqrt{124}>\sqrt{100}=10\\\Rightarrow2\sqrt{31}>10\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
Suy ra y = 21o48'
=> x = 90o - y = 68o12' (x, y là hai góc phụ nhau)
Vậy x – y = 68o12' - 21o48' = 46o24'
![](https://rs.olm.vn/images/avt/0.png?1311)
\(i,\sqrt{12,1.360}=\sqrt{12,1}.6\sqrt{10}=6.\sqrt{12,1.10}=6.\sqrt{121}=6.\sqrt{11^2}=6.11=66\)
\(k,\sqrt{0,4}.\sqrt{6,4}=\sqrt{0,4.6,4}=\sqrt{\dfrac{64}{25}}=\dfrac{\sqrt{8^2}}{\sqrt{5^2}}=\dfrac{8}{5}\)
\(l,-0,4.\sqrt{\left(-0,4\right)^2}=-0,4.0,4=-0,16\)
\(m,\sqrt{2^4.\left(-7\right)^2}=\sqrt{4^2}.\sqrt{\left(-7\right)^2}=4.7=28\)
i, \(\sqrt{12,1\cdot360}=\sqrt{4356}=\sqrt{66^2}=66\)
k, \(\sqrt{0,4}.\sqrt{6,4}=\sqrt{0,4\cdot6,4}=\sqrt{\dfrac{64}{25}}=\sqrt{\dfrac{2^6}{5^2}}=\dfrac{2^3}{5}=\dfrac{8}{5}\)
l, \(-0,4\sqrt{\left(-0,4\right)^2}=-0,4\cdot\left|-0,4\right|=-0,4\cdot0,4=-\dfrac{4}{25}\)
m, \(\sqrt{2^4\cdot\left(-7\right)^2}=2^2\cdot\left|-7\right|=4\cdot7=28\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: \(\dfrac{-4}{3}\cdot\sqrt{\left(-0.4\right)^2}\)
\(=-\dfrac{4}{3}\cdot0.4\)
\(=\dfrac{-1.6}{3}=-\dfrac{8}{15}\)
b) Ta có: \(\sqrt[3]{\dfrac{3}{4}}\cdot\sqrt[3]{\dfrac{9}{16}}\)
\(=\sqrt[3]{\dfrac{27}{64}}=\dfrac{3}{4}\)
c) Ta có: \(\dfrac{1}{3+\sqrt{2}}+\dfrac{1}{3-\sqrt{2}}\)
\(=\dfrac{3-\sqrt{2}+3+\sqrt{2}}{7}\)
\(=\dfrac{6}{7}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\dfrac{-4}{3}\cdot\sqrt{\left(-0.4\right)^2}=\dfrac{-4}{3}\cdot\dfrac{2}{5}=\dfrac{-8}{15}\)
b: \(\sqrt[3]{\dfrac{3}{4}}\cdot\sqrt[3]{\dfrac{9}{16}}=\dfrac{3}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,6x^2-2x+0,4=0\)
\(\Leftrightarrow\dfrac{8}{5}x^2-2x+\dfrac{2}{5}=0\)
\(\Leftrightarrow\dfrac{8}{5}x^2-\dfrac{8}{5}x-\dfrac{2}{5}x+\dfrac{2}{5}=0\)
\(\Leftrightarrow\dfrac{8}{5}x\left(x-1\right)-\dfrac{2}{5}\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(\dfrac{8}{5}x-\dfrac{2}{5}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\dfrac{8}{5}x-\dfrac{2}{5}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\dfrac{8}{5}x=\dfrac{2}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy \(S=\left\{1;\dfrac{1}{4}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(\sqrt{\left(0,1\right)^2}=\left|0,1\right|=0,1\)do \(0,1>0\)
b, \(\sqrt{\left(-0,3\right)^2}=\sqrt{\left(0,3\right)^2}=\left|0,3\right|=0,3\)do \(0,3>0\)
c, \(-\sqrt{\left(-1,3\right)^2}=-\sqrt{\left(1,3\right)^2}=-\left|1,3\right|=-1,3\)do \(1,3>0\)
d, \(-0,4\sqrt{\left(-0,4\right)^2}=-0,4\sqrt{\left(0,4\right)^2}=-0,4.\left|0,4\right|=-0,4.0,4=-0,14\)
do \(0,4>0\)
\(\sqrt{\left(0,1\right)^2}=\left|0,1\right|=0,1\)
\(\sqrt{\left(-0,3\right)^2}=\left|-0,3\right|=0,3\)
\(-\sqrt{\left(-1,3\right)^2}=-\left|-1,3\right|=-1,3\)
\(-0,4\sqrt{\left(-0,4\right)^2}=-0,4\cdot\left|-0,4\right|=-0,16\)
Chọn đáp án B