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\(i,\sqrt{12,1.360}=\sqrt{12,1}.6\sqrt{10}=6.\sqrt{12,1.10}=6.\sqrt{121}=6.\sqrt{11^2}=6.11=66\)
\(k,\sqrt{0,4}.\sqrt{6,4}=\sqrt{0,4.6,4}=\sqrt{\dfrac{64}{25}}=\dfrac{\sqrt{8^2}}{\sqrt{5^2}}=\dfrac{8}{5}\)
\(l,-0,4.\sqrt{\left(-0,4\right)^2}=-0,4.0,4=-0,16\)
\(m,\sqrt{2^4.\left(-7\right)^2}=\sqrt{4^2}.\sqrt{\left(-7\right)^2}=4.7=28\)
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)
\(a,\sqrt{0,1^2}=0,1\)
\(b,\sqrt{\left(-0,4\right)^2}=|-0,4|=0,4\)
\(c,-\sqrt{\left(-1,7\right)^2}=-|-1,7|=-1,7\)
\(d,-0,5\sqrt{\left(-0,5\right)^4}=\frac{-1}{2}\sqrt{[\left(\frac{-1}{2}\right)^2]^2}=-\frac{1}{2}.\left(\frac{1}{2}\right)^2=\frac{-1}{2}.\frac{1}{4}=\frac{-1}{8}\)
\(e,\sqrt{\left(1-\sqrt{2}\right)^2}=|1-\sqrt{2}|=\sqrt{2}-1\)
\(g,\sqrt{\left(\sqrt{3}-1\right)^2}=|\sqrt{3}-1|=\sqrt{3}-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có : \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-3\sqrt{0,4}\right)\)
\(=\left(\sqrt{10}-\sqrt{2}\right)\left(\sqrt{2}-\dfrac{3\sqrt{10}}{5}\right)\)
\(=2\sqrt{5}-6-2+\dfrac{6\sqrt{5}}{5}=\dfrac{16}{5}\sqrt{5}-8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: \(=\sqrt{2}-\dfrac{2}{5}\sqrt{2}+2\sqrt{2}+2\sqrt{2}=\dfrac{23}{5}\sqrt{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(\sqrt{8}-3\sqrt{2}+10\right)\left(\sqrt{2}-3\sqrt{0.4}\right)=\sqrt{16}-\frac{12\sqrt{5}}{5}+\sqrt{20}-6\sqrt{10}-6+\frac{18\sqrt{5}}{5}\)
\(A=-2+\frac{16\sqrt{5}}{5}-6\sqrt{10}\)
b)\(B=\frac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}-\frac{\sqrt{5}-1}{2}=\frac{\sqrt{6+2\sqrt{5}}}{2}-\frac{\sqrt{5}-1}{2}=\frac{\sqrt{\left(\sqrt{5}+1\right)^2}}{2}-\frac{\sqrt{5}-1}{2}=\frac{\sqrt{5}+1}{2}-\frac{\sqrt{5}-1}{2}=1\)
Lời giải:
\(\sqrt[-0,4]{(-0,4)^2}=\sqrt[-0,4]{\frac{4}{25}}=(\frac{4}{25})^{\frac{1}{-0,4}}=(\frac{4}{25})^{\frac{-5}{2}}=\frac{1}{(\frac{4}{25})^{\frac{5}{2}}}=\frac{1}{\sqrt{(\frac{4}{25})^5}}=\frac{1}{(\frac{2}{5})^5}=\frac{3125}{32}\)
\(\sqrt[-0.4]{\left(-0.4\right)^2}=\sqrt[-0.4]{\dfrac{4}{25}}=\left(\dfrac{4}{25}\right)^{\dfrac{1}{-0.4}}=\dfrac{3125}{32}\)