1).15,3-21,5-3.1,5 = -10.7

2).2(4  ^2-2.4,1)+1,25:5 = 0.300370048 ( cái nài mik bấm máy ra vậy nếu sai thì bảo mik nha )

a) \(\left(\dfrac{3}{4}\right)^{-2}\cdot3^2\cdot12^0=16\)

b) \(\left(\dfrac{1}{12}\right)^{-1}\cdot\left(\dfrac{2}{3}\right)^{-2}=27\)

c) \(\left(2^{-2}\cdot5^2\right)^{-2}:\left(5\cdot5^{-5}\right)=16\)

a)

$16^{\alpha }+16^{-\alpha } = (4^2)^{\alpha }+(4^2)^{-\alpha } = 4^{2\alpha }+4^{-2\alpha }$

$4^{2\alpha }+4^{-2\alpha } = 4^{2\log_4{\frac{1}{5}}}+4^{-2\log_4{\frac{1}{5}}} = \left(\frac{1}{5}\right)^2+\left(\frac{1}{5}\right)^{-2} = \frac{1}{25}+25 = \frac{26}{25}$

b)

$\left(2^{\alpha }+2^{-\alpha }\right)^2 = \left(\sqrt{4}\right)^{\alpha }+\left(\sqrt{4}\right)^{-\alpha } = 4^{\frac{\alpha}{2}}+4^{-\frac{\alpha}{2}}$

$4^{\frac{\alpha}{2}}+4^{-\frac{\alpha}{2}} = 4^{\frac{\log_4{\frac{1}{5}}}{2}}+4^{-\frac{\log_4{\frac{1}{5}}}{2}} = \left(\frac{1}{5}\right)^{\frac{1}{2}}+\left(\frac{1}{5}\right)^{-\frac{1}{2}} = \sqrt{\frac{1}{5}}+\frac{1}{\sqrt{5}} = \frac{2}{\sqrt{5}}$

a) \(\left(-5\right)^{-1}=-\dfrac{1}{5}\)

b) \(2^0\cdot\left(\dfrac{1}{2}\right)^{-5}=1\cdot32=32\)

c) \(6^{-2}\cdot\left(\dfrac{1}{3}\right)^{-3}:2^{-2}\)

\(=\dfrac{1}{36}\cdot27:\dfrac{1}{4}\)

\(=\dfrac{27\cdot4}{36}=3\)

\(b.\)

\(=\sqrt{\left(3a\right)^2\cdot\left(b-2\right)^2}\)

\(=\left|3a\right|\cdot\left|b-2\right|\)

Với : \(a=2,b=-\sqrt{3}\)

\(2\cdot3\cdot\left(-\sqrt{3}-2\right)=6\cdot\left(-\sqrt{3}-2\right)\)

\(a.\)

\(=\sqrt{4\cdot\left(3x+1\right)^2}=2\cdot\left|3x+1\right|\)

Với : \(x=-\sqrt{2}\)

\(2\cdot\left|3\cdot-\sqrt{2}+1\right|=2\cdot\left|1-\sqrt{6}\right|\)

a) Ta có:\(\sqrt{4\left(9x^2+6x+1\right)^2}\)

\(=2\left(3x+1\right)^2\)

\(=2\cdot\left(-3\cdot\sqrt{2}+1\right)^2\)

\(=2\left(19-6\sqrt{2}\right)\)

\(=38-12\sqrt{2}\)

b) Ta có: \(\sqrt{9a^2\left(b^2-4b+4\right)}\)

\(=3\left|a\right|\left|b-2\right|\)

\(=3\cdot\left|2\right|\cdot\left|-\sqrt{3}-2\right|\)

\(=6\left(2+\sqrt{3}\right)=12+6\sqrt{3}\)

a)

 \(\begin{array}{l}\left( {0,25 - \frac{5}{6}} \right).1,6 + \frac{{ - 1}}{3}\\ =(\frac{25}{100}-\frac{5}{6}).\frac{16}{10}+\frac{-1}{3}\\= \left( {\frac{1}{4} - \frac{5}{6}} \right).\frac{8}{5} + \frac{{ - 1}}{3}\\ = \left( {\frac{6}{{24}} - \frac{{20}}{{24}}} \right).\frac{8}{5} + \frac{{ - 1}}{3}\\ = \frac{{ - 14}}{{24}}.\frac{8}{5} + \frac{{ - 1}}{3}\\ = \frac{{ - 14}}{{15}} + \frac{{ - 1}}{3}\\ = \frac{{ - 14}}{{15}} + \frac{{ - 5}}{{15}}\\ = \frac{{ - 19}}{{15}}\end{array}\)

b)

\(\begin{array}{l}3 - 2.\left[ {0,5 + \left( {0,25 - \frac{1}{6}} \right)} \right]\\ = 3 - 2.\left[ {\frac{1}{2} + \left( {\frac{1}{4} - \frac{1}{6}} \right)} \right]\\ = 3 - 2.\left( {\frac{1}{2} + \frac{1}{{12}}} \right)\\ =3-2.(\frac{6}{12}+\frac{1}{12})\\= 3 - 2.\frac{7}{{12}}\\ = 3 - \frac{7}{6}\\=\frac{18}{6}-\frac{7}{6}\\ = \frac{{11}}{6}\end{array}\)

a) \(\sqrt[4]{\dfrac{1}{16}}=\dfrac{1}{2}\)

b) \(\left(\sqrt[6]{8}\right)^2=\sqrt[\dfrac{6}{2}]{8}=\sqrt[3]{8}=2\)

c) \(\sqrt[4]{3}\cdot\sqrt[4]{27}=\sqrt[4]{3\cdot27}=\sqrt[4]{81}=3\)

Bài 1: 

a) Ta có: \(A=-1.7\cdot2.3+1.7\cdot\left(-3.7\right)-1.7\cdot3-0.17:0.1\)

\(=1.7\cdot\left(-2.3\right)+1.7\cdot\left(-3.7\right)+1.7\cdot\left(-3\right)+1.7\cdot\left(-1\right)\)

\(=1.7\cdot\left(-2.3-3.7-3-1\right)\)

\(=-10\cdot1.7=-17\)

b) Ta có: \(B=2\dfrac{3}{4}\cdot\left(-0.4\right)-1\dfrac{2}{3}\cdot2.75+\left(-1.2\right):\dfrac{4}{11}\)

\(=\dfrac{11}{4}\cdot\left(-0.4\right)-\dfrac{5}{3}\cdot\dfrac{11}{4}+\left(-1.2\right)\cdot\dfrac{11}{4}\)

\(=\dfrac{11}{4}\left(-0.4-\dfrac{5}{3}-1.2\right)\)

\(=-\dfrac{539}{60}\)

c) Ta có: \(C=\dfrac{\left(2^3\cdot5\cdot7\right)\cdot\left(5^2\cdot7^3\right)}{\left(2\cdot5\cdot7^2\right)^2}\)

\(=\dfrac{2^3\cdot5^3\cdot7^4}{2^2\cdot5^2\cdot7^4}\)

\(=10\)