\(=\left(\dfrac{1}{2}-\dfrac{6}{5}\right):\dfrac{21}{20}-\dfrac{25}{4}+\dfrac{1}{2}=\dfrac{-7}{10}\cdot\dfrac{20}{21}-\dfrac{23}{4}\)

\(=\dfrac{-2}{3}-\dfrac{23}{4}=\dfrac{-8-69}{12}=-\dfrac{77}{12}\)

\(=\left(\dfrac{1}{2}-\dfrac{6}{5}\right):\dfrac{21}{20}-\dfrac{25}{4}+\dfrac{1}{2}\)

\(=\dfrac{-7}{10}\cdot\dfrac{20}{21}-\dfrac{25}{4}+\dfrac{2}{4}\)

\(=\dfrac{-2}{3}-\dfrac{23}{4}=\dfrac{-8-69}{12}=\dfrac{-77}{12}\)

\(\left(\sqrt{\dfrac{1}{4}}-1,2\right):1\dfrac{1}{20}-\left(-\dfrac{5}{2}\right)^2+\left|1,25-\dfrac{3}{4}\right|\)

\(=-\dfrac{7}{10}:\dfrac{21}{20}-\dfrac{25}{4}+\left|\dfrac{1}{2}\right|\)

\(=-\dfrac{7}{10}.\dfrac{20}{21}-\dfrac{25}{4}+\dfrac{1}{2}\)

\(=-\dfrac{2}{3}-\dfrac{25}{4}+\dfrac{1}{2}\)

\(=-\dfrac{77}{12}\)

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

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

\(=10\)

a, \(\dfrac{5}{7}+\left(\dfrac{3}{5}+\dfrac{-5}{7}\right)\)
\(=\dfrac{5}{7}+\dfrac{-5}{7}+\dfrac{3}{5} =0+\dfrac{3}{5}=\dfrac{3}{5}\)

b, \(=\dfrac{-3}{4}-\dfrac{15}{14}:\dfrac{-5}{7}+\left(-1\right)^2=\dfrac{-3}{4}-\dfrac{-3}{2}+1=\dfrac{-3}{4}-\dfrac{-6}{4}+1=\dfrac{3}{4}+1=\dfrac{7}{4}\)

 c, \(\dfrac{-5}{9}+\left(\dfrac{-2}{3}\right)^2.\left(20\%-1.2\right)=\dfrac{-5}{9}+\dfrac{4}{9}x\left(\dfrac{1}{5}-\dfrac{6}{5}\right)=\dfrac{-5}{9}+\dfrac{4}{9}x\left(-1\right)=\dfrac{-5}{9}+\dfrac{-4}{9}=-1\)

Bài 1:

a) \(\dfrac{5}{7}+\left(\dfrac{3}{5}+\dfrac{-5}{7}\right)\)\(=\left(\dfrac{5}{7}+\dfrac{-5}{7}\right)+\dfrac{3}{5}\)\(=0+\dfrac{3}{5}=\dfrac{3}{5}\)

b) \(\dfrac{-3}{4}-\dfrac{15}{14}:\dfrac{-5}{7}+\left(-1\right)^2\)\(=\dfrac{-3}{4}-\dfrac{15}{14}:\dfrac{-5}{7}+1\)\(=\dfrac{-3}{4}-\dfrac{-3}{2}+1\)

                                            \(=\dfrac{3}{4}+1\)\(=\dfrac{7}{4}\)

`(-2)^3 . [-1] / 24 + ( 4 / 5 - 1,2 ) :  2/ 15`

`= -8 . [-1] / 24 + ( 4 / 5 - 6 / 5 ) . 15 / 2`

`= 1 / 3 + [-2] / 5 . 15 / 2`

`= 1 / 3 + (-3)`

`= [-8] / 3`

2: Thay \(x=\dfrac{1}{2}\) và y=2 vào M, ta được:

\(M=\dfrac{2\cdot\left(\dfrac{1}{2}\right)^2\cdot2-1.2\cdot\left(3\cdot\dfrac{1}{2}-2\cdot2\right)}{\dfrac{1}{2}\cdot2}\)

\(=4\cdot\dfrac{1}{4}-1.2\left(\dfrac{3}{2}-4\right)\)

\(=1-1.8+4.8\)

\(=4\)

1: Ta có: \(\left(-\dfrac{2}{3}x^3y^2\right)z\cdot5xy^2z^2\)

\(=\left(-\dfrac{2}{3}\cdot5\right)\cdot\left(x^3\cdot x\right)\cdot\left(y^2\cdot y^2\right)\cdot\left(z\cdot z^2\right)\)

\(=\dfrac{-10}{3}x^4y^4z^3\)

a) \(\left( { - 12,5} \right).1,2 =  - \left( {12,5.1,2} \right) =  - 15\)

b) \(\left( { - 12,5} \right).\left( { - 1,2} \right) = 12,5.1,2 = 15\)

Lời giải:
\(-\sin 2n\leq 1< 1,2^n\) khi $n\to +\infty$ nên $\(|\frac{-\sin 2n}{1,2^n}|< 1\Rightarrow \lim\frac{-\sin 2n}{1,2^n}=0\)

g: \(=\left(-\sqrt{5}-2\right)\left(\sqrt{5}-2\right)\)

=-(căn 5+2)(căn 5-2)

=-(5-4)=-1

h: \(=\left(\dfrac{4}{3}\sqrt{3}+\sqrt{2}+\dfrac{\sqrt{30}}{3}\right)\left(\dfrac{\sqrt{30}}{5}+\sqrt{2}-\dfrac{4}{5}\sqrt{5}\right)\)

=4/5*căn 10+4/3*căn 6-16/15*căn 15+2/5*căn 15+2-4/5*căn 10+30/15+2/3*căn 15-4/3*căn 6

=4