a)\(25\frac{3}{5}:\left(\frac{-2}{3}\right)-15\frac{3}{5}:\left(\frac{-2}{3}\right)\)

\(=\left(25\frac{3}{5}-15\frac{3}{5}\right):\left(-\frac{2}{3}\right)\)

\(=10:\left(\frac{-2}{3}\right)\)

\(=-15\)

b)\(9.\left(\frac{-2}{3}\right)^3+\frac{1}{2}:5\)

\(=9.\frac{-8}{27}+\frac{1}{10}\)

\(=\frac{-8}{3}+\frac{1}{10}\)

\(=\frac{-77}{30}\)

c)\(\left[10\left(\frac{-1}{5}\right)^2+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)

\(=\frac{2}{5}:\left(\frac{-6}{5}\right)\)

\(=\frac{-1}{3}\)

\(a.25\frac{3}{5}:\left(-\frac{2}{3}\right)-15\frac{3}{5}:\left(-\frac{2}{3}\right)\)

\(=\frac{128}{5}:\left(-\frac{2}{3}\right)-\frac{75}{5}:\left(-\frac{2}{3}\right)\)

\(=\left(-\frac{192}{5}\right)-\left(-\frac{117}{5}\right)\)

\(=\frac{\left(-192\right)-\left(-117\right)}{5}\)

\(=-15\)

\(b.9.\left(-\frac{2}{3}\right)^3+\frac{1}{2}:5\)

\(=9.\left(-\frac{8}{27}\right)+\frac{1}{2}:5\)

\(=-\frac{8}{3}+\frac{1}{10}\)

\(=-\frac{77}{30}\)

\(c.\left[10\left(\frac{-1}{5}\right)^2+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)

\(=\left[10\left(\frac{-1}{25}\right)+5\left(\frac{-1}{5}\right)+1\right]:\left(\frac{-1}{5}-1\right)\)

\(=\left[\frac{-2}{5}+\left(-1\right)+1\right]:\left(-\frac{6}{5}\right)\)

\(=\left(-\frac{2}{5}\right):\left(-\frac{6}{5}\right)\)

\(=\frac{1}{3}\)

bai2:

a.x=3/5 hoacx=3/5

Bài 2 

a. \(-1\frac{2}{3}-|2x-1|:\frac{3}{5}=-2\)

\(|2x-1|:\frac{3}{5}=\frac{5}{3}-2\)

\(|2x-1|:\frac{3}{5}=-\frac{1}{3}\)

\(|2x-1|=-\frac{1}{5}\)

Vì giá trị tuyệt đối luôn \(\ge0\)với mọi x

mà \(-\frac{1}{5}< 0\)

=> \(x\in\varnothing\)

bài 1 nữa bạn

a) A = \(\frac{15}{7}:\left(\frac{1}{15}-\frac{7}{5}\right)-\frac{15}{7}:\left(\frac{17}{15}+\frac{11}{5}\right)=\frac{15}{7}:\frac{-20}{15}-\frac{15}{7}:\frac{50}{15}\)

A = \(\frac{15}{7}.\frac{15}{-20}-\frac{15}{7}.\frac{15}{50}=\frac{15}{7}.\left(\frac{-15}{20}-\frac{15}{50}\right)=\frac{15}{7}.\frac{-105}{100}=-\frac{9}{4}\)

b) B = \(\frac{1}{\left(-\frac{2}{3}\right)^4}.\left(-4\right)^2-1^{2016}-10\frac{1}{3}=\frac{1}{\frac{16}{81}}.16-1-10\frac{1}{3}=\frac{81}{16}.16-1-10\frac{1}{3}\)

B = \(81-1-10-\frac{1}{3}=70-\frac{1}{3}=\frac{209}{3}\)

b)\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{100}-1\right)=\frac{-1}{2}.\frac{-2}{3}.\frac{-3}{4}....\frac{-99}{100}=\frac{-1.\left(-2\right).\left(-3\right)...\left(-99\right)}{2.3.4...100}=-\frac{1}{100}\)

a, Tự chép đề bài ((:

\(=\frac{1}{9}\cdot1+\left(-\frac{1}{243}\right)\cdot\frac{9}{2}\)

\(=\frac{1}{9}-\frac{1}{54}\)

\(=\frac{5}{54}\)

b, 1. \(\left(\frac{2^2\cdot2^3}{4^2\cdot16}\right)^{15}\)

\(=\left(\frac{2^5}{2^4\cdot2^4}\right)^5=\left(\frac{2^5}{2^8}\right)^5=\left(\frac{1}{2^3}\right)^5=\left(\frac{1}{8}\right)^5=\frac{1}{8^5}\)(Để vậy đi :v)

     2. \(\left(\frac{2^6}{16^2}\right)^{10}\)

\(=\left(\frac{2^6}{2^8}\right)^{10}=\left(\frac{1}{2^2}\right)^{10}=\frac{1}{2^{20}}\)

c, \(\frac{2^{15}\cdot9^4}{6^6\cdot8^3}\)

\(=\frac{2^{15}\cdot\left(3^2\right)^4}{\left(2\cdot3\right)^6\cdot\left(2^3\right)^3}=\frac{2^{15}\cdot3^8}{2^6\cdot3^6\cdot2^9}=\frac{2^{15}\cdot3^8}{2^{15}\cdot3^6}=\frac{3^2}{1}=3^2=9\)