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12 tháng 10 2019

\(-1\frac{1}{3}:\left(\frac{1}{3}-\frac{1}{2}\right)^2-\left(-0,25\right)^2:\left(\frac{1}{4}-\frac{1}{2}\right)^3\)

\(=\frac{-4}{3}:\left(\frac{2}{6}-\frac{3}{6}\right)^2-\frac{1}{16}:\left(\frac{1}{4}-\frac{2}{4}\right)^2\)

\(=\frac{-4}{3}:\left(-\frac{1}{6}\right)^2-\frac{1}{16}:\left(-\frac{1}{4}\right)^2\)

\(=\frac{-4}{3}:\frac{1}{36}-\frac{1}{16}:\frac{1}{16}\)

\(=\frac{-4}{3}.36-1\)

\(=-48-1=-49\)

Chúc em học tốt nhé!

14 tháng 2 2020

Có:

\(A=2010\cdot\left(\frac{\frac{1}{6}+0.25-\frac{1}{8}}{1+1\frac{1}{2}-\frac{3}{4}}+\frac{0.4-\frac{2}{9}+\frac{2}{11}}{3-\frac{15}{9}+1\frac{4}{11}}\right)\)

\(=2010\cdot\left(\frac{\frac{1}{6}+\frac{1}{4}-\frac{1}{8}}{\frac{3}{3}+\frac{3}{2}-\frac{3}{4}}+\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{15}{5}-\frac{15}{9}+\frac{15}{11}}\right)\)

\(=2010\cdot\left[\frac{\frac{1}{2}\cdot\left(\frac{1}{3}+\frac{1}{2}-\frac{1}{4}\right)}{3\cdot\left(\frac{1}{3}+\frac{1}{2}-\frac{1}{4}\right)}+\frac{2\cdot\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{15\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}\right]\)

\(=2010\cdot\left(\frac{\frac{1}{2}}{3}+\frac{2}{15}\right)\)

\(=2010\cdot\left(\frac{\frac{5}{2}}{15}+\frac{2}{15}\right)\\ =2010\cdot\left(\frac{\frac{5}{2}+2}{15}\right)\)

\(=2010\cdot\frac{\frac{9}{2}}{15}\\ =\frac{2010\cdot\frac{9}{2}}{15}\\ =\frac{1005\cdot9}{15}\\ =201\cdot3\\ =603\)

6 tháng 7 2016

với phân số thứ nhất bạn rút 2 ở mẫu ra . còn phân số thứ 2 bạn đổi ra ạ

5 tháng 5 2019

\(\frac{1}{4}.\frac{2}{3}-\frac{3}{2}.\frac{1}{6}+\frac{1}{12}\)

\(=\frac{1}{6}+\frac{-1}{4}+\frac{1}{12}\)

\(=\frac{2}{12}+\frac{-3}{12}+\frac{1}{12}=\frac{0}{12}=0\)

~ Hok tốt ~

A, \(=\frac{1}{6}-\frac{5}{3}+\frac{1}{12}\)

\(=\frac{2}{12}-\frac{20}{12}+\frac{1}{12}\)

\(=\frac{2-20+1}{12}=-\frac{17}{12}\)

Hok tốt

Trả lời

\(A=\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{2.\left(\frac{1}{6}-\frac{1}{8}-\frac{1}{10}\right)}{\frac{7}{6}-\frac{7}{8}-\frac{7}{10}}\right):\left(1^2+2^2+...+2015^2\right).\)

\(A=\left(\frac{2}{7}-\frac{2}{7}\right):\left(1^2+2^2+3^2+...+2015^2\right)\)

\(A=0:\left(1^2+2^2+3^2+.....+2015^2\right)\)

A=0

Study well 

\(A=...\)

\(=\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\right):\left(1^2+2^2+...+2015^2\right)\)

\(=\left[\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\right]:\left(1^2+2^2+...+2015^2\right)\)

\(=\left(\frac{2}{7}-\frac{1}{\frac{7}{2}}\right):\left(1^2+2^2+...+2015^2\right)\)

\(=\left(\frac{2}{7}-\frac{2}{7}\right):\left(1^2+2^2+...+2015^2\right)\)

\(=0:\left(1^2+2^2+...+2015^2\right)=0\)

5 tháng 8 2017

\(\left(\frac{1}{9}\right)^{2015}.9^{2015}-96^2:24^2=1^{2015}-4^2=1-16=-15\)

\(16\frac{2}{7}:\left(\frac{-3}{5}\right)-28\frac{2}{7}:\left(\frac{-3}{5}\right)=\left(16\frac{2}{7}-28\frac{2}{7}\right):\left(\frac{-3}{5}\right)=-12.\frac{-5}{3}=20\)

\(\left(-2\right)^3.\left(\frac{3}{4}-0,25\right):\left(2\frac{1}{4}-1\frac{1}{6}\right)=-8.\frac{1}{2}:\frac{13}{12}=-8.\frac{1}{2}.\frac{12}{13}=\frac{-48}{13}\)

18 tháng 2 2021

b) Ta có: \(3+\left(x-5\right)=2\left(3x-2\right)\)

\(\Leftrightarrow3+x-5=6x-4\)

\(\Leftrightarrow x-2-6x+4=0\)

\(\Leftrightarrow-5x+2=0\)

\(\Leftrightarrow-5x=-2\)

\(\Leftrightarrow x=\dfrac{2}{5}\)

Vậy: \(S=\left\{\dfrac{2}{5}\right\}\)

c) Ta có: \(2\left(x-0.5\right)+3=0.25\left(4x-1\right)\)

\(\Leftrightarrow2x-1+3=x-\dfrac{1}{4}\)

\(\Leftrightarrow2x+2-x+\dfrac{1}{4}=0\)

\(\Leftrightarrow x+\dfrac{9}{4}=0\)

\(\Leftrightarrow x=-\dfrac{9}{4}\)

Vậy: \(S=\left\{-\dfrac{9}{4}\right\}\)

d) Ta có: \(2\left(x-\dfrac{1}{4}\right)-4=-6\left(-\dfrac{1}{3}x+0.5\right)+2\)

\(\Leftrightarrow2x-\dfrac{1}{2}-4=2x-3+2\)

\(\Leftrightarrow2x-\dfrac{9}{2}=2x-1\)

\(\Leftrightarrow2x-2x=-1+\dfrac{9}{2}\)

\(\Leftrightarrow0x=\dfrac{7}{2}\)(vô lý)

Vậy: \(S=\varnothing\)