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6 tháng 7 2017

\(a.\left(\frac{6}{11}+\frac{5}{11}\right).\frac{3}{7}=1\cdot\frac{3}{7}=\frac{3}{7}b.\frac{3}{5}\cdot\frac{7}{9}+\frac{3}{5}\cdot\frac{2}{9}=\frac{3}{5}\cdot\left(\frac{7}{9}+\frac{2}{9}\right)=\frac{3}{5}\cdot1=\frac{3}{5}\)

5 tháng 7 2017

\(a,\left(\frac{6}{11}+\frac{5}{11}\right)\times\frac{3}{7}\)

Cách 1: \(\left(\frac{6}{11}+\frac{5}{11}\right)\times\frac{3}{7}=1\times\frac{3}{7}=\frac{10}{7}\)

Cách 2: \(\left(\frac{6}{11}+\frac{5}{11}\right)\times\frac{3}{7}=\frac{6}{11}\times\frac{3}{7}+\frac{5}{11}\times\frac{3}{7}=\frac{18}{77}+\frac{15}{77}=\frac{33}{77}=\frac{3}{7}\)

\(b,\frac{3}{5}\times\frac{7}{9}+\frac{3}{5}\times\frac{2}{9}\)

Cách 1: \(\frac{3}{5}\times\frac{7}{9}+\frac{3}{5}\times\frac{2}{9}=\frac{7}{15}+\frac{2}{15}=\frac{9}{15}\)

Cách 2: \(\frac{3}{5}\times\frac{7}{9}+\frac{3}{5}\times\frac{2}{9}=\frac{3}{5}\times1=\frac{3}{5}\)

P/s: Ý B có vấn đề thì phải

\(a,\left(\frac{6}{11}+\frac{5}{11}\right)x\frac{3}{7}\)

\(=\frac{11}{11}=1x\frac{3}{7}\)

\(=\frac{3}{7}\)

cach 2 

\(\frac{6}{11}x\frac{3}{7}+\frac{5}{11}x\frac{3}{7}\)

\(=\frac{18}{77}+\frac{15}{77}\)

\(=\frac{33}{77}=\frac{3}{7}\)

\(b,\frac{3}{5}x\frac{7}{9}+\frac{3}{5}x\frac{2}{9}\)

\(=\frac{21}{45}+\frac{6}{45}\)

\(=\frac{27}{45}=\frac{3}{5}\)

cách 2 : 

\(\frac{3}{5}x\left(\frac{7}{9}+\frac{2}{9}\right)\)

\(=\frac{3}{5}x\frac{9}{9}\)

\(=\frac{27}{45}=\frac{3}{5}\)

a) Ta có: \(\frac{3x-11}{11}-\frac{x}{3}=\frac{3x-5}{7}-\frac{5x-3}{9}\)

\(\Leftrightarrow\frac{63\left(3x-11\right)}{693}-\frac{231x}{693}-\frac{99\left(3x-5\right)}{693}+\frac{77\left(5x-3\right)}{693}=0\)

\(\Leftrightarrow189x-693-231x-297x+495+385x-231=0\)

\(\Leftrightarrow46x-429=0\)

\(\Leftrightarrow46x=429\)

hay \(x=\frac{429}{46}\)

Vậy: \(x=\frac{429}{46}\)

b) Ta có: \(\frac{9x-0,7}{4}-\frac{5x-1,5}{7}=\frac{7x-1,1}{6}-\frac{5\left(0,4-2x\right)}{5}\)

\(\Leftrightarrow\frac{9x-0,7}{4}-\frac{5x-1,5}{7}-\frac{7x-1,1}{6}+\frac{5\left(0,4-2x\right)}{5}=0\)

\(\Leftrightarrow105\left(9x-0,7\right)-60\left(5x-1,5\right)-70\left(7x-1,1\right)+420\left(0,4-2x\right)=0\)

\(\Leftrightarrow945x-\frac{147}{2}-300x+90-490x+77+168-840x=0\)

\(\Leftrightarrow-685x+261.5=0\)

\(\Leftrightarrow-685x=-261.5\)

hay \(x=\frac{523}{1370}\)

Vậy: \(x=\frac{523}{1370}\)

c) Ta có: \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x-1\right)}{7}-5\)

\(\Leftrightarrow\frac{14\left(5x-3\right)}{84}-\frac{21\left(7x-1\right)}{84}-\frac{24\left(2x-1\right)}{84}+\frac{420}{84}=0\)

\(\Leftrightarrow70x-42-147x+21-48x+24+420=0\)

\(\Leftrightarrow-125x+423=0\)

\(\Leftrightarrow-125x=-423\)

hay \(x=\frac{423}{125}\)

Vậy: \(x=\frac{423}{125}\)

d) Ta có: \(14\frac{1}{2}-\frac{2\left(x+3\right)}{5}=\frac{3x}{2}-\frac{2\left(x-7\right)}{3}\)

\(\Leftrightarrow\frac{435}{30}-\frac{12\left(x+3\right)}{30}-\frac{45x}{30}+\frac{20\left(x-7\right)}{30}=0\)

\(\Leftrightarrow435-12x-36-45x+20x-140=0\)

\(\Leftrightarrow-37x+259=0\)

\(\Leftrightarrow-37x=-259\)

hay \(x=7\)

Vậy: x=7

25 tháng 3 2017

a) = -3/7 . 5/11 + -3/7 . 6/11 + 9/7

   = -3/7. ( 5/11 + 6/11 ) + 9/7

  = -3/7. 1 + 9/7

  = -3/7 + 9/7

  = 6/7

b) = 4/13 + 9/13 + -11/5 + 6/5 - 3/4

    = 13/13 + -5/5 - 3/4

    = 1 + (-1) - 3/4

    = 0 - 3/4

    = -3/4

c) = -19/17. 4/7 + 19/17. -3/7 + 19/17

    = 19/17. -4/7 + 19/17. -3/7 + 19/17.1

    = 19/17.( -4/7 + -3/7 + 19/17

    = 19/17. -7/7 + 19/17

    = 19/17. (-1) + 19/17

    = -19/17 + 19/17

    = 0

tk mk nha,thanks

30 tháng 3 2015

a) 11 1/4- 2 5/7 - 5 1/4

11 1/4 - 5 1/4 - 2 5/7

( 11 -5) + ( 1/4 -1/4) -2 5/7

6 +0 -2 5/7

6- 2 5/7

5+1 - 2 5/7

( 5-2) + ( 1-5/7)

3+( 7/7 -5/7)

3 + 2/7

3 2/7

3+

a: \(=\dfrac{17}{7}+\dfrac{2}{9}-\dfrac{10}{7}-\dfrac{5}{3}\cdot9=1+\dfrac{2}{9}-15=-14+\dfrac{2}{9}=-\dfrac{126}{9}+\dfrac{2}{9}=-\dfrac{124}{9}\)

b: \(=\dfrac{-11}{23}\left(\dfrac{6}{7}+\dfrac{8}{7}\right)-\dfrac{1}{23}=\dfrac{-22}{23}-\dfrac{1}{23}=-1\)

c: \(=\left(\dfrac{377}{-231}-\dfrac{123}{89}+\dfrac{34}{791}\right)\cdot\dfrac{4-3-1}{24}=0\)

d: \(=\dfrac{12}{7}\left(19+\dfrac{5}{8}-15-\dfrac{1}{4}\right)=\dfrac{12}{7}\cdot\dfrac{35}{8}=\dfrac{15}{2}\)

\(\left(\frac{1}{4}-x\right)\left(x+\frac{2}{5}\right)=0\)

Ta xét 2 trường hợp 

\(\begin{cases}\frac{1}{4}-x=0\\x+\frac{2}{5}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=-\frac{2}{5}\end{cases}}\)

tớ mới làm bài 1 thôi bài 2 3 tớ ko có thời gian