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8 tháng 1 2020

1.

\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)

\(MC:12\)

Quy đồng :

\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)

\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)

\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)

\(\Leftrightarrow6x+9-3x=-4-9+16\)

\(\Leftrightarrow-7x=3\)

\(\Leftrightarrow x=\frac{-3}{7}\)

2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)

\(MC:20\)

Quy đồng :

\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)

\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)

\(\Leftrightarrow30x+15-20=15x-2\)

\(\Leftrightarrow15x=3\)

\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)

22 tháng 8 2019

1,\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\left(7-\frac{1}{6}\right)+\frac{1}{3}\)

  \(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{3}{7}.\frac{41}{6}+\frac{1}{3}\)

\(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{41}{14}+\frac{1}{3}\)

 \(\frac{2}{9}.\left(x-\frac{9}{4}\right)+\frac{1}{2}=\frac{137}{42}\)

\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{137}{42}-\frac{1}{2}\)

\(\frac{2}{9}.\left(x-\frac{9}{4}\right)=\frac{58}{21}\)

 \(\left(x-\frac{9}{4}\right)=\frac{5}{2}:\frac{2}{9}\)

\(\left(x-\frac{9}{4}\right)=\frac{45}{4}\)

\(x=\frac{45}{4}+\frac{9}{4}\)

\(x=\frac{27}{2}\)

22 tháng 8 2019

Bước cưối 58/21 minh man viết nhầm nên sai 

11 tháng 3 2020

1) \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)

<=> \(\frac{21x}{24}-\frac{100\left(x-9\right)}{24}=\frac{80x+6}{24}\)

<=> 21x - 100x + 900 = 80x + 6

<=> -79x - 80x = 6 - 900

<=> -159x = -894

<=> x = 258/53

Vậy S = {258/53}

2) \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x+1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)

<=> \(\frac{3\left(4x^2+4x+1\right)}{15}-\frac{5\left(x^2+2x+1\right)}{15}=\frac{7x^2-14x-5}{15}\)

<=> 12x2 + 12x + 3 - 5x2 - 10x - 5 = 7x2 - 14x - 5

<=> 7x2 + 2x - 7x2 + 14x = -5 + 2

<=> 16x = 3

<=> x = 3/16

Vậy S  = {3/16}

11 tháng 3 2020

3) 4(3x - 2) - 3(x - 4) = 7x+  10

<=> 12x - 8 - 3x + 12 = 7x + 10

<=> 9x - 7x = 10 - 4

<=> 2x = 6

<=> x = 3

Vậy S = {3}

4) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)

<=> \(\frac{x^2+14x+40}{12}+\frac{3\left(x^2+2x-8\right)}{12}=\frac{4\left(x^2+8x-20\right)}{12}\)

<=> x2 + 14x + 40 + 3x2 + 6x - 24 = 4x2 + 32x - 80

<=> 4x2 + 20x - 4x2 - 32x = -80 - 16

<=> -12x = -96

<=> x = 8

Vậy S = {8}

khỏi chép lại đề ha

  • 2 - 4x - 5x + \(\frac{3}{2}\)= \(\frac{7}{4}\)

          \(\frac{7}{2}\)- 9x = \(\frac{7}{4}\)

           -9x = \(\frac{7}{2}-\frac{7}{4}\)

           -9x = \(\frac{7}{4}\)

            x = \(\frac{7}{4}:\left(-9\right)\)

            x = \(\frac{-7}{36}\)

  • 3 - 2x - \(\frac{1}{3}=7x-\frac{1}{4}\)

          -2x - 7x = \(\frac{-1}{4}-3+\frac{1}{3}\)

         -9x = \(\frac{-35}{12}\)

          x = \(\frac{-35}{12}:\left(-9\right)\)

          x = \(\frac{35}{108}\)

  • \(\frac{-15}{2}\)+ \(\frac{1}{4}\)+ 4x -2 = 1

            4x = 1 + \(\frac{15}{2}-\frac{1}{4}+2\)

            4x = \(\frac{41}{4}\)

            x = \(\frac{41}{4}:4\)

            x = \(\frac{41}{16}\)

b) Ta có: \(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\)

\(\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)-\left(x+1\right)^3=0\)

\(x^3-6x^2+12x-8+9x^2-1-\left(x^3+3x^2+3x+1\right)=0\)

\(x^3+3x^2+12x-9-x^3-3x^2-3x-1=0\)

\(9x-10=0\)

hay 9x=10

\(x=\frac{10}{9}\)

Vậy: \(x=\frac{10}{9}\)

c) \(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{5}\)

\(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{5}=0\)

\(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}-\frac{3\left(x+7\right)}{15}=0\)

\(3\left(2x-1\right)-5\left(x-2\right)-3\left(x+7\right)=0\)

\(6x-3-5x+10-3x-21=0\)

\(-2x-14=0\)

\(-2x=14\)

hay x=-7

Vậy: x=-7

d) \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}=\frac{13x+4}{21}\)

\(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)

\(\frac{6\left(x-3\right)}{21}+\frac{7\left(x-5\right)}{21}-\frac{13x+4}{21}=0\)

\(6x-18+7x-35-13x-4=0\)

\(-21\ne0\)

Vậy: x∈∅

e) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)

\(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}-\frac{\left(x+10\right)\left(x-2\right)}{3}=0\)

\(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{3\left(x+4\right)\left(2-x\right)}{12}-\frac{4\left(x+10\right)\left(x-2\right)}{12}=0\)

\(x^2+14x+40-\left(3x+12\right)\left(2-x\right)-\left(4x+40\right)\left(x-2\right)=0\)

\(x^2+14x+40-\left(24-6x-3x^2\right)-\left(4x^2+32x-80\right)=0\)

\(x^2+14x+40-24+6x+3x^2-4x^2-32x+80=0\)

\(-12x+96=0\)

\(-12x=-96\)

hay x=8

Vậy: x=8

24 tháng 7 2018

Giải:

a) \(\frac{1}{5}-\frac{2}{3}+2x=\frac{1}{2}\)

\(\Leftrightarrow2x=\frac{1}{2}-\left(\frac{1}{5}-\frac{2}{3}\right)\)

\(\Leftrightarrow2x=\frac{1}{2}-\frac{-7}{15}\)

\(\Leftrightarrow2x=\frac{11}{15}\)

\(\Leftrightarrow x=\frac{11}{15}:2\)

\(\Leftrightarrow x=\frac{11}{30}\)

b) \(4\left(\frac{1}{3}-3\right)+\frac{1}{2}=\frac{5}{6}+x\)

\(\Leftrightarrow\frac{-61}{6}=\frac{5}{6}+x\)

\(\Leftrightarrow x=\frac{-61}{6}-\frac{5}{6}\)

\(\Leftrightarrow x=\frac{-66}{6}=-11\)

6 tháng 8 2015

Bạn xem ở đây nè: http://olm.vn/hoi-dap/question/158643.html

31 tháng 1 2016

2(\(\frac{3}{4}\)-5x)=\(\frac{4}{5}\)-3x

=> \(\frac{6}{4}-10x=\frac{4}{5}-3x\)

=>\(-10x+3x=\frac{4}{5}-\frac{6}{4}\)

=> \(x=\frac{1}{10}\)

2 .

\(\frac{3}{2}-4\left(\frac{1}{4}-x\right)=\frac{2}{3}-7x\)

=>\(\frac{3}{2}-1+4x=\frac{2}{3}-7x\)

=>\(11x=\frac{1}{6}\)

=>x=\(\frac{1}{66}\)

3.

\(3\left(\frac{1}{2}-x\right)+\frac{1}{3}=\frac{7}{6}-x\)

=>\(\frac{3}{2}-3x+\frac{1}{3}=\frac{7}{6}-x\)

=>\(-2x=\frac{-2}{3}\)

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

4. câu 4 ko hiểu bạn ơi

 

11 tháng 7 2016

a)  \(\Leftrightarrow\frac{x+7}{2003}+1+\frac{x+4}{2006}+1-\frac{x-1}{2011}-1-\frac{x-5}{2015}-1=0\)

     \(\Leftrightarrow\frac{x+2010}{2003}+\frac{x+2010}{2006}-\frac{x+2010}{2011}-\frac{x+2010}{2015}=0\)

     \(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2003}+\frac{1}{2006}-\frac{1}{2011}-\frac{1}{2015}\right)=0\)

     \(\Leftrightarrow x+2010=0\) ( vì 1/2003  +  1/2006  --  1/2011  -- 1/2015   \(\ne\)0)

    \(\Leftrightarrow x=-2010\)

câu b làm tương tự (có gì không hiểu hỏi mk nha) >v<