2:

a: x=2,4-0,4=2

b: =>2x=-1,5+0,8=-0,7

=>x=-0,35

c: =>x-16=-15

=>x=1

`#040911`

`a)`

`3 1/3 x + 16 3/4 = -13,25`

`=> 3 1/3 x = -13,25 - 16 3/4`

`=> 3 1/3 x = -30`

`=> x = -30 \div 3 1/3`

`=> x =-9`

Vậy, `x = -9`

`b)`

`3 2/7*x - 1/8 = 2 3/4`

`=> 3 2/7x = 2 3/4 + 1/8`

`=> 3 2/7x = 23/8`

`=> x = 23/8 \div 3 2/7`

`=> x = 7/8`

Vậy, `x = 7/8`

`c)`

`x \div 4 1/3 = -2,5`

`=> x = -2,5 * 4 1/3`

`=> x = -65/6`

Vậy, `x = -65/6`

`d)`

`( (3x)/7 + 1) \div (-4) = (-1)/28`

`=> (3x)/7 +1 = (-1)/28 * (-4)`

`=> (3x)/7 + 1 = 1/7`

`=> (3x)/7 = 1/7 - 1`

`=> (3x)/7 = -6/7`

`=> 3x = -6`

`=> x = -6 \div 3`

`=> x = -2`

Vậy, `x = -2.`

a

=>10/3 . x + 16 + 3/4 = -13,25

=>10/3 x + 3/4 = -29,25

=>10/3 x = -30

=>x=-30 : 10/3

=>x=-30 . 3/10

=>x=-9

b.

=>23/7 x - 1/8 = = 11/4

=>23/7 x = 11/4 + 1/8

=>23/7 x= 22/8 + 1/8

=>23/7 x= 23/8

=>x=23/8 : 23/7

=>x=23/8 . 7/23

=>x=7/8

c.

=>x : 13/3 =-5/2

=>x=-5/2 . 13/3

=>x=-65/6

d.

=>3x/7 +1 = (-1/28) . (-4)

=>3x/7 + 1 = 1/7

=>3x/7 = -6/7

=>3x=-6

=>x=-2

a) \(3\dfrac{1}{3}x+16\dfrac{3}{4}=-13,25\)

\(\Rightarrow\dfrac{10}{3}x+\dfrac{67}{4}=-\dfrac{53}{4}\)

\(\Rightarrow\dfrac{10}{3}x=-30\)

\(\Rightarrow x=-30:\dfrac{10}{3}\)

\(\Rightarrow x=-9\)

b) \(3\dfrac{2}{7}x-\dfrac{1}{8}=2\dfrac{3}{4}\)

\(\Rightarrow\dfrac{23}{7}x-\dfrac{1}{8}=\dfrac{11}{4}\)

\(\Rightarrow\dfrac{23}{7}x=\dfrac{11}{4}+\dfrac{1}{8}\)

\(\Rightarrow\dfrac{23}{7}x=\dfrac{23}{8}\)

\(\Rightarrow x=\dfrac{23}{8}:\dfrac{23}{7}\)

\(\Rightarrow x=\dfrac{7}{8}\)

c) \(x:4\dfrac{1}{3}=-2,5\)

\(\Rightarrow x:\dfrac{13}{3}=-\dfrac{5}{2}\)

\(\Rightarrow x=-\dfrac{5}{2}\cdot\dfrac{13}{3}\)

\(\Rightarrow x=-\dfrac{65}{6}\)

d) \(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{28}\)

\(\Rightarrow\dfrac{3x}{7}+1=\dfrac{-1}{28}\cdot-4\)

\(\Rightarrow\dfrac{3x}{7}+1=\dfrac{1}{7}\)

\(\Rightarrow\dfrac{3x}{7}=-\dfrac{6}{7}\)

\(\Rightarrow x=-\dfrac{6}{7}:\dfrac{3}{7}\)

\(\Rightarrow x=-2\)

giúp mình giải bài toán trên với. Mình cảm ơn rất nhiều

a: =>1/2x-3/4x=-5/6+7/3

=>-1/4x=14/6-5/6=3/2

=>x=-3/2*4=-6

b: =>4/5x-3/2x=1/2+6/5

=>-7/10x=17/10

=>x=-17/7

c: =>6/5x+6/20=6/5-1/3x

=>6/5x+1/3x=6/5-3/10=12/10-3/10=9/10

=>x=27/46

d: =>6x+3/2+4/5=1/2-2x

=>8x=1/2-3/2-4/5=-1-4/5=-9/5

=>x=-9/40

Giải:

a)-3/10-(-1/5)+x)=-3/2

             -1/5+x   =-3/10-(-3/2)

              -1/5+x   =6/5

                       x   =6/5-(-1/5)

                       x   =7/5

b)-(-x+3/4)-12/8.(-32/15)=-(-1/2)

           x-3/4+16/5          =1/2

           x-3/4                =1/2-16/5

           x                           =-27/10

           x                           =-27/10+3/4

           x                           =-39/20

c)x-3/x+5=4/3

=>(x-3).3=4.(x+5)

     3x-9   =4x+20

     3x-4x =20+9

     -1x     =29

        x     =-29

Câu b cậu nên tính lại cho kĩ nhé, ấn máy tính dễ nhầm lắm đấy!

Mk phải ấn: -(39/20+3/4)-12/8.-32/15=1/2

Vì x là số âm mà đằng trước x là dấu ''-'' nên -(-39/20)=39/20 ; -(-1/2)=1/2

Chúc bạn học tốt!

a)

\(\left|x-2\right|-\dfrac{3}{5}=\dfrac{1}{2}\\ \left|x-2\right|=\dfrac{1}{2}+\dfrac{3}{5}\\ \left|x-2\right|=\dfrac{11}{10}\\ =>\left[{}\begin{matrix}x-2=\dfrac{11}{10}\\x-2=-\dfrac{11}{10}\end{matrix}\right.\left[{}\begin{matrix}x=\dfrac{31}{10}\\x=\dfrac{9}{10}\end{matrix}\right.\)

b)

\(\left(x-\dfrac{7}{3}\right):\dfrac{-1}{3}=0,4\\ x-\dfrac{7}{3}=0,4\cdot\dfrac{-1}{3}\\ x-\dfrac{7}{3}=-\dfrac{2}{15}\\ x=-\dfrac{2}{15}+\dfrac{7}{3}\\ x=\dfrac{11}{5}\)

c)

\(\left|x-3\right|=5\\ =>\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\left[{}\begin{matrix}x=5+3\\x=-5+3\end{matrix}\right.\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)

d)

\(\left(2x+3\right)^2=25\\ =>\left[{}\begin{matrix}2x+3=5\\2x+3=-5\end{matrix}\right.\left[{}\begin{matrix}2x=2\\2x=-8\end{matrix}\right.\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)

e)

\(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}\)

\(\dfrac{1}{4}:x=-\dfrac{7}{20}\)

\(x=\dfrac{1}{4}:\dfrac{-7}{20}\\ x=-\dfrac{5}{7}\)

f)

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{1}{27}\\ =>x-\dfrac{1}{2}=\dfrac{1}{3}\\ x=\dfrac{1}{3}+\dfrac{1}{2}\\ x=\dfrac{5}{6}\)

\(a,-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+1}{6}=\dfrac{8}{3}\)

\(\Rightarrow-\dfrac{6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{8}{3}\)

\(\Rightarrow\dfrac{-6x+8x+3x+3+4x+2}{12}=\dfrac{8}{3}\)

\(\Rightarrow\dfrac{9x+5}{12}=\dfrac{8}{3}\)

\(\Rightarrow27x+15=96\)

\(\Rightarrow27x=81\)

\(\Rightarrow x=3\left(tm\right)\)

\(b,\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3+5-2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow2x+1=13\)

\(\Rightarrow2x=12\)

\(\Rightarrow x=6\left(tm\right)\)

#Toru

a) \(-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+2}{6}=\dfrac{8}{3}\) 

\(\Rightarrow\dfrac{-6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{4\cdot8}{12}\)

\(\Rightarrow-6x+8x+3x+3+4x+2=32\)

\(\Rightarrow9x+5=32\)

\(\Rightarrow9x=32-5\)

\(\Rightarrow9x=27\)

\(\Rightarrow x=\dfrac{27}{9}\)

\(\Rightarrow x=3\)

b) \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\) (ĐK: \(x\ne-\dfrac{1}{2}\)) 

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)

\(\Rightarrow2x+1=13\)

\(\Rightarrow2x=12\)

\(\Rightarrow x=\dfrac{12}{2}\)

\(\Rightarrow x=6\left(tm\right)\)

a/ \(x-\dfrac{3}{7}=\dfrac{2}{5}\cdot\dfrac{1}{4}\)

\(x-\dfrac{3}{7}=\dfrac{1}{10}\)

\(x=\dfrac{1}{10}+\dfrac{3}{7}=\dfrac{37}{70}\)

Vậy....

b/ \(x+\dfrac{4}{5}=-\dfrac{5}{12}\cdot\dfrac{3}{25}\)

\(x+\dfrac{4}{5}=-\dfrac{1}{20}\)

\(x=-\dfrac{1}{20}-\dfrac{4}{5}=-\dfrac{17}{20}\)

Vậy....

c/ \(\dfrac{x}{182}=-\dfrac{6}{12}\cdot\dfrac{35}{91}\)

\(\dfrac{x}{182}=-\dfrac{5}{26}\)

\(=>x\cdot26=-5\cdot182\)

\(26x=-910\)

\(x=-910:26=-35\)

Vậy....

a) Ta có: \(x-\dfrac{3}{7}=\dfrac{2}{5}\cdot\dfrac{1}{4}\)

\(\Leftrightarrow x-\dfrac{3}{7}=\dfrac{1}{10}\)

\(\Leftrightarrow x=\dfrac{1}{10}+\dfrac{3}{7}=\dfrac{7}{70}+\dfrac{30}{70}\)

hay \(x=\dfrac{37}{70}\)

Vậy: \(x=\dfrac{37}{70}\)

a) Ta có: \(\dfrac{1}{7}+x=-\dfrac{2}{3}\)

\(\Leftrightarrow x=-\dfrac{2}{3}-\dfrac{1}{7}=\dfrac{-14}{21}-\dfrac{3}{21}\)

hay \(x=-\dfrac{17}{21}\)

Vậy: \(x=-\dfrac{17}{21}\)

b) Ta có: \(\dfrac{-2}{3}:x=\dfrac{-5}{6}\)

\(\Leftrightarrow x=\dfrac{-2}{3}:\dfrac{-5}{6}=\dfrac{-2}{3}\cdot\dfrac{6}{-5}=\dfrac{-12}{-15}=\dfrac{4}{5}\)

Vậy: \(x=\dfrac{4}{5}\)

c) Ta có: \(\left(\dfrac{3}{5}-2x\right)\cdot\dfrac{5}{8}=1\)

\(\Leftrightarrow\left(\dfrac{3}{5}-2x\right)=1:\dfrac{5}{8}=\dfrac{8}{5}\)

\(\Leftrightarrow-2x=\dfrac{8}{5}-\dfrac{3}{5}=1\)

hay \(x=-\dfrac{1}{2}\)

Vậy: \(x=-\dfrac{1}{2}\)

d) Ta có: \(\dfrac{3}{4}+\dfrac{2}{5}x=\dfrac{29}{60}\)

\(\Leftrightarrow x\cdot\dfrac{2}{5}=\dfrac{29}{60}-\dfrac{3}{4}=\dfrac{29}{60}-\dfrac{45}{60}=\dfrac{-16}{60}=\dfrac{-4}{15}\)

hay \(x=\dfrac{-4}{15}:\dfrac{2}{5}=\dfrac{-4}{15}\cdot\dfrac{5}{2}=\dfrac{-20}{30}=-\dfrac{2}{3}\)

Vậy: \(x=-\dfrac{2}{3}\)

e) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)

hay \(x=-\dfrac{1}{4}:\dfrac{7}{20}=\dfrac{-1}{4}\cdot\dfrac{20}{7}=\dfrac{-20}{28}=\dfrac{-5}{7}\)

Vậy: \(x=-\dfrac{5}{7}\)

f) Ta có: \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\Leftrightarrow-x+\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=0\)

\(\Leftrightarrow-x+\dfrac{55}{60}-\dfrac{24}{60}-\dfrac{40}{60}=0\)

\(\Leftrightarrow-x-\dfrac{9}{60}=0\)

\(\Leftrightarrow-x=\dfrac{9}{60}=\dfrac{3}{20}\)

hay \(x=-\dfrac{3}{20}\)

Vậy: \(x=-\dfrac{3}{20}\)

g) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=\dfrac{-1}{2}\)

\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{-1}{2}+4=\dfrac{-1}{2}+\dfrac{8}{2}=\dfrac{7}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{7}{2}\\x+\dfrac{1}{3}=-\dfrac{7}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{21}{6}-\dfrac{2}{6}=\dfrac{19}{6}\\x=-\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{-21}{6}-\dfrac{2}{6}=\dfrac{-23}{6}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{19}{6};-\dfrac{23}{6}\right\}\)

Bài 1:

a.

$|x+\frac{7}{4}|=\frac{1}{2}$

\(\Leftrightarrow \left[\begin{matrix} x+\frac{7}{4}=\frac{1}{2}\\ x+\frac{7}{4}=-\frac{1}{2}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-5}{4}\\ x=\frac{-9}{4}\end{matrix}\right.\)

b. $|2x+1|-\frac{2}{5}=\frac{1}{3}$
$|2x+1|=\frac{1}{3}+\frac{2}{5}$

$|2x+1|=\frac{11}{15}$

\(\Leftrightarrow \left[\begin{matrix} 2x+1=\frac{11}{15}\\ 2x+1=\frac{-11}{15}\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{-2}{15}\\ x=\frac{-13}{15}\end{matrix}\right.\)

c.

$3x(x+\frac{2}{3})=0$

\(\Leftrightarrow \left[\begin{matrix} 3x=0\\ x+\frac{2}{3}=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=0\\ x=\frac{-3}{2}\end{matrix}\right.\)

d.

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

$\Leftrightarrow x=\frac{2}{5}$

Bài 2:

$\frac{1}{100}-A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}$

$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{100-99}{99.100}$

$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}$

$=\frac{99}{100}$

$\Rightarrow A=\frac{1}{100}-\frac{99}{100}=-\frac{98}{100}=\frac{-49}{50}$

Bài 1: 

a) Ta có: \(\left|x+\dfrac{7}{4}\right|=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{7}{4}=\dfrac{1}{2}\\x+\dfrac{7}{4}=\dfrac{-1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{4}\\x=\dfrac{-9}{4}\end{matrix}\right.\)

b) Ta có: \(\left|2x+1\right|-\dfrac{2}{5}=\dfrac{1}{3}\)

\(\Leftrightarrow\left|2x+1\right|=\dfrac{11}{15}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{11}{15}\\2x+1=\dfrac{-11}{15}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{-4}{15}\\2x=\dfrac{-26}{15}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2}{15}\\x=\dfrac{-13}{15}\end{matrix}\right.\)

c) Ta có: \(3x\left(x+\dfrac{2}{3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+\dfrac{2}{3}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-2}{3}\end{matrix}\right.\)