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!

1: Ta có: \(\dfrac{3}{x-3}+\dfrac{4}{x+3}=\dfrac{3x-7}{x^2-9}\)

\(\Leftrightarrow\dfrac{3x+9}{\left(x-3\right)\left(x+3\right)}+\dfrac{4x-12}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-7}{\left(x-3\right)\left(x+3\right)}\)

Suy ra: \(3x+9+4x-12=3x-7\)

\(\Leftrightarrow4x=-7+12-9=-4\)

hay \(x=-1\left(nhận\right)\)

2: Ta có: \(\dfrac{3}{x-4}-\dfrac{4}{x+4}=\dfrac{3x-4}{x^2-16}\)

\(\Leftrightarrow\dfrac{3x+12}{\left(x-4\right)\left(x+4\right)}-\dfrac{4x-16}{\left(x+4\right)\left(x-4\right)}=\dfrac{3x-4}{\left(x-4\right)\left(x+4\right)}\)

Suy ra: \(3x+12-4x+16=3x-4\)

\(\Leftrightarrow28-4x=-4\)

\(\Leftrightarrow4x=32\)

hay \(x=8\left(tm\right)\)

3: Ta có: \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)

Suy ra: \(5x^2-12+3x+3=5x^2-5x\)

\(\Leftrightarrow3x-9+5x=0\)

\(\Leftrightarrow8x=9\)

hay \(x=\dfrac{9}{8}\left(nhận\right)\)

c: Ta có: \(\dfrac{1}{3}-\dfrac{7}{8}x=\dfrac{1}{4}\)

\(\Leftrightarrow x\cdot\dfrac{7}{8}=\dfrac{1}{12}\)

\(\Leftrightarrow x=\dfrac{1}{12}\cdot\dfrac{8}{7}=\dfrac{2}{21}\)

d: Ta có: \(\dfrac{3}{2}x+\dfrac{1}{7}=\dfrac{7}{8}\cdot\dfrac{64}{49}\)

\(\Leftrightarrow x\cdot\dfrac{3}{2}=1\)

hay \(x=\dfrac{2}{3}\)

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: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3

=>x=-1/3+3/4=-4/12+9/12=5/12

b: =>x(1/2-5/6)=7/2

=>-1/3x=7/2

hay x=-21/2

c: (4-x)(3x+5)=0

=>4-x=0 hoặc 3x+5=0

=>x=4 hoặc x=-5/3

d: x/16=50/32

=>x/16=25/16

hay x=25

e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4

=>2x=-7/4+3=5/4

hay x=5/8

Lời giải:

a. $\frac{2-x}{4}=\frac{3x-1}{3}$

$\Rightarrow 3(2-x)=4(3x-1)$

$\Rightarrow 6-3x=12x-4$

$\Rightarrow 6+4=12x+3x$

$\Rightarrow 10=15x$

$\Rightarrow x=\frac{10}{15}=\frac{2}{3}$

b.

$\frac{x}{7}=\frac{x+16}{35}$

$\Rightarrow \frac{5x}{35}=\frac{x+16}{35}$

$\Rightarrow 5x=x+16$

$\Rightarrow 4x=16$

$\Rightarrow x=4$

c.

$\sqrt{x^2+1}=3$

$\Rightarrow x^2+1=9$

$\Rightarrow x^2=8\Rightarrow x=\pm \sqrt{8}=\pm 2\sqrt{2}$

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 4: 

a) \(\dfrac{4}{3}+\left(1,25-x\right)=2,25\)

\(1,25-x=2,25-\dfrac{4}{3}=\dfrac{9}{4}-\dfrac{4}{3}\)

\(1,25-x=\dfrac{11}{12}\)

\(x=1,25-\dfrac{11}{12}=\dfrac{5}{4}-\dfrac{11}{12}\)

\(x=\dfrac{1}{3}\)

b) \(\dfrac{17}{6}-\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(x-\dfrac{7}{6}=\dfrac{17}{6}-\dfrac{7}{4}=\dfrac{34}{12}-\dfrac{21}{12}\)

\(x-\dfrac{7}{6}=\dfrac{13}{12}\)

\(x=\dfrac{13}{12}+\dfrac{7}{6}=\dfrac{13}{12}+\dfrac{14}{12}\)

\(x=\dfrac{27}{12}=\dfrac{9}{4}\)

c) \(4-\left(2x+1\right)=3-\dfrac{1}{3}=\dfrac{9}{3}-\dfrac{1}{3}\)

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

\(2x+1=\dfrac{8}{3}+4=\dfrac{8}{3}+\dfrac{12}{3}\)

\(2x+1=\dfrac{20}{3}\)

\(2x=\dfrac{20}{3}-1=\dfrac{20}{3}-\dfrac{3}{3}\)

\(2x=\dfrac{17}{3}\)

\(x=\dfrac{17}{3}.\dfrac{1}{2}=\dfrac{17}{6}\)

Bài 15:

a) \(\left(\dfrac{-2}{3}\right)^9:x=\dfrac{-2}{3}\)

\(x=\left(\dfrac{-2}{3}\right)^9:\dfrac{-2}{3}=\left(\dfrac{-2}{3}\right)^{9-1}\)

\(=>x=\left(\dfrac{-2}{3}\right)^8\)

b) \(x:\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^4\)

\(x=\left(\dfrac{4}{9}\right)^4.\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^{4+5}\)

\(=>x=\left(\dfrac{4}{9}\right)^9\)

c) \(\left(x+4\right)^3=-125\)

\(\left(x+4\right)^3=\left(-5\right)^3\)

\(=>x+4=-5\)

\(x=-5-4\)

\(=>x=-9\)

d) \(\left(10-5x\right)^3=64\)

\(\left(10-5x\right)^3=4^3\)

\(=>10-5x=4\)

\(5x=10-4\)

\(5x=6\)

\(=>x=\dfrac{6}{5}\)

e) \(\left(4x+5\right)^2=81\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}4x=-14\\4x=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-14}{4}\\x=1\end{matrix}\right.\)

Bài 16:

a) \(4-1\dfrac{2}{5}-\dfrac{8}{3}\)

\(=4-\dfrac{7}{5}-\dfrac{8}{3}\)

\(=\dfrac{60-21-40}{15}=\dfrac{-1}{15}\)

b) \(-0,6-\dfrac{-4}{9}-\dfrac{16}{15}\)

\(=\dfrac{-3}{5}+\dfrac{4}{9}-\dfrac{16}{15}\)

\(=\dfrac{\left(-27\right)+20-48}{45}=\dfrac{-55}{45}=\dfrac{-11}{9}\)

c) \(-\dfrac{15}{4}.\left(\dfrac{-7}{15}\right).\left(-2\dfrac{2}{5}\right)\)

\(=\dfrac{7}{4}.\dfrac{-12}{5}\)

\(=\dfrac{-21}{5}\)

\(#Wendy.Dang\)

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