\(x\) x\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{5}\)= 55

\(x\) x \(\dfrac{11}{20}\)       = 55

\(x\)                = 55 : \(\dfrac{11}{20}\)

\(x\)                =100

\(x\left(1-3x\right)\left(4-3x\right)-\left(x-4\right)\left(3x+5\right)=4x-15x^2+9x^3-3x^2+7x+20=9x^3-18x^2+11x+20\)

x(1 - 3x)(4 - 3x) - (x - 4)(3x + 5)

= (x - 3x²)(4 - 3x) - 3x² - 5x + 12x + 20

= 4x - 3x² - 12x² + 9x³ - 3x² - 5x + 12x + 20

= 9x³ - 18x² + 11x + 20

a)\(\frac{1}{2}+\frac{3}{4}.x=\frac{1}{4}\)

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

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

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

                      \(x=\frac{-1}{3}\)

Vậy \(x=\frac{-1}{3}\)

b)\(|x-5|-\frac{1}{3}=0,5\)

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

                \(|x-5|=\frac{5}{6}\)

\(\Rightarrow\orbr{\begin{cases}x-5=\frac{5}{6}\\x-5=\frac{-5}{6}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{6}+5\\x=\frac{-5}{6}+5\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{35}{6}\\x=\frac{25}{6}\end{cases}}\)

Vậy\(x=\frac{35}{6}\)hoặc\(x=\frac{25}{6}\)

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

\(\Rightarrow5\left(x+3\right)=-4\left(1-x\right)\)

\(\Leftrightarrow5x+15=-4+4x\)

\(\Leftrightarrow5x-4x=-4-15\)

\(\Leftrightarrow x=-19\)

Vậy \(x=-19\).

`(x+2)/3=(x-4)/5`

`<=> (x+2)*5=3(x-4)`

`<=> 5x+10=3x-12`

`<=> 5x-3x=-12-10`

`<=> 2x=-22`

`<=> x=-11`

cảm ơn bn nhiều nhé

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

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

=>\(x=1:\dfrac{4}{5}=\dfrac{5}{4}\)

b; \(-\dfrac{3}{7}-\dfrac{4}{7}:x=-2\)

=>\(\dfrac{4}{7}:x+\dfrac{3}{7}=2\)

=>\(\dfrac{4}{7}:x=2-\dfrac{3}{7}=\dfrac{11}{7}\)

=>\(x=\dfrac{4}{7}:\dfrac{11}{7}=\dfrac{4}{11}\)

(0,3+0,5):x =1,6

<=> 0,8 : x =1,6

<=> x= 0,5

\(\dfrac{0,3}{x}\) + \(\dfrac{0,5}{x}\) = 1,6

          \(\dfrac{0,8}{x}\) = 1,6

            \(x\) = 0,8 : 1,6

           \(x\) = 0,5 

\(\sqrt{x^2-2x+4}+\sqrt{x^2+5}=9-2x\left(đk:x\le\dfrac{9}{2}\right)\)

\(\Leftrightarrow x^2-2x+4+x^2+5+2\sqrt{\left(x^2-2x+4\right)\left(x^2+5\right)}=81-36x+4x^2\)

\(\Leftrightarrow2\sqrt{\left(x^2-2x+4\right)\left(x^2+5\right)}=2x^2-34x+72\)

\(\Leftrightarrow4\left(x^2-2x+4\right)\left(x^2+5\right)=4x^4+1156x^2+5184-136x^3+288x^2-4896x\)

\(\Leftrightarrow4x^4-8x^3+36x^2-40x+80=4x^4-136x^3+1444x^2-4896x+5184\)

\(\Leftrightarrow128x^3-1408x^2+4856x-5104=0\)

\(\Leftrightarrow128x^2\left(x-2\right)-1152x\left(x-2\right)+2552\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(128x^2-1152x+2552\right)=0\)

\(\Leftrightarrow x=2\left(tm\right)\)(do \(128x^2-1152x+2552>0\))

cảm mơn bn ạ

a: =>4/3x=7/9-4/9=1/3

=>x=1/4

b: =>5/2-x=9/14:(-4/7)=-9/8

=>x=5/2+9/8=29/8

c: =>3x+3/4=8/3

=>3x=23/12

hay x=23/36

d: =>-5/6-x=7/12-4/12=3/12=1/4

=>x=-5/6-1/4=-10/12-3/12=-13/12

em moi lop 4 mà

|2.x+4|=6

TH1: 2.x+4 = 6                

x = 1

TH2: 2.x+4 = - 6                

x = -5

Vậy x thuộc 1 và -5

|2-3.x|=5

Th1: 2-3.x=5

x = -1

Th2: 2-3.x= -5

x = 7/3

Vậy x thuộc -1 và 7/3

|7-x|=9

TH1: 7-x =9

x = -2

TH2: 7-x = -9

x = 16

Vậy.........

a) Ta có: \(\left|2x+4\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+4=6\\2x+4=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6-4=2\\2x=-6-4=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)

Vậy: \(x\in\left\{1;-5\right\}\)

b) Ta có: \(\left|2-3x\right|=5\)

\(\Leftrightarrow\left[{}\begin{matrix}2-3x=5\\2-3x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=3\\-3x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{7}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{-1;\dfrac{7}{3}\right\}\)

c) Ta có: \(\left|7-x\right|=9\)

\(\Leftrightarrow\left[{}\begin{matrix}7-x=9\\7-x=-9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=2\\-x=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=16\end{matrix}\right.\)

Vậy: \(x\in\left\{-2;16\right\}\)