2 ( x - 3 ) - 3 ( 2 - 3x ) = 4 [ ( 1 - 2x ) + 15 ]

 2x - 6 - 6 + 9x = 4 [ 1 - 2 x + 15 ]

2x - 6 -6 + 9x =   4 - 8x + 60

2x + 9x + 8x  = 4 + 60 + 6 + 6 

 19x                 = 76

=> x                 = 76 : 19 

=> x                 = 4

Vậy x = 4

\(2\left(x-3\right)-3\left(2-3x\right)=4\left[\left(1-2x\right)+15\right]\)

\(\Rightarrow2x-6-6+9x=4\left[1-2x+15\right]\)

\(\Rightarrow2x-6-6+9x=4-8x+60\)

\(\Rightarrow2x+9x+8x=4+60+6+6\)

\(\Rightarrow19x=76\)

\(\Rightarrow x=76:19=4\)

Vậy x = 4

đề bài yêu cầu gì vậy pn

`a)f(x)-g(x)`

`=x^3-2x^2+3x+1-(x^3+x-1)`

`=x^3-2x^2+3x+1-x^3-x+1`

`=(x^3-x^3)+(3x-x)-2x^2+2`

`=-2x^2+2x+2=0`

`b)f(x)-g(x)+h(x)=0`

`<=>-2x^2+2x+2+2x^2-1=0`

`<=>2x+1=0`

`<=>2x=-1`

`<=>x=-1/2`

Vậy `x=-1/2` thì `f(x)-g(x)+h(x)=0`

a) f(x) - g(x)=-2x²+2x+2

b) f(x) - g(x) + h(x) =2x-1=0

=> 2x=1

=>x=\(\dfrac{1}{2}\)

cảm ơn các bạn nha

Ko cần đâu bn à mk mong bn đấy

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

\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5-\frac{1}{2}x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{3}\\x=10\end{cases}}\)

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

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

    \(\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{8}\)

\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{8}\\\frac{1}{2}x-\frac{1}{3}=-\frac{7}{8}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{29}{12}\\x=-\frac{13}{12}\end{cases}}\)

a)\(\left(3x-1\right)\left(\frac{-1}{2}x+5\right)=0\)
\(\Leftrightarrow\)3x - 1 = 0      hay      \(\frac{-1}{2}\)x + 5 = 0
\(\Leftrightarrow\)3x     = 1         I\(\Leftrightarrow\)\(\frac{-1}{2}\)x     = -5
\(\Leftrightarrow\)  x     = \(\frac{1}{3}\)  I\(\Leftrightarrow\)            x     = 10

b) 2 I \(\frac{1}{2}x-\frac{1}{3}\)I - \(\frac{3}{2}\)=\(\frac{1}{4}\)
\(\Leftrightarrow\) 2 I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{4}\)
\(\Leftrightarrow\)    I\(\frac{1}{2}x-\frac{1}{3}\)I = \(\frac{7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{7}{8}\)          hay     \(\frac{1}{2}x-\frac{1}{3}\)= \(\frac{-7}{8}\)
\(\Leftrightarrow\)\(\frac{1}{2}x\)           = \(\frac{29}{24}\)        I\(\Leftrightarrow\)\(\frac{1}{2}x\)           = \(\frac{-13}{24}\)
\(\Leftrightarrow\)      x              = \(\frac{29}{12}\)        I\(\Leftrightarrow\)      x              = \(\frac{-13}{12}\)

c) (2x +\(\frac{3}{5}\))² - \(\frac{9}{25}\)= 0
\(\Leftrightarrow\)(2x +\(\frac{3}{5}\))²       = \(\frac{9}{25}\)
\(\Leftrightarrow\) 2x +\(\frac{3}{5}\)         = \(\frac{3}{5}\)    hay      2x +\(\frac{3}{5}\)= \(\frac{-3}{5}\)
\(\Leftrightarrow\) 2x                    = 0           I \(\Leftrightarrow\)2x           = \(\frac{-6}{5}\)
\(\Leftrightarrow\)   x                    = 0           I \(\Leftrightarrow\) x           = \(\frac{-3}{5}\)

d) 3(x -\(\frac{1}{2}\)) - 5(x +\(\frac{3}{5}\)) = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)3x - \(\frac{3}{2}\)- 5x - 3 = -x + \(\frac{1}{5}\)
\(\Leftrightarrow\)-2x + x - \(\frac{9}{2}\)- \(\frac{1}{5}\)= 0
\(\Leftrightarrow\)-x = \(\frac{-47}{10}\)
\(\Leftrightarrow\) x = \(\frac{47}{10}\)

-2x - 11 = 3x +2

-2x -11 - 2 = 3x

-2x - 13 = 3x

2x + 13 = 3x

13 = x

bài này yêu cầu tìm x thuộc Z nha mấy bn.

\(-2x-11⋮3x+2\)

=>\(-\frac{2}{3}\left(3x+2\right)+\frac{4}{3}⋮3x+2\)

=>\(3\left(-\frac{2}{3}\left(3x+2\right)+\frac{4}{3}\right)⋮3x+2\)

=>\(-2\left(3x+2\right)+4⋮3x+2\)

=>\(4⋮3x+2\)

=>\(3x+2\inƯ\left(4\right)=\left\{\pm1\pm2\pm4\right\}\)

tự làm nốt nha

\(\frac{6}{2x+2}=\frac{3}{3x-15}ĐK:x\ne-1;3\)

\(\Leftrightarrow\frac{18x-90}{\left(2x+2\right)\left(3x-15\right)}=\frac{6x+6}{\left(3x-15\right)\left(2x+2\right)}\)

\(\Leftrightarrow18x-90=6x+6\Leftrightarrow12x-96=0\Leftrightarrow x=8\)TM điều kiện 

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

\(\Rightarrow A=\left|2x-1\right|+\left|3-2x\right|\ge\left|2x-1+3-2x\right|\)

\(\Rightarrow A=\left|2x-1\right|+\left|3-2x\right|\ge\left|2\right|=2\)

dấu "="xảy ra khi \(\left(2x-1\right).\left(3-2x\right)\ge0\)

\(\Rightarrow\frac{1}{2}\le x\le\frac{3}{2}\)

vậy min A=2 khi \(\frac{1}{2}\le x\le\frac{3}{2}\)