1/ (2x+3)(x-4)+(x+5)(x-2)=(3x-5)(x-4)

<=> 2x² - 8x + 3x - 12 + x² - 2x + 5x  - 10 - 3x² + 12x + 5x - 20 = 0

<=> 15x - 20 = 0

<=> 15x = 20

<=> x = 4/3

a) \(\left(x-1\right)^3+3\left(x+1\right)^2=\left(x^2-2x+4\right)\)

\(\Leftrightarrow x^3+9x+2=x^3+8\)

\(\Leftrightarrow x^3+9x=x^3+8-2\)

\(\Leftrightarrow x^3+9x=x^3+6\)

\(\Leftrightarrow x^3+9x=x^3+6x-x^3\)

\(\Leftrightarrow\frac{2}{3}\)

b) \(x^2-4=8\left(x-2\right)\)

\(\Leftrightarrow x^2-4=8x-16\)

\(\Leftrightarrow x^4-4=8x-16+16\)

\(\Leftrightarrow x^2+12=8x\)

\(\Leftrightarrow x^2+12=8x-8x\)

\(\Leftrightarrow x^2-8x+12=0\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=6\end{cases}}\)

a. \(8x\left(x-2007\right)-2x+4034=0\)

\(\Rightarrow\left(x-2017\right)\left(4x-1\right)\)

\(\Rightarrow\left[{}\begin{matrix}x-2017=0\\4x-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2017\\4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)

Vậy x=2017 hoặc x=1/4

b.\(\dfrac{x}{2}+\dfrac{x^2}{8}=0\)

\(\Rightarrow\dfrac{x}{2}\left(1+\dfrac{x}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}=0\\1+\dfrac{x}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{x}{4}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

Vậy x=0 hoặc x=-4

c.\(4-x=2\left(x-4\right)^2\)

\(\Rightarrow\left(4-x\right)-2\left(x-4\right)^2=0\)

\(\Rightarrow\left(4-x\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4-x=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{2}\end{matrix}\right.\)

Vậy x=4 hoặc x=7/2

d.\(\left(x^2+1\right)\left(x-2\right)+2x=4\)

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

Nxet: (x²+3)>0 với mọi x

=> x-2=0 <=>x=2

Vậy x=2

a, 8\(x\).(\(x-2007\)) - 2\(x\) + 4034 = 0

     4\(x\)(\(x\) - 2007) - \(x\) + 2017 = 0

     4\(x^2\) - 8028\(x\) - \(x\) + 2017 = 0

     4\(x^2\) - 8029\(x\) + 2017 = 0

     4(\(x^2\) - 2. \(\dfrac{8029}{8}\) \(x\) +( \(\dfrac{8029}{8}\))²) - (\(\dfrac{8029}{4}\))²  + 2017 = 0

    4.(\(x\) + \(\dfrac{8029}{8}\))² = (\(\dfrac{8029}{4}\))² - 2017

       \(\left[{}\begin{matrix}x=-\dfrac{8029}{8}+\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\\x=-\dfrac{8029}{8}-\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\end{matrix}\right.\) 

ĐKXĐ : X khác 1

pt <=> X^2+X+1/(X-1).(X^2+X+1) - 3X^2/(X-1).(X^2+X+1) = 2X.(X-1)/(X-1).(X^2+X+1)

<=> X^2+X+1/(X-1).(X^2+X+1) - 3X^2/(X-1).(X^2+X+1) - 2X^2-2X/(X-1).(X^2+X+1) = 0

<=> X^2+X+1-3X^2-2X^2+2X/(X-1).(X^2+X+1) = 0

<=> X^2+X+1-3X^2-2X^2+2X=0

<=> -4X^2+3X+1=0

<=> 4X^2-3X-1=0

<=> (X-1).(4X+1) = 0

<=> 4X+1=0 ( vì X khác 1 nên X-1 khác 0 )

<=> X = -1/4 (tm)

Vậy pt có tập nghiệm S = {-1/4}

Tk mk nha

a,y=2/3

b,y=1/24

c,y=3/7

mình cần đầy đủ nhé bn 

a,        y - 1/6 = 1/2 + 1/3

          y - 1/6 =     5/6

             y     =      5/6 + 1/6

            y      =          6/6

           y       =            1

b,      y + 1/8 = 1/2 - 1/3

         y + 1/8 =     1/6

            y      =     1/6 -1/8

            y      =       2/48

            y      =        1/14

c,     y : 1/2 = 2/3 x 3/4

       y  : 1/2 =    1/2

           y     =     1/2 x 1/2

           y     =          1/4

mk nhanh nhất đó mk nha mk lại cho bn ko mk mk ko chịu đâu mk khóc đó

7: Ta có: \(\left(3x+4\right)\left(2x-1\right)+6x\left(1-x\right)=0\)

\(\Leftrightarrow6x^2-3x+8x-4+6x-6x^2=0\)

\(\Leftrightarrow11x=4\)

hay \(x=\dfrac{4}{11}\)

8: Ta có: \(2x\left(x^2-1\right)+x\left(-2x^2-3x+1\right)=-x-27\)

\(\Leftrightarrow2x^3-2x-2x^3-3x^2+x+x+27=0\)

\(\Leftrightarrow x^2=9\)

hay \(x\in\left\{3;-3\right\}\)