(3x+2).(x+1)=3x.(5+x)                                                                               

\(\Rightarrow\)\(3x^2+3x+2x+2=15x+3x^2\)                               

\(\Rightarrow3x^2+5x+2=15x+3x^2\)

\(\Rightarrow5x-15x+2=3x^2-3x^2\)

\(\Rightarrow-10x+2=0\)

\(-10x=-2\)

    \(x=\frac{1}{5}\)

a) 4x²  - 2x + 3 - 4x.(x - 5) = 7x - 3

--> 4x² - 2x + 3 - 4x² + 20x = 7x - 3

--> 4x² - 2x - 4x² + 20x - 7x = -3 - 3

--> 11x = -6

--> x = \(\frac{-6}{11}\)

b) -3x.(x - 5) + 5.(x - 1) + 3x² = 4x

--> -3x² + 15x + 5x - 5 + 3x² = 4x

--> -3x²  + 15x + 5x + 3x² - 4x = 5 

--> 16x = 5

--> x = \(\frac{5}{16}\)

c) 7x.(x - 2) - 5.(x - 1) = 21x² - 14x² + 3

--> 7x² - 14x - 5x + 5 = 7x² + 3 

--> 7x²  - 14x - 5x - 7x²  = -5 + 3 

--> -19x = -2 

--> x = \(\frac{2}{19}\)

d) 3.(5x - 1) - x.(x - 2) + x² - 13x = 7

--> 15x - 3 - x² + 2x + x² - 13x = 7

--> 15x - x² + 2x + x² - 13x = 3 + 7

--> 4x = 10

--> x = \(\frac{5}{2}\)

e) \(\frac{1}{5}\)x.(10x - 15) - 2x.(x - 5) = 12

--> 2x² - 3x - 2x² + 10x = 12

--> 7x = 12

--> x = \(\frac{12}{7}\)

~ Học tốt ~

a) 4x² - 2x + 3 - 4x(x - 5) = 7x - 3

=> 4x² - 2x + 3 - 4x² + 20x = 7x - 3

=> 18x + 3 = 7x - 3

=> 18x - 7x = -3 - 3

=> 11x = -6

=>  x = -6/11

b) -3x(x - 5) + 5(x - 1) + 3x² = 4x

=> -3x² + 15x + 5x - 5 + 3x² = 4x

=> 20x - 5 = 4x

=> 20x - 4x = 5

=> 16x = 5

=> x = 5/16

\(c,7x\left(x-2\right)-5\left(x-1\right)=21x^2-14x^2+3\)

\(\Leftrightarrow7x^2-14x-5x+5=7x^2+3\)

\(\Leftrightarrow7x^2-7x^2-19x=3-5\)

\(\Leftrightarrow-19x=-2\)

\(\Leftrightarrow x=\frac{2}{19}\)

a: \(\Leftrightarrow3x^3-2x^2+15x^2-10x+3x-2+7⋮3x-2\)

\(\Leftrightarrow3x-2\in\left\{1;-1;7;-7\right\}\)

hay \(x\in\left\{3;1\right\}\)

b: \(\Leftrightarrow2x^5-7x^3+4x^4-14x^2+14x^2-49x+49x-44⋮2x^2-7\)

\(\Leftrightarrow2401x^2-1936⋮2x^2-7\)

\(\Leftrightarrow4802x^2-3872⋮2x^2-7\)

\(\Leftrightarrow2x^2-7\inƯ\left(12935\right)\)

\(\Leftrightarrow2x^2-7\in\left\{1;5;13;65;199;995;2587;12935;-1;-5\right\}\)

\(\Leftrightarrow2x^2\in\left\{8;72;2\right\}\)

hay \(x\in\left\{2;-2;6;-6;1;-1\right\}\)

Bài 3:

a: =>3x^2-6x-x-3x^2=14

=>-7x=14

=>x=-2

b: \(\Leftrightarrow2x^2+10x-x-5-2x^2-9x-x-4.5=3.5\)

=>-x-9,5=3,5

=>-x=12

=>x=-12

c: =>\(3x-3x^2+9x=36\)

=>-3x^2+12x-36=0

=>x^2-6x+12=0(loại)

d: \(\Leftrightarrow3x^2-3x+x-1+4x-3x^2=5\)

=>2x=6

=>x=3

a) ĐKXĐ: \(x\notin\left\{-1;0\right\}\)

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

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

Suy ra: \(x^2+3x+x^2-3x+2=2x^2+2x\)

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

\(\Leftrightarrow-2x+2=0\)

\(\Leftrightarrow-2x=-2\)

hay x=1(nhận)

Vậy: S={1}

b) ĐKXĐ: \(x\notin\left\{-7;\dfrac{3}{2}\right\}\)

Ta có: \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)

\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(6x+1\right)\left(x+7\right)\)

\(\Leftrightarrow6x^2-9x-4x+6=6x^2+42x+x+7\)

\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)

\(\Leftrightarrow-56x-1=0\)

\(\Leftrightarrow-56x=1\)

hay \(x=-\dfrac{1}{56}\)(nhận)

Vậy: \(S=\left\{-\dfrac{1}{56}\right\}\)

c) ĐKXĐ: \(x\ne-\dfrac{2}{3}\)

Ta có: \(\dfrac{5}{3x+2}=2x-1\)

\(\Leftrightarrow5=\left(3x+2\right)\left(2x-1\right)\)

\(\Leftrightarrow6x^2-3x+4x-2-5=0\)

\(\Leftrightarrow6x^2+x-7=0\)

\(\Leftrightarrow6x^2-6x+7x-7=0\)

\(\Leftrightarrow6x\left(x-1\right)+7\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(6x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\6x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\6x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{7}{6}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{1;-\dfrac{7}{6}\right\}\)

d) ĐKXĐ: \(x\ne\dfrac{2}{7}\)

Ta có: \(\left(2x+3\right)\cdot\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)

\(\Leftrightarrow\left(2x+3\right)\cdot\left(\dfrac{3x+8+2-7x}{2-7x}\right)-\left(x-5\right)\left(\dfrac{3x+8+2-7x}{2-7x}\right)=0\)

\(\Leftrightarrow\left(2x+3-x+5\right)\cdot\dfrac{-4x+6}{2-7x}=0\)

\(\Leftrightarrow\left(x+8\right)\cdot\left(-4x+6\right)=0\)(Vì \(2-7x\ne0\forall x\) thỏa mãn ĐKXĐ)

\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\-4x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\-4x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\left(nhận\right)\\x=\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{-8;\dfrac{3}{2}\right\}\)

1) 4x² - 2x + 3 - 4x(x - 5) = 7x - 3

<=> 4x² - 9x - 4x² + 20x = -6

<=> 11x = -6

<=> x = -6/11

2. -3x(x - 5) + 5(x - 1) + 3x² = 4 - x

<=> -3x² + 15x + 5x - 5 + 3x² + x = 4

<=> 21x = 9

<=> x = 3/7

1.\(4x^2-2x+3-4x\left(x-5\right)=7x-3\)

\(\Leftrightarrow4x^2-2x+3-4x^2+20x=7x-3\)

\(\Leftrightarrow4x^2-2x+3-4x^2+20x-7x+3=0\)

\(\Leftrightarrow11x+6=0\)

\(\Leftrightarrow11x=-6\)

\(\Leftrightarrow x=-\frac{6}{11}\)

Vậy .....

2.\(-3x\left(x-5\right)+5\left(x-1\right)+3x^2=4-x\)

\(\Leftrightarrow-3x^2+15x+5x-5+3x^2-4+x=0\)

\(\Leftrightarrow21x-9=0\)

\(\Leftrightarrow x=\frac{3}{7}\)