a)x.(5-2x)-2x.(1-x)=15
   x [ 5 - 2x -2.(1-x) ] = 15
   x ( 5 - 2x -2 + 2x ) =15
   x . 3 =15
   x = 5
b)(3x+2)²+(1+3x).(1-3x)=2
   9x²+12x+4+1-9x²=2
   12x + 5 = 2
    12x = -3
        x = -1/4

a)\(\Leftrightarrow\)\(5x-2x^2-2x+2x^2=15\)

\(\Leftrightarrow\)\(3x=15\)

\(\Leftrightarrow\)\(x=5\)

b)\(\Leftrightarrow\)\(9x^2+12x+4+1-9x^2-2=0\)

\(\Leftrightarrow\)\(12x+3=0\)

\(\Leftrightarrow\)\(x=-0,25\)

a: Ta có: \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

`a,x(x-1)-(x+2)^2=1`

`<=>x^2-x-x^2-4x-4=1`

`<=>-5x=5`

`<=>x=-1`

`b,(x+5)(x-3)-(x-2)^2=-1`

`<=>x^2+2x-15-x^2+4x-4+1=0`

`<=>6x-18=0`

`<=>x-3=0`

`<=>x=3`

`c,x(2x-4)-(x-2)(2x+3)=0`

`<=>2x(x-2)-(x-2)(2x+3)=0`

`<=>(x-2)(2x-2x-3)=0`

`<=>-3(x-2)=0`

`<=>x-2=0`

`<=>x=2`

`d,x(3x+2)+(x+1)^2-(2x-5)(2x+5)=-12`

`<=>3x^2+2x+x^2+2x+1-4x^2+25=-12`

`<=>4x+26=-12`

`<=>4x=-38`

`<=>x=-19/2`

Rảnh rỗi thật sự .-.

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

\(\Leftrightarrow2x^3-2x^2+2x-x^2+x-1-2x^3+3x^2-2=0\)

\(\Leftrightarrow3x=3\)

hay x=1

Vậy: S={1}

b) Ta có: \(\left(x+1\right)\left(x^2+2x+4\right)-x^3-3x^2+16=0\)

\(\Leftrightarrow x^3+2x^2+4x+x^2+2x+4-x^3-3x^2+16=0\)

\(\Leftrightarrow6x=-20\)

hay \(x=-\dfrac{10}{3}\)

c) Ta có: \(\left(x+1\right)\cdot\left(x+2\right)\left(x+5\right)-x^3-8x^2=27\)

\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-x^3-8x^2-27=0\)

\(\Leftrightarrow x^3+5x^2+3x^2+15x+2x+10-x^3-8x^2-27=0\)

\(\Leftrightarrow17x=17\)

hay x=1

a: \(\Leftrightarrow\left(x+2\right)\left(12-x\right)=0\)

\(\Leftrightarrow x\in\left\{-2;12\right\}\)

b: \(\Leftrightarrow\left(2x+5\right)\left(x-1\right)=0\)

\(\Leftrightarrow x\in\left\{-\dfrac{5}{2};1\right\}\)

Bài 1:

a) \(x^2-6x+15=\left(x^2-6x+9\right)+6=\left(x-3\right)^2+6\ge6\)

Dấu "=" xảy ra \(\Leftrightarrow x=3\)

b) \(3x^2-15x+4=3\left(x^2-5x+\dfrac{25}{4}\right)-\dfrac{59}{4}=3\left(x-\dfrac{5}{2}\right)^2-\dfrac{59}{4}\ge-\dfrac{59}{4}\)

Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{5}{2}\)

Bài 2:

a) \(\Rightarrow\left(x-5\right)\left(x+5\right)+2\left(x+5\right)=0\)

\(\Rightarrow\left(x+5\right)\left(x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=3\end{matrix}\right.\)

c) \(\Rightarrow x^2\left(x-2\right)+7\left(x-2\right)=0\)

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

\(\Rightarrow x=2\left(do.x^2+7\ge7>0\right)\)

\(a,\left(x+2\right)^2+\left(x+3\right)^2-2\left(x-2\right)\left(x-3\right)=19\\ \Leftrightarrow x^2+4x+4+x^2+6x+9-2x^2+10x-12=19\\ \Leftrightarrow20x=20\\ \Leftrightarrow x=1\\ b,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-5\right)=15\\ \Leftrightarrow x^3+8-x^3+5x=15\\ \Leftrightarrow5x=7\\ \Leftrightarrow x=\dfrac{7}{5}\\ c,\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\\ \Leftrightarrow x^3-3x^2+3x+1+8-x^3+3x^2+6x=17\\ \Leftrightarrow9x=8\\ \Leftrightarrow x=\dfrac{8}{9}\)

a. (x + 2)² + (x + 3)² - 2(x - 2)(x - 3) = 19

<=> (x² + 4x + 4) + (x² + 6x + 9) - (2x + 4)(x - 3) = 19

<=> x² + 4x + 4 + x² + 6x + 9 - 2x² + 6x - 4x + 12 = 19

<=> x² + x² - 2x² + 4x + 6x + 6x - 4x + 9 + 4 + 12 - 19 = 0

<=> 12x + 6 = 0

<=> 6(2x + 1) = 0

<=> 2x + 1 = 0

<=> 2x = -1

<=> x = \(\dfrac{-1}{2}\)  

b. (x + 2)(x² - 2x + 4) - x(x² - 5) = 15

<=> x³ - 2x² + 4x + 2x² - 4x + 8 - x³ + 5x - 15 = 0

<=> x³ - x³ - 2x² + 2x² + 4x - 4x + 5x + 8 - 15 = 0

<=> 5x - 7 = 0

<=> 5x = 7

<=> x = \(\dfrac{7}{5}\)

a: Ta có: \(4\left(2-x\right)+x\left(x+6\right)=x^2\)

\(\Leftrightarrow8-4x+x^2+6x-x^2=0\)

\(\Leftrightarrow2x=-8\)

hay x=-4

b: Ta có: \(x\left(x-7\right)-\left(x-2\right)\left(x+5\right)=0\)

\(\Leftrightarrow x^2-7x-x^2-3x+10=0\)

\(\Leftrightarrow-10x=-10\)

hay x=1

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

\(\Leftrightarrow9-4x^2+4x^2-4x+1=2\)

\(\Leftrightarrow-4x=-8\)

hay x=2

a, <=> x² -2x +1 + 5x -x² =8

<=> 3x +1 =8 

<=> 3x = 7

<=> x= 7/3

b, thiếu đề

c, <=> 2x³ -1 + 2x(4 -x²) = 7

<=> 2x³ + 8x -2³ = 8

<=> 8x =8

<=> x=1