A=(12x³y⁴-15x²y³-9x³y²):3x²y

  =(12x³y⁴:3x²y)-(15x²y³:3x²y)- (9x³y²:3x²y)

  =4x²y³-5y²-3x²

B, sai đề bạn ơi

Bài 2 : 

a, \(x^2-4x=0\Leftrightarrow x\left(x-4\right)=0\Leftrightarrow x=0;4\)

b, \(5x\left(x-2020\right)-x+2020=0\)

\(\Leftrightarrow5x\left(x-2020\right)-\left(x-2020\right)=0\Leftrightarrow\left(5x-1\right)\left(x-2020\right)=0\)

\(\Leftrightarrow x=\frac{1}{5};2020\)

c, \(\left(4x+5\right)^2-\left(2x-1\right)^2=0\)

\(\Leftrightarrow16x^2+40x+25-\left(4x^2-4x+1\right)=0\)

\(\Leftrightarrow12x^2+44x+24=0\Leftrightarrow4\left(x+3\right)\left(3x+2\right)=0\)

\(\Leftrightarrow x=-3;-\frac{2}{3}\)

a,x²-4x=0

= x.(x-4)=0

=> x=0 hoặc x-4=0

=>x=0 hoặc x=4

Bài 3 : 

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

b, \(x^2+2x-y^2+1=\left(x+1\right)^2-y^2=\left(x+1-y\right)\left(x+1+y\right)\)

c, \(x^2+y^2-z^2+2xy=\left(x+y\right)^2-z^2=\left(x+y-z\right)\left(x+y+z\right)\)

d, \(x^2-7x+12=x^2-3x-4x+12=\left(x-4\right)\left(x-3\right)\)

e, \(x^2-4x+xy-4y=x\left(x-4\right)+y\left(x-4\right)=\left(x+y\right)\left(x-4\right)\)

g, \(5x^2-10xy+5y^2-20z^2=5\left(x^2-2xy+y^2-4z^2\right)=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)

\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)

f, \(4x^2-4xy+y^2-9z^2=\left(2x+y\right)^2-\left(3z\right)^2=\left(2x+y-3z\right)\left(2x+y+3z\right)\)

n, \(\left(x+y\right)^3-\left(z-t\right)^3=\left(x+y-z+t\right)\left[\left(x+y\right)^2+\left(x+y\right)\left(z-t\right)+\left(z-t\right)^2\right]\)

Làm nốt nhé, ko phải đi học thì t giải hết cho cậu r :)) 

Ta có: \(4x\left(x+y\right)\left(x+y+z\right)\left(x+z\right)+y^2z^2\)

\(=4\left[x\left(x+y+z\right)\right]\left[\left(x+y\right)\left(x+z\right)\right]+y^2z^2\)

\(=4\left(x^2+xy+zx\right)\left(x^2+xy+yz+zx\right)+y^2z^2\) \(\left(1\right)\)

Đặt \(\hept{\begin{cases}x^2+xy+zx=a\\yz=b\end{cases}}\)

Khi đó: \(\left(1\right)=4a\left(a+b\right)+b^2\)

\(=4a^2+4ab+b^2\)

\(=\left(2a+b\right)^2\)

\(=\left(2x^2+2xy+2zx+yz\right)^2\ge0\left(\forall x,y,z\right)\)

=> đpcm

Ta có:\(4x\left(x+y\right)\left(x+y+z\right)\left(x+z\right)+y^2z^2=4x\left(x+y+z\right)\left(x+y\right)\left(x+z\right)+y^2z^2=4\left(x^2+xy+xz\right)\left(x^2+xy+yz+zx\right)+y^2z^2\)Đặt \(x^2+xy+xz=t\)thì biểu thức trên trở thành \(4t\left(t+yz\right)+y^2z^2=4t^2+4yzt+y^2z^2=\left(2t+yz\right)^2=\left(2x^2+2xy+2xz+yz\right)^2\ge0\forall x,y,z\left(đpcm\right)\)

       \(x^2-4x+6\)

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

\(=\left(x-2\right)^2+2\)

\(\ge\left(3-2\right)^2+2\)

\(\ge1+2\)

\(\ge3\)

Dấu "=" xảy ra <=> x=3

Vậy min của biểu thức bằng 3 khi x=3

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

\(=2x^2-x-4x+2-3\left(x^2+2x+1\right)-4x^2-8x\)

\(=2x^2-5x+2-3x^2-6x-3-4x^2-8x\)

\(=-5x^2-19x-1\)

Ta có :

x³ - 3x² + 5x - 6

= x³ - 2x² - x² + 2x + 3x - 6

= x²( x - 2 ) - x( x - 2 ) + 3( x - 2 )

= ( x - 2 )( x² - x + 3 )

=> ( x³ - 3x² + 5x - 6 ) : ( x - 2 ) = x² - x + 3 

x^3-3x^2+5x-6 x-2 x^2-x+3 x^3-2x^2 -x^2+5x -x^2+2x 3x-6 3x-6 0

   4x(x+y)(x+y+z)(x+z) + y^2.z^2

= 4(x^2 + xy + xz)( x^2 + xy + xz + yz) + y^2.z^2

Đặt x^2 + yz + xz = t

=>  4x(x+y)(x+y+z)(x+z) + y^2.z^2 = 4t( t + yz) + y^2.z^2 = 4t^2 + 4tyz +y^2.z^2 = ( 2t + yz)^2 \(\ge\)0(ĐPCM)

Vậy 4t^2 + 4tyz +y^2.z^2 = ( 2t + yz)^2 \(\ge\)0 với moji x,y,z

???????????? what !