\(4x^2-4xy+y-25a^2+10a-136\)

\(\text{Phân tích thành nhân tử}\)

\(-\left(4xy-y-4x^2+25a^2-10a+136\right)\)

k nhé !

Bài 1: 

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

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

\(4x^2-4xy+2x-y+y^2\\=[(2x)^2-2\cdot 2x\cdot y+y^2]+2x-y\\=(2x-y)^2+(2x-y)\\=(2x-y)(2x-y+1)\)

\(\left(2x-y\right)\left(4x^2-4xy+y^2\right)-8x^2\left(x-y\right)\)

\(=8x^3-y^3-8x^3+8x^2y\)

\(=8x^2y-y^3\)

\(a,=x\left(x^2+2\right)\\ b,=2021\left(x+y\right)+x\left(x+y\right)=\left(x+2021\right)\left(x+y\right)\\ c,=49-\left(2x+y\right)^2=\left(7-2x-y\right)\left(7+2x+y\right)\)

a) x³+2x
  = x(x²+2)
b) 2021x+2021y+x²+xy
  = (2021x+2021y)+(x²+xy)
  = 2021(x+y)+x(x+y)
  = (2021+x)(x+y)
c) 49-4x²-4xy-y²  
  = -[(2x)²+2.2x.y+y²] + 7²
  = -(2x-y)²+7²
  = 7²-(2x-y)²
  = (7-2x+y)(7+2x-y)

\(A=\left(6x-3y\right)+\left(4x^2-4xy+y^2\right)=3\left(2x-y\right)+\left(2x-y\right)^2=\left(2x-y\right)\left(2+2x-y\right)\)

\(B=9x^2-\left(y^2-4y+4\right)=9x^2-\left(y-2\right)^2=\left(3x-y+2\right)\left(3x+y-2\right)\)

\(C=-25x^2+y^2-6y+9=\left(y^2-6y+9\right)-25x^2=\left(y-3\right)^2-\left(5x\right)^2=\left(y-3-5x\right)\left(y-3+5x\right)\)\(D=x^2-4x-y^2-8y-12=\left(x^2-4x+4\right)-\left(y^2+8y+16\right)=\left(x-2\right)^2-\left(y+4\right)^2=\left(x-2-y-4\right)\left(x-2+y+4\right)=\left(x-y-6\right)\left(x+y+2\right)\)

a: Ta có: \(A=\left(6x-3y\right)+\left(4x^2-4xy+y^2\right)\)

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

\(=\left(2x-y\right)\left(2x-y+3\right)\)

b: Ta có: \(B=9x^2-\left(y^2-4y+4\right)\)

\(=9x^2-\left(y-2\right)^2\)

\(=\left(3x-y+2\right)\left(3x+y-2\right)\)

b) \(\left(4x^2+4xy+y^2\right):\left(2x+y\right)=\dfrac{\left(2x+y\right)^2}{2x+y}=2x+y\)

c) \(\left(x^2-6xy+9y^2\right):\left(3y-x\right)=\dfrac{\left(3y-x\right)^2}{3y-x}=3y-x\)

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

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