\(a,4x^2-4x+1\\ =\left(2x\right)^2-2.2x+1^2=\left(2x-1\right)^2\\ c,x^2-6xy-25z^2+9y^2\\ =\left(x^2-2.x.3y+9y^2\right)-\left(5z\right)^2\\ =\left(x-3y\right)^2-\left(5z\right)^2\\ =\left(x-3y-5z\right)\left(x-3y+5z\right)\)

Xem lại đề ý b

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

\(x^2-6xy-4z^2+9y^2=\left(x-3y\right)^2-\left(2z\right)^2=\left(x-3y-2z\right)\left(x-3y+2z\right)\)

b: Ta có: \(xy-3x-2y+6\)

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

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

c: Ta có: \(x^2-6xy+9y^2-4z^2\)

\(=\left(x-3y\right)^2-4z^2\)

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

b: Ta có: \(xy-3x-2y+6\)

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

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

\(A=x^2-y^2+7x+7y\)

\(=\left(x-y\right)\left(x+y\right)+7\left(x+y\right)\)

\(=\left(x+y\right)\left(x-y+7\right)\)

\(B=4x^3-4x^2+x\)

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

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

\(C=x^2-6xy+9y^2-9\)

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

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

A=\(x^2+7x+7y-y^2=\left(x^2-y^2\right)+\left(7x+7y\right)=\left(x-y\right)\left(x+y\right)+7\left(x+y\right)=\left(x+y\right)\left(x-y+7\right)\)

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

C=\(x^2+9y^2-9-6xy=\left(x^2-6xy+9y^2\right)-9=\left(x-3y\right)^2-3^2=\left(x-3y-3\right)\left(x-3y+3\right)\)

a) = (x - 4y)(x + 1)

b) = (x - 3y)^2 - 2^2

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

c) = x^2(x + 3) - 7x(x + 3) + 9(x + 3)

= (x + 3)(x^2 - 7x + 9)

a: \(x^2-4xy+x-4y\)

\(=x\left(x-4y\right)+\left(x-4y\right)\)

\(=\left(x-4y\right)\left(x+1\right)\)

b: \(x^2-6xy+9y^2-4\)

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

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

Em xem gõ lại đề cho đúng nha

`#\text {Kr.Ryo}`

`a)`

`4x^2 - 4x + 1`

`= (2x)^2 - 2*2x*1 + 1^2`

`= (2x - 1)^2`

`b)`

Xem lại đề

`c)`

`2x^2 + 7x + 5`

`= 2x^2 + 2x + 5x + 5`

`= (2x^2 + 2x) + (5x + 5)`

`= 2x(x + 1) + 5(x + 1)`

`= (2x + 5)(x + 1)`

`d)`

`x^2 - 6xy - 25z^2 + 9y^2`

`= (x^2 - 6xy + 9y^2) - 25z^2`

`= [ (x)^2 - 2*x*3y + (3y)^2] - (5z)^2`

`= (x + 3y)^2 - (5z)^2`

`= (x + 3y - 5z)(x + 3y + 5z)`

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

đíu bt làm nka

\(x^2+x+\dfrac{1}{4}-\dfrac{1}{4}+4=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\)(vô lí)

Vậy pt vô nghiệm 

\(x^2+x+4=\left(x^2+4x+4\right)-3x=\left(x+2\right)^2-3x=\left(x+2-\sqrt{3}\right)\left(x+2+\sqrt{3}\right)\)

Giải thích: Sử dụng hằng đẳng thức 1 và 3 

⚽

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

\(x^2+5x+4\)

\(=\left(x^2+x\right)+\left(4x+4\right)\)

\(=x\left(x+1\right)+4\left(x+1\right)\)

\(=\left(x+1\right)\left(x+4\right)\)