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

Đề sai nhé .Sửu lại

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

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

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

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

\(A=3x^2-14x^2+4x+3\)

Giả sử:

\(A=\left(3x+a\right)\left(x^2+bx+c\right)\)

\(=3x^3+3bx^2+3cx+ax^{2\:}+abx+ac\)

\(=3x^3+\left(3b+a\right)x^2+\left(3c+ab\right)x+ac\)

Ta có:

\(\begin{cases}3b+a=-14\\3c+ab=4\\ac=3\end{cases}\)\(\Rightarrow\begin{cases}a=1\\b=-5\\c=3\end{cases}\)

Vậy \(A=\left(3x+1\right)\left(x^2-5x+3\right)\)

x² - 4x -5

= x² - 5x + x -5

= x(x-5) + (x-5)

=(x+1)(x-5)

x^2-4x-4=x²-2.2.x-2²

           =[x-2]²

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

\(3x^2+4x-7\)

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

\(=\left(3x^2-3x\right)+\left(7x-7\right)\)

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

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

       \(3x^2+4x-7\)

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

\(=\left(3x^2-3x\right)+\left(7x-7\right)\)

\(=3x\times\left(x-1\right)+7\times\left(x-1\right)\)

\(=\left(3x+7\right)\times\left(x-1\right)\)

Đặt: x - y = a ; 3x + y - z = b ; -4x + z = c 

Ta có: a + b +  c  = x - y + 3x + y - z - 4x + z = 0 

Khi đó: \(\left(x-y\right)^3+\left(3x+y-z\right)^3+\left(-4x+z\right)^3\)

= \(a^3+b^3+c^3\)

= \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc+ac\right)+3abc\)

= \(0.\left(a^2+b^2+c^2-ab-bc+ac\right)+3abc\)

= \(3abc\)

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

cảm ơn ạ 

tôi mới học lớp 2

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

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

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

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

3x^3+x^2 -15x^2-5x+9x+3

= (3x+1)(x^2-5x+3)

\(3x^3-14x^2+4x+3\)

\(=3x^3+x^2-15x^2-5x+9x+3\)

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

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

b. 5x + 5y - 3x - 3y

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

= 2x + 2y

= 2(x + y)