\(\Rightarrow x^2+7x-8=x^2-x+8x-8\)

\(\Rightarrow x.\left(x-1\right)+8.\left(x-1\right)\)

\(\Rightarrow\left(x-1\right).\left(x+8\right)\)

= (x-1)(x+8)

a) -7x² + 5xy +12y²

=-7x²-7xy+12xy+12y²

=-7x(x+y)+12y(x+y)

=(x+y)(12y-7x)

b) x⁸ + 3x⁴ + 4

=x⁸+4x⁴+4-x⁴

=(x⁴+2)²-x⁴

=(x⁴-x²+2)(x⁴+x²+2)

=(x⁴+x²-2x²+2)(x⁴+x²+2)

=[x²(x+1)-2(x+1)](x⁴+x²+1)

=(x+1)(x²-2)(x⁴+x²+1)

\(x^8+3x^4+4\)

\(=x^8+4x^4+4-x^4\)

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

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

        \(x^4+3x^2+36\)

\(=\left(x^2\right)^2+2.x^2.6+6^2-9x^2\)

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

      \(2x^4-3x^3-7x^2+6x+8\)

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

\(=2x^3\left(x+1\right)-5x^2\left(x+1\right)-2x\left(x+1\right)+8\left(x+1\right)\)

\(=\left(x+1\right)\left(2x^3-5x^2-2x+8\right)\)

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

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

Chúc bạn học tốt.

a) \(x^2-5x+6=x^2-2x-3x+6=\left(x-2\right)\left(x-3\right)\)

b)\(3x^2+9x-30=3x^2-6x+15x-30=3\left(x-2\right)\left(x+5\right)\)

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

d)\(x^2-7x+10=x^2-2x-5x+10=\left(x-2\right)\left(x-5\right)\)

a) \(x^2-5x+6=x^2-2x-3x+6=\left(x^2-2x\right)-\left(3x-6\right)\)

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

b) \(3x^2+9x-30=3\left(x^2+3x-10\right)=3\left(x^2-2x+5x-10\right)\)

\(=3\left[\left(x^2-2x\right)+\left(5x-10\right)\right]=3\left[x\left(x-2\right)+5\left(x-2\right)\right]\)

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

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

\(=x\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x-4\right)\)

d) \(x^2-7x+10=x^2-2x-5x+10=\left(x^2-2x\right)-\left(5x-10\right)\)

\(=x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(x-5\right)\)

Ta có :  \(M=\left(x^2+3x+2\right)\left(x^2+7x+12\right)+1=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)

\(=\left[\left(x+1\right)\left(x+4\right)\right].\left[\left(x+2\right)\left(x+3\right)\right]+1=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)

Đặt \(t=x^2+5x+5\) \(\Rightarrow M=\left(t-1\right)\left(t+1\right)+1=t^2-1+1=t^2\)

Vậy \(M=\left(x^2+5x+5\right)^2\)

\(x^2-1x-6x+6=x\left(x-1\right)-6\left(x-1\right)=\left(x-6\right)\left(x-1\right)\)

\(x^2-7x+6\)

\(=x^2-x-6x+6\)

\(=x\left(x-1\right)-6\left(x-1\right)\)

\(=\left(x-1\right)\left(x-6\right)\)

 kb voi tui nha

Phân tích thành nhân tử

x² + 7x +12

=(x²+3x)+(4x+12)

=x(x+3)+4(x+3)

=(x+3)(x+4)

Bạn ghi sai đề nha phải là x² + 7x +12 ,nhớ cho mk 1 dấu đúng đấy!

x²-5x-2x+10

=x(x-5)-2(x-5)

=(x-5)(x-2)

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