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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) \(x^2-7x+10=x^2-2x-5x+10=\left(x^2-2x\right)-\left(5x-10\right)\)
\(=x\left(x-2\right)-5\left(x-2\right)=\left(x-2\right)\left(x-5\right)\)
a, x2 + 7x + 10
= x2 + 2x + 5x + 10
= x(x + 2) + 5(x + 2)
= (x + 2)(x + 5)
b, 3x2 + 12x + 9
= 3x2 + 9x + 3x + 9
= 3x(x + 3) + 3(x + 3)
= (3x + 3)(x + 3)
Dễ mà ^_^: 3x2-7x+2=3x2 -x-6x+2=(3x2-x)-(6x-2)=x(3x-1)-2(3x-1)=(3x-1)(x-2)
\(3x^3-7x^2+17x-5\)
\(=3x^3-6x^2-x^2+15x+2x-5\)
\(=\left(-6x^2+2x\right)+\left(3x^3+15x\right)-\left(x^2+5\right)\)
\(=-2x\left(3x-1\right)+3x\left(x^2+5\right)-\left(x^2+5\right)\)
\(=-2x\left(3x-1\right)+\left(x^2+5\right)\left(3x-1\right)\)
\(=\left(3x-1\right)\left(-2x+x^2+5\right)\)
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 \(M=\left(x^2+5x+5\right)^2\)
\(3x^2+7x+2\)
\(=3x^2+6x+x+2\)
\(=3x\left(x+2\right)+\left(x+2\right)\)
\(=\left(3x+1\right)\left(x+2\right)\)
\(3x^2+7x+2\)
\(=3x^2+x+6x+2\)
\(=x\left(3x+1\right)+2\left(3x+1\right)\)
\(=\left(3x+1\right).\left(x+2\right)\)
3x3 - 7x2 + 17x - 5x = (3x3 - x2) + (- 6x2 + 2x) + (15x - 5)
= (3x - 1)(x2 - 6x + 5)
\(3x^3-7x^2+17x-5.\)
\(=3x^3-x^2-6x^2+2x-15x+5\)
\(=\left(3x^3-x^2\right)-\left(6x^2-2x\right)-\left(15x-5\right)\)
\(=x^2\left(3x-1\right)-2x\left(3x-1\right)-5\left(3x-1\right)\)
\(=\left(3x-1\right)\left(x^2-2x-5\right)\)
3x2 -7x -10 = 3x2 +3x - 10x -10
= 3x(x+1) - 10(x +1)
=(x+1)(3x-10)