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a)x2-5x+6=(x2-2x)-(3x-6)=x(x-2)-3(x-2)(=(x-2)(x-3)
b)x2+5x+6=(x2+2x)+(3x+6)=x(x+2)+3(x+2)=(x+2)(x+3)
c)x2-7x+12=(x2-3x)-(4x-12)=x(x-3)-4(x-3)=(x-3)(x-4)
d)x2+7x+12=(x2+3x)+(4x+12)=x(x+3)+4(x+3)=(x+3)(x+4)
a, x^2 + 5x +4
= x^2 + 1x + 4x + 4
= (x^2 + 1x) + (4x + 4)
= x ( x + 1 ) + 4 ( x + 1 )
= (x + 1) (x + 4)
b, x^2 - 6x + 5
= x^2 - 1x - 5x + 5
= (x^2 - 1x) - (5x - 5)
= x (x - 1) - 5 (x - 1)
= (x - 1) (x - 5)
c, x^2 + 7x + 12
= x^2 + 3x + 4x + 12
= (x^2 + 3x) + (4x + 12)
= x (x + 3) + 4 (x + 3)
= (x + 3) (x + 4)
d, 2x^2 - 5x + 3
= 2^x2 - 2x - 3x + 3
= 2x (x - 1) - 3 (x - 1)
= (x-1) (2x - 3)
e, 7x - 3x^2 - 4
= 3x + 4x - 3x^2 - 4
= (3x - 3x^2) + (4x - 4)
= 3x (1 - x) + 4 (x - 1)
= 3x (1-x) - 4 (1 - x)
= (1 - x) (3x - 4)
f, x^2 - 10x + 16
= x^2 - 2x - 8x + 16
= (x^2 - 2x) - (8x - 16)
= x (x - 2) - 8 (x - 2)
= (x - 2) (x - 8)
a, (x+1)(x+4)
b,(x-5)(x-1)
c,(x+3)(x+4)
d,(2x-3)(x-1)
e,(-3x+4)(x-1)
f, (x-8)(x-2)
f)\(x^2-5x-14=x^2-7x+2x-14=x\left(x-7\right)+2\left(x-7\right)=\left(x-7\right)\left(x+2\right)\)
i)\(x^2-7x+10=x^2-2x-5x+10=x\left(x-2\right)-5\left(x-2\right)=\left(x-5\right)\left(x-2\right)\)
h)\(x^2-7x+12=x^2-3x-4x+12=x\left(x-3\right)-4\left(x-3\right)=\left(x-4\right)\left(x-3\right)\)
g)\(x^2+6x+5=x^2+x+5x+5=x\left(x+1\right)+5\left(x+1\right)=\left(x+1\right)\left(x+5\right)\)
f)\(x^2-5x-14=x^2-7x+2x-14\)
\(=\left(x+2\right)\left(x-7\right)\)
i)\(x^2-7x+10=x^2-5x-2x+10\)
\(=\left(x-2\right)\left(x-5\right)\)
h)\(x^2-7x+12=x^2-4x-3x+12\)
\(=\left(x-3\right)\left(x-4\right)\)
g)\(x^2+6x+5=x^2+x+5x+5\)
\(=\left(x+5\right)\left(x+1\right)\)
a: =(x+6)(x-1)
n: \(=4x^4+36x^2+81-36x^2\)
\(=\left(2x^2+9-6x\right)\left(2x^2+9+6x\right)\)
\(x^2-7x+12\)
\(=\left(x^2-4x\right)-\left(3x-12\right)\)
\(=x\left(x-4\right)-3\left(x-4\right)\)
\(=\left(x-4\right)\left(x-3\right)\)
Vô đây xem: bài 1:phân tích đa thức thành nhân tửa)7x^3y-14x^2y+7xy^3b)3x^2-3xy-5x+5yc)x^2+7x+12giúp mình với - Hoc24