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c ) \(x^2-7x+12\)
\(=\left(x^2-3x\right)-\left(4x-12\right)\)
\(=x\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-4\right)\left(x-3\right)\)
d ) \(x^2+7x+12\)
\(=\left(x^2+3x\right)+\left(4x+12\right)\)
\(=x\left(x+3\right)+4\left(x+3\right)\)
\(=\left(x+4\right)\left(x+3\right)\)
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)
\(x^2-7x+12\)
\(=\left(x^2-4x\right)-\left(3x-12\right)\)
\(=x\left(x-4\right)-3\left(x-4\right)\)
\(=\left(x-4\right)\left(x-3\right)\)
(x^2+3x+2)(x^2+7x+12)
=(x2+x+2x+2)(x2+3x+4x+12)
=[x.(x+1)+2.(x+1)][x.(x+3)+4.(x+3)]
=(x+1)(x+2)(x+3)(x+4)
a) (x^2+2xy+y^2)-9=(x+y)^2-9=(x+y-3)(x+y+3)
b) 5(x^2-2xy+y^2-4z^2)=5[(x-y)^2-4z^2]=5[(x-y-2z)(x-y+2z)
c)x^2-2x-5x+10=x(x-2)-5(x-2)=(x-5)(x-2)
d)2x^2-4x-3x+6=2x(x-2)-3(x-2)=(2x-3)(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)\(x^2+6x+5=x^2+x+5x+5\)
\(=\left(x+5\right)\left(x+1\right)\)
(x^2+3x+2)(x^2+7x+12)+1
=(x2+x+2x+2)(x2+3x+4x+12)+1
=[x.(x+1)+2.(x+1)][x.(x+3)+4.(x+3)]+1
=(x+1)(x+2)(x+3)(x+4)+1
=[(x+1)(x+4)][(x+2)(x+3)]+1
=(x2+5x+4)(x2+5x+6)+1
=(x2+5x+4)[(x2+5x+4)+2]+1
=(x2+5x+4)2+2(x2+5x+4)+1
=(x2+5x+4+1)2
=(x2+5x+5)2
(x^2+3x+2)(x^2+7x+12)-24
=(x2+x+2x+2)(x2+3x+4x+12)-24
=[x.(x+1)+2.(x+1)][x.(x+3)+4.(x+3)]-24
=(x+1)(x+2)(x+3)(x+4)-24
=(x+1)(x+4)(x+2)(x+3)-24
=(x2+5x+4)(x2+5x+6)-24
Đặt t=x2+5x+4 ta được:
t.(t+2)-24
=t2+2t-24
=t2-4t+6t-24
=t.(t-4)+6.(t-4)
=(t-4)(t+6)
thay t=x2+5x+4 ta được:
(x2+5x+4-4)(x2+5x+4+6)
=(x2+5x)(x2+5x+10)
=x.(x+5)(x2+5x+10)
Vậy (x^2+3x+2)(x^2+7x+12)-24=x.(x+5)(x2+5x+10)
\(x^2+7x+12=x\left(x+3\right)+4\left(x+3\right)=\left(x+3\right)\left(x+4\right)\)
\(=x^2+3x+4x+12\)
\(=x\left(x+3\right)+4\left(x+3\right)\)
\(=\left(x+3\right)\left(x+4\right)\)