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a) x3-2x2-x+2
=x(x2-1)+2(-x2+1)
=x(x2-1)-2(x2-1)
=(x2-1)(x-2)
b)
x2+6x-y2+9
=x2+6x+9-y2
=(x+3)2-y2
=(x+3-y)(x+3+y)
a) xy+3x-7y-21
=x(y+3)-7(x+3)
=(x-7)(y+3)
b)2xy-15-6x-5y
=2x(y-3)-5(-3+y)
=(2x-5)(y-3)
c)2x^2y+2xy^2-2x-2y
=2x(xy-1)+2y(xy-1)
=(2x+2y)(xy-1)
x(x+3)-5x(x-5)-5(x+3)
=(x-5)(x+3)-5x(x-5)
=(x-5)(x+3-5x)
Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn
#)Giải :
\(x^3-2x-4\)
\(=x^3+2x^2-2x^2+2x-4x-4\)
\(=x^3+2x^2+2x-2x^2-4x-4\)
\(=x\left(x^2+2x+2\right)-2\left(x^2+2x+2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
\(x^4+2x^3+5x^2+4x-12\)
\(=x^4+x^3+6x^2+x^3+x^2+6x-2x^2-2x-12\)
\(=x^2\left(x^2+x+6\right)+x\left(x^2+x+6\right)-2\left(x^2+x+6\right)\)
\(=\left(x^2+x+6\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+6\right)\left(x-1\right)\left(x+2\right)\)
Câu 1.
Đoán được nghiệm là 2.Ta giải như sau:
\(x^3-2x-4\)
\(=x^3-2x^2+2x^2-4x+2x-4\)
\(=x^2\left(x-2\right)+2x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+2\right)\)
x4 + x3 + 2x2 + x + 1
= (x4 + 2x2 + 1) + (x3 + x)
= (x2 + 1)2 + x (x2 + 1)
= (x2 + 1) ( x2 + 1 + x)
= (x2 + 1) (x + 1)2
\(-2x^4-7x^3-x^2+7x+3\)
\(=-2x^3\left(x+1\right)-5x^2\left(x+1\right)+4x\left(x+1\right)+3\left(x+1\right)\)
\(=-\left(x+1\right)\left(2x^3+5x^2-4x-3\right)\)
\(=-\left(x+1\right)\left[2x^2\left(x-1\right)+7x\left(x-1\right)+3\left(x-1\right)\right]\)
\(=-\left(x+1\right)\left(x-1\right)\left(2x^2+7x+3\right)\)
\(=-\left(x+1\right)\left(x-1\right)\left(x+3\right)\left(2x+1\right)\)
(x2+2x)2-2(x2+2x)-3
=(x2+2x)(x2+2x-2)-3
Đặt t=x2+2x ta có:
t(t-2)-3=t2-2t-3
=(t-3)(t+1)=(x2+2x-3)(x2+2x+1)
=(x-1)(x+3)(x+1)2
(x^2+2x)^2-2(x^2+2x)-3
=(x^2+2x)(x^2+2x-2)-3
=(x^2+2x)(x^2+2x-5)
\(=x^2+x-3x-3.=x\times\left(x+1\right)-3\times\left(x+1\right)=\left(x+1\right).\left(x-3\right)\)
\(x^2-2x-3\)
\(=x^2-3x+x-3\)
\(=x\left(x-3\right)+\left(x-3\right)\)
\(=\left(x-3\right)\left(x+1\right)\)
Trả lời:
\(-x^2-2x+3\)
\(=-\left(x^2+2x-3\right)\)
\(=-\left(x^2+3x-x-3\right)\)
\(=-\left[\left(x^2+3x\right)-\left(x+3\right)\right]\)
\(=-\left[x\left(x+3\right)-\left(x+3\right)\right]\)
\(=-\left(x+3\right)\left(x-1\right)\)
TL
x=-3,
x=1
HT