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a, 7x^3 + 5 ( x - y )^2 v- 7y^3
= 7 ( x^3 - y^3 ) + 5 ( x-y )^2
= 7 ( x - y )^3 + 5 ( x-y ) ^2
= [ 7 ( x- y ) + 5 ] ( x-y) ^2
x6+3x4y2-8x3y3+3x2y4+y6= x6+3x4y2+3x2y4+y6-8x3y3=(x2+y2)3-(2xy)3
= (x2+y2-2xy)[(x2+y2)2+2xy(x2+y2)+(2xy)2]= (x-y)2(x4+6x2y2+y4+2x3y+2xy3)
(x2+y2-5)2-4x2y2-16xy-16=(x2+y2-5)2-(4x2y2+16xy+16)=(x2+y2-5)2-(2xy+4)2
=(x2+y2-5+2xy+4)(x2+y2-5-2xy-4)=(x2+2xy+y2-1)(x2-2xy+y2-9)=[(x+y)2-1][(x-y)2-32]=(x+y-1)(x+y+1)(x-y-3)(x-y+3)
x4+324=x4+36x2+324-36x2=(x2+18)2-(6x)2=(x2+18-6x)(x2+18+6x)
\(b,9x^2+90x+225-\left(x-y\right)^2\)
\(=\left(3x+15\right)^2-\left(x-y\right)^2\)
\(=\left(3x+15-x+y\right)\left(3x+15+x-y\right)\)
\(=\left(2x+y+15\right)\left(4x-y+15\right)\)
1) \(x^2-x-y^2-y=\left(x^2-y^2\right)-\left(x+y\right)=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\)
\(x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
2)\(5x-5y+ax-ay=5\left(x-y\right)+a\left(x-y\right)=\left(x-y\right)\left(a+5\right)\)
\(a^3-a^2x-ay+xy=a^2\left(a-x\right)-y\left(a-x\right)=\left(a-x\right)\left(a^2-y\right)\)
Bài 1:
a) \(x^2-y^2+10x+25\)
\(=\left(x^2+10x+25\right)-y^2\)
\(=\left(x+5\right)^2-y^2\)
\(=\left(x+y+5\right)\left(x-y+5\right)\)
b) \(x^3-x^2-5x+125\)
\(=x^3+5x^2-6x^2-30x+25x+125\)
\(=x^2\left(x+5\right)-6x\left(x+5\right)+25\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
c) \(x^4+4y^4\)
\(=\left(x^2\right)^2+2x^22y^2+\left(2y^2\right)^2-2x^22y^2\)
\(=\left(x^2+2y^2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2+2y^2-2xy\right)\left(x^2+2y^2+2xy\right)\)
d)Sửa đề \(a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)\)
\(=a\left(b^2-c^2\right)-b\left[\left(b^2-c^2\right)+\left(a^2-b^2\right)\right]+c\left(a^2-b^2\right)\)
\(=a\left(b^2-c^2\right)-b\left(b^2-c^2\right)-b\left(a^2-b^2\right)+c\left(a^2-b^2\right)\)
\(=\left(a-b\right)\left(b^2-c^2\right)-\left(b-c\right)\left(a^2-b^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(b+c\right)-\left(b-c\right)\left(a-b\right)\left(a+b\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(b+c-a-b\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
e) \(7x^2-10xy+3y^2\)
\(=\left(\sqrt{7}x\right)^2-2.\sqrt{7}x.\sqrt{3}y+\left(\sqrt{3}y\right)^2\)
\(=\left(\sqrt{7}x-\sqrt{3}y\right)^2\)
f) Sửa đề \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b\right)-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc+2ab-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ac-bc-ab\right)\)
h) \(xy\left(x+y\right)-yz\left(y+z\right)+xz\left(x-z\right)\)
\(=x^2y+xy^2-y^2z-yz^2+x^2z-xz^2\)
\(=\left(x^2y+x^2z\right)+\left(xy^2-xz^2\right)-yz\left(y+z\right)\)
\(=x^2\left(y+z\right)+x\left(y^2-z^2\right)-yz\left(y+z\right)\)
\(=x^2\left(y+z\right)+x\left(y+z\right)\left(y-z\right)-yz\left(y+z\right)\)
\(=\left(y+z\right)\left[x^2+x\left(y-z\right)-yz\right]\)
\(=\left(y+z\right)\left(x^2+xy-xz-yz\right)\)
\(=\left(y+z\right)\left[x\left(x+y\right)-z\left(x+y\right)\right]\)
\(=\left(y+z\right)\left(x+y\right)\left(x-z\right)\)
a) \(\left(x+y+z\right)\left(xy+yz+xz\right)-xyz\)
\(=\left(y+z\right)\left(xy+yz+zx\right)+x^2y+x^2z+xyz-xyz\)
\(=\left(y+z\right)\left(xy+yz+zx\right)+x^2\left(y+z\right)\)
\(=\left(y+z\right)\left(xy+yz+zx+x^2\right)\)
\(=\left(y+z\right)\left[y\left(x+z\right)+x\left(z+x\right)\right]\)
\(=\left(y+z\right)\left(x+z\right)\left(x+y\right)\)
b) \(\left(x^2+y^2+5\right)^2-4x^2y^2-16xy-16\)
\(=\left(x^2+y^2+5\right)^2-\left(4x^2y^2+16xy+16\right)\)
\(=\left(x^2+y^2+5\right)^2-\left(2xy+4\right)^2\)
\(=\left(x^2+y^2+5-2xy-4\right)\left(x^2+y^2+5+2yx+4\right)\)
\(=\left(x^2+y^2+5-2xy-4\right)\left(x^2+y^2+5+2yx+4\right)\)
c)sai đề.
đặt \(x^2+x+1=t\)
\(\Rightarrow\left(x^2+x+1\right)^2+\left(x^2+x+2\right)-12\)
\(=t^2+t+1-12\)
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mình sửa đề không biết có đúng hay không nên mình chỉ nêu hướng làm thôi. bạn thông cảm.
d) \(x^2-x-2001.2002\)
\(=x\left(x+2001\right)-2002\left(x+2001\right)\)
\(=\left(x-2002\right)\left(x+2001\right)\)