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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)
a: \(=3x^2-3y^2=3\left(x-y\right)\left(x+y\right)\)
b: \(=\left(4x^2-7x-50\right)^2-\left(16x^4+56x^3+49x^2\right)\)
\(=\left(4x^2-7x-50\right)^2-\left(4x^2+7x\right)^2\)
\(=\left(4x^2-7x-50-4x^2-7x\right)\left(4x^2-7x-50+4x^2+7x\right)\)
\(=\left(-14x-50\right)\left(8x^2-50\right)\)
\(=-4\left(7x+25\right)\left(2x-5\right)\left(2x+5\right)\)
d: \(=\left(x^2+y^2\right)^3-8x^3y^3\)
\(=\left(x^2+y^2-2xy\right)\left[x^4+2x^2y^2+y^4+2x^3y^2+2x^2y^3+4x^2y^2\right]\)
\(=\left(x-y\right)^2\cdot\left[x^4+y^4+6x^2y^2+2x^3y^2+2x^2y^3\right]\)
a: \(=3x^2-3y^2=3\left(x-y\right)\left(x+y\right)\)
c: \(=\left(x^2-y^2\right)^2-10\left(x^2-y^2\right)+25-4\left(x^2y^2+4xy+4\right)\)
\(=\left(x^2-y^2-5\right)^2-4\left(xy+2\right)^2\)
\(=\left(x^2-y^2-5-2xy-4\right)\left(x^2-y^2-5+2xy+4\right)\)
\(=\left(x^2-y^2-2xy-9\right)\left(x^2+2xy-y^2-1\right)\)
tớ lm câu a thui nha , tại khó quá ^^
a/ \(=3x^6+3x^5+6x^4+3x^3+3x^2-7x^5-7x^4-14x^3-7x^2-7x+3x^4+3x^3+6x^2+3x+1\)
\(=3x^2\left(x^4+x^3+2x^2+x+1\right)-7x\left(x^4+x^3+2x^2+x+1\right)+3\left(x^4+x^3+2x^2+x+1\right)\)
\(=\left(3x^2-7x+3\right)\left(x^4+x^3+x^2+x^2+x+1\right)\)
\(=\left(3x^2-7x+3\right)\left[x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]\)
\(=\left(3x^2-7x+3\right)\left(x^2+1\right)\left(x^2+x+1\right)\)
a.\(3x^2-11x+6\)
= \(3x^2-9x-2x+6\)
=\(3x\left(x-3\right)-2\left(x-3\right)\)
=\(\left(x-3\right)\left(3x-2\right)\)
b\(8x^2+10x-3\)
=.\(8x^2-2x+12x-3\)
=\(2x\left(4x-1\right)+3\left(4x-1\right)\)
=\(\left(4x-1\right)\left(2x+3\right)\)
d.\(x^2-y^2+10x-6y+16\)
=\(\left(x^2+10x+25\right)-\left(y^2+6y+9\right)\)
=\(\left(x+5\right)^2-\left(y+3\right)^2\)
=\(\left(x+5-y-3\right)\left(x+5+y+3\right)\)
=\(\left(x-y+2\right)\left(x+y+8\right)\)
e.\(x^4+x^2y^2+y^4\)
=\(x^4+2x^2y^2+y^4-x^2+y^2\)
=\(\left(x^2+y^2\right)^2-x^2y^2\)
=\(\left(x^2+y^2-xy\right)\left(x^2+y^2+xy\right)\)
a)
\(=3x^2-9x-2x+6=3x\left(x-3\right)-2\left(x-3\right)=\left(x-3\right)\left(3x-2\right)\)
a, \(3x^2-9x-2x+6=3x\left(x-3\right)-2\left(x-3\right)=\left(x-3\right)\left(3x-2\right)\)
b. \(8x^2-2x+12x-3=2x\left(4x-1\right)+3\left(4x-1\right)=\left(4x-1\right)\left(2x+3\right)\)
c. đề kiểu gì vậy? -2x-x để thành -3x à? xem lại đi nha
d. \(\left(x^2+10x+25\right)-\left(y^2+6y+9\right)=\left(x+5\right)^2-\left(y+3\right)^2=\left(x+5-y-3\right)\left(x+5+y+3\right)=\left(x-y+2\right)\left(x+y+8\right)\)
e. \(=x^4+2x^2y^2+y^4-x^2y^2=\left(x^2+y^2\right)^2-x^2y^2=\left(x^2+y^2-xy\right)\left(x^2+y^2+xy\right)\)
nhớ L I K E
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) \(x^2-x-2001.2002\)
\(=x\left(x+2001\right)-2002\left(x+2001\right)\)
\(=\left(x-2002\right)\left(x+2001\right)\)