Phân tích đa thức thành nhân tử, các siêu sao giúp em ạ <3
a^4 + b^4 + c^4 - 2*a^2*b^2 - 2*b^2*c^2 - 2*c^2*a^2
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\(4a^2b^2-\left(a^2+b^2-c^2\right)^2\)
\(=4a^2b^2-2ab\left(a^2+b^2-c^2\right)+2ab\left(a^2+b^2-c^2\right)-\left(a^2+b^2-c^2\right)^2\)
\(=2ab\left[2ab-\left(a^2+b^2-c^2\right)\right]+\left(a^2+b^2-c^2\right)\left[2ab-\left(a^2+b^2-c^2\right)\right]\)
\(=\left(2ab+a^2+b^2-c^2\right)\left(2ab-a^2-b^2+c^2\right)\)
\(=\left(a^2+ab+ab+b^2-c^2\right)\left[c^2-\left(a^2-ab-ab+b^2\right)\right]\)
\(=\left[a\left(a+b\right)+b\left(a+b\right)-c^2\right]\left[c^2-\left(a\left(a-b\right)-b\left(a-b\right)\right)\right]\)
\(=\left[\left(a+b\right)^2-c^2\right]\left[c^2-\left(a-b\right)^2\right]\)
\(=\left[\left(a+b\right)^2-c\left(a+b\right)+c\left(a+b\right)-c^2\right]\left[c^2+c\left(a-b\right)-c\left(a-b\right)-\left(a-b\right)^2\right]\)
\(=\left[\left(a+b\right)\left(a+b-c\right)+c\left(a+b-c\right)\right]\left[c\left(c+a-b\right)-\left(a-b\right)\left(c+a-b\right)\right]\)
\(=\left(a+b+c\right)\left(a+b-c\right)\left(c+a-b\right)\left(c-a+b\right)\)
o: x^4+x^3+x^2-1
=x^3(x+1)+(x-1)(x+1)
=(x+1)(x^3+x-1)
q: \(=\left(x^3-y^3\right)+xy\left(x-y\right)\)
=(x-y)(x^2+xy+y^2)+xy(x-y)
=(x-y)(x^2+2xy+y^2)
=(x-y)(x+y)^2
s: =(2xy)^2-(x^2+y^2-1)^2
=(2xy-x^2-y^2+1)(2xy+x^2+y^2-1)
=[1-(x^2-2xy+y^2]+[(x+y)^2-1]
=(1-x+y)(1+x-y)(x+y-1)(x+y+1)
u: =(x^2-y^2)-4(x+y)
=(x+y)(x-y)-4(x+y)
=(x+y)(x-y-4)
x: =(x^3-y^3)-(3x-3y)
=(x-y)(x^2+xy+y^2)-3(x-y)
=(x-y)(x^2+xy+y^2-3)
z: =3(x-y)+(x^2-2xy+y^2)
=3(x-y)+(x-y)^2
=(x-y)(x-y+3)
o) \(x^4+x^3+x^2-1\)
\(=\left(x^4+x^3\right)+\left(x^2-1\right)\)
\(=x^3\left(x+1\right)+\left(x+1\right)\left(x-1\right)\)
\(=\left(x+1\right)\left(x^3+x-1\right)\)
q) \(x^3+x^2y-xy^2-y^3\)
\(=\left(x^3+x^2y\right)-\left(xy^2+y^3\right)\)
\(=x^2\left(x+y\right)-y^2\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-y^2\right)\)
\(=\left(x+y\right)^2\left(x-y\right)\)
s) \(4x^2y^2-\left(x^2+y^2-1\right)^2\)
\(=\left(2xy\right)^2-\left(x^2+y^2-1\right)^2\)
\(=\left(2xy-x^2-y^2+1\right)\left(2xy+x^2+y^2-1\right)\)
\(=-\left(x^2-2xy+y^2-1\right)\left(x^2+2xy+y^2-1\right)\)
\(=-\left(x-y-1\right)\left(x-y+1\right)\left(x+y+1\right)\left(x+y-1\right)\)
u) \(x^2-y^2-4x-4y\)
\(=\left(x^2-y^2\right)-\left(4x+4y\right)\)
\(=\left(x+y\right)\left(x-y\right)-4\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-4\right)\)
x) \(x^3-y^3-3x+3y\)
\(=\left(x^3-y^3\right)-\left(3x-3y\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)-3\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2-3\right)\)
z) \(3x-3y+x^2-2xy+y^2\)
\(=\left(3x-3y\right)+\left(x^2-2xy+y^2\right)\)
\(=3\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left(3+x-y\right)\)
Tham khảo:https://hoc247.net/hoi-dap/toan-8/phan-tich-da-thuc-x-7-x-2-1-thanh-nhan-tu-faq417522.html
\(=x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2+x^2-x^2+x-x+1\\ =\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^4+x^3+x^2\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)\left(x^5-x^4+x^2-x+1\right)\)
a) \(2xy-y+6x-3=\left(2xy+6x\right)-\left(y+3\right)=2x\left(y+3\right)-\left(y+3\right)=\left(2x-1\right)\left(y+3\right)\)
b) \(x^2-2xy-x+2y=\left(x^2-2xy\right)-\left(x-2y\right)=x\left(x-2y\right)-\left(x-2y\right)=\left(x-1\right)\left(x-2y\right)\)
\(64x^4+1\)
\(=64x^4+16x^2+1-16x^2\)
\(=\left(8x^2-4x+1\right)\left(8x^2+4x+1\right)\)
ap dung :(a-b-c)^2=a^2+b^2+c^2-2ab-2bc-2ca
ta dc:A=(a^2)^2+(b^2)^2+(c^2)^2-2.a^2.b^2-2.b^2-c^2-2.c^2.a^a
=>a=(a^2-b^2-c^2)^2