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\(a.\left(b^2+c^2+bc\right)+b.\left(c^2+a^2+ac\right)+c.\left(a^2+b^2+ab\right)\)
\(=ab^2+ac^2+abc+bc^2+ba^2+bac+ca^2+cb^2+cab\)
\(=\left(ab^2+ba^2+abc\right)+\left(ac^2+ca^2+bac\right)+\left(bc^2+cb^2+cab\right)\)
\(=ab.\left(b+a+c\right)+ac.\left(c+a+b\right)+bc.\left(c+b+a\right)\)
\(=\left(a+b+c\right).\left(ab+ac+bc\right)\)
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ta có: \(\left(a+b+c\right)^2+\left(a+b-c\right)^2-4c^2=\left(a+b+c\right)^2+\left(a+b-c-2c\right)\left(a+b-c+2c\right).\)
\(=\left(a+b+c\right)^2+\left(a+b-3c\right)\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a+b+c+a+b-3c\right)\)
\(=2\left(a+b+c\right)\left(a+b-c\right)\)
(a+b+c)^2+(a+b-c)^2-4c^2
=(a^2+b^2+c^2+2ab+2bc+2ac)+(a^2-2ab+b^2-2ac+c^2-abc)-4c^2
=a^2+b^2+c^2+2ab+2bc+2ac+a^2-2ab+b^2-2ac+c^2-abc-4c^2
=(a^2+a^2)+(b^2+b^2)+(c^2+c^2)+(2ab-2ab)+(2bc-2bc)+(2ac-2ac)-4c^2
=2a^2+2b^2+2c^2-4c^2
=(2a^2+2b^2)+(2c^2-4c^2)
=2*(a^2+b^2)+2c^2*(1-2)
Câu a) dễ, ko làm
b) \(x^2y^2+1-x^2-y^2\)
\(=x^2\left(y^2-1\right)-\left(y^2-1\right)\)
\(=\left(x^2-1\right)\left(y^2-1\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(y+1\right)\left(y-1\right)\)
Câu c) đề sai
Câu c) ,đề đúng nek
\(bc\left(b+c\right)+ac\left(c-a\right)-ab\left(a+b\right)\)
\(=bc\left(b+c\right)+ac\left[\left(b+c\right)-\left(a+b\right)\right]-ab\left(a+b\right)\)
\(=bc\left(b+c\right)+ac\left(b+c\right)-ac\left(a+b\right)-ab\left(a+b\right)\)
\(=\left(b+c\right)\left(bc+ac\right)-\left(a+b\right)\left(ac+ab\right)\)
\(=\left(b+c\right)c\left(a+b\right)-\left(a+b\right)a\left(b+c\right)\)
\(=\left(b+c\right)\left(a+b\right)\left(c-a\right)\)
mình chỉ biết làm câu b thôi
x4 - 2x3 + 2x - 1
= (x4 - 1) - (2x3 - 2x)
= (x2 - 1) (x2 + 1) - 2x (x2 - 1)
(x2 - 1) (x2 + 1 - 2x)
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2+xy+yz+zx\right)\)
\(\left(a+b+c\right)^2-\left(ab+ac\right)\left(a+b+c\right)-a-b-c\)
\(\left(a+b+c\right)^2-\left(ab+ac\right)\left(a+b+c\right)-\left(a+b+c\right)\)
\(\left(a+b+c\right)\left(a+b+c-ab-ac-1\right)\)