Phân tích đa thức sau thành nhân tử:
a2(b-2c)+b2(c-a)+2c2(a-b)+abc
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\(=\left[a^2+b^2+2-2\left(ab-1\right)\right]\left[a^2+b^2+2+2\left(ab-1\right)\right]\\ =\left(a^2+b^2-2ab+4\right)\left(a^2+b^2+2ab\right)\\ =\left(a+b\right)^2\left(a^2+b^2-2ab+4\right)\)
\(abc-\left(ab+bc+ac\right)+\left(a+b+c\right)-1=\left(abc-ab\right)-\left(bc-b\right)-\left(ac-a\right)+\left(c-1\right)=ab\left(c-1\right)-b\left(c-1\right)-a\left(c-1\right)+\left(c-1\right)=\left(c-1\right)\left(ab-b-a+1\right)=\left(c-1\right)\left[b\left(a-1\right)-\left(a-1\right)\right]=\left(a-1\right)\left(b-1\right)\left(c-1\right)\)
(a+b)3+(a-b)3=(a3+3a2b+3ab2+b3)+(a3-3a2b+3ab2-b3)
=a6+6a2b4
b) \(a^6-b^3\)
\(=\left(a^2\right)^3-b^3\)
\(=\left(a^2-b\right)\left(a^4+a^2b+b^2\right)\)
c) \(x^4-1\)
\(=\left(x^2\right)^2-1^2\)
\(=\left(x^2-1\right)\left(x^2+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\)
\(=4ab\left(ab+ax+bx+x^2\right)=4a^2b^2+4a^2bx+4ab^2x+4abx^2\)
\(\left(ab-1\right)^2+\left(a+b\right)^2=a^2b^2-2ab+1+a^2+2ab+b^2=a^2+b^2+a^2b^2+1=a^2\left(b^2+1\right)+\left(b^2+1\right)=\left(a^2+1\right)\left(b^2+1\right)\)
\(B=a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)\)
\(B=ab^2-ac^2+bc^2-a^2b+a^2c-b^2c\)
\(B=\left(ab^2-a^2b\right)-\left(ac^2-c^2b\right)+\left(a^2c-b^2c\right)\)
\(B=-ab\left(a-b\right)-c^2\left(a-b\right)+c\left(a-b\right)\left(a+b\right)\)
\(B=\left(a-b\right)\left(-ab-c^2+ac+bc\right)\)
\(B=\left(a-b\right)\left[a\left(c-b\right)-c\left(c-b\right)\right]\)
\(B=\left(a-b\right)\left(c-b\right)\left(a-c\right)\)
\(x^2-xy\left(a+b\right)+aby^2=x^2-xya-xyb+aby^2=x\left(x-ya\right)-yb\left(x-ya\right)=\left(x-ya\right)\left(x-yb\right)\)
\(x^2-xy\left(a+b\right)+aby^2\)
\(=x^2-axy-bxy+aby^2\)
\(=x\left(x-ay\right)-by\left(x-ay\right)\)
\(=\left(x-ay\right)\left(x-by\right)\)
Lời giải:
\(a^2(b-2c)+b^2(c-a)+2c^2(a-b)+abc\)
\(=a^2(b-c)+b^2(c-a)+c^2(a-b)-a^2c+c^2(a-b)+abc\)
\(=a^2(b-c)+b^2(c-a)+c^2(a-b)+c^2(a-b)-ac(a-b)\)
\(=a^2(b-c)-b^2[(b-c)+(a-b)]+c^2(a-b)+(a-b)(c^2-ac)\)
\(=(a^2-b^2)(b-c)-(b^2-c^2)(a-b)+c(a-b)(c-a)\)
\(=(a-b)(b-c)(a+b-b-c)+c(a-b)(c-a)\)
\(=(a-b)(b-c)(a-c)-c(a-b)(a-c)\)
\(=(a-b)(a-c)(b-2c)\)