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a)\(2a^2-3ab+b^2\)
=\(a^2+a^2-2ab-ab+b^2\)
=\(\left(a-b\right)^2+a\left(a-b\right)\)
=\(\left(a-b\right)\left(2a-b\right)\)
b)\(x^2-7x-30\)
=\(x^2-10x+3x-30\)
=\(x\left(x-10\right)+3\left(x-10\right)\)
=\(\left(x-10\right)\left(x+3\right)\)
c)\(6a^2-5ab-6b^2\)
=\(6a^2-9ab+4ab-6b^2\)
=\(3a\left(2a-3b\right)+2b\left(2a-3b\right)\)
=\(\left(2a-3b\right)\left(3a+2b\right)\)
d)\(a^4+a^2+1\)
=\(a^4+2a^2-a^2+1\)
=\(\left(a^2+1\right)^2-a^2\)
=\(\left(a^2+1-a\right)\left(a^2+1+a\right)\)
e)\(x^3+6x^2+11x+6\)
=\(x\left(x^2+6x+9+2\right)+6\)
\(=x\left(\left(x+3\right)^2+2\right)+6\)
=\(x\left(x+3\right)^2+2x+6\)
=\(x\left(x+3\right)^2+2\left(x+3\right)\)
=\(\left(x+3\right)\left(x^2+3x+2\right)\)
a, Ta có: \(=4a^2b^2-\left(2a^2b^2-2b^2c^2-2c^2a^2+a^4+b^4+c^4\right)\)
\(=\left(2ab\right)^2-\left(a^2+b^2-c^2\right)^2\)
\(=\left(2ab-a^2-b^2+c^2\right)\left(2ab+a^2+b^2-c^2\right)\)
b, ta có: \(=\left(x^8+2x^4+1\right)-x^4\)
\(=\left(x^4+1\right)^2-\left(x^2\right)^2\)
\(=\left(x^4+1-x^2\right)\left(x^4+1+x^2\right)\)
CHÚC BẠN HỌC TỐT....
a) \(4a^3b^3c^2x+12a^3b^4c^2-16a^4b^5cx\)
\(=4a^3b^3c\left(cx+3bc-4ab^2x\right)\)
b) \(\left(b-2c\right)\left(a-b\right)-\left(a+b\right)\left(2c-b\right)\)
\(=\left(b-2c\right)\left(a-b+a+b\right)=2a\left(b-2c\right)\)
c) \(3a\left(a+5\right)-2\left(5+a\right)=\left(a+5\right)\left(3a-2\right)\)
d) \(\left(x+1\right)^2-3\left(x+1\right)=\left(x+1\right)\left(x+1-3\right)\)
a)=(a2+2ab+b2) +(b2-c2) +(ab+ac)-c2
=(a+b)2 -c2 +(b+c)(b-c) +a(b+c)
=(a+b-c)(a+b+c)+(b+c)(a+b-c)
=(a+b-c)(a+2b+2c)
c)a4+2a3+1
=a4 +a3+a3+a2-a2-a+a+1
=a3(a+1)+a2(a+1)-a(a+1)+(a+1)
=(a+1)(a3+a2-a+1)
d)x5+x+1
=(x5+x4+x3)-x4-x3-x2+x2+x+1
=x3(x2+x+1) -x2(x2+x+1) +(x2+x+1)
=(x2+x+1)((x3-x2+1)
e)x8+x4+1
=(x4)2 +2x4+1-x4
=(x4+1)2 -x4
=(x4+1+x2)(x4+1-x2)
=(x4+2x2+1-x2)(x4-x2+1)
=[(x2+1)2-x2 ](x4-x2+1)
=(x2+1-x)(x2+1 )(x4-x2+1)
\(x^8+x^4+1\)
\(=x^8+2x^4+1-x^4\)
\(=\left(x^4+1\right)^2-x^4\)
\(=\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)\)
\(=\left(x^2-x+1\right)\left(x^2+x+1\right)\left(x^4-x^2+1\right)\)