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a. = \(\left(x^3+x^2\right)+\left(7x^2+7x\right)+\left(10x+10\right)\)
= \(x^2\left(x+1\right)+7x\left(x+1\right)+10x\left(x+1\right)\)
= \(\left(x+1\right)\left(x^2+7x+10x\right)\)
= \(\left(x+1\right)\left(x+2\right)\left(x+5\right)\)
c: Ta có: \(a\left(a+2b\right)^3-b\left(2a+b\right)^3\)
\(=a^4+6a^3b+12a^2b^2+8ab^3-8a^3b-12a^2b^2-6ab^3-b^4\)
\(=a^4-2a^3b+2ab^3-b^4\)
\(=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)-2ab\left(a^2-b^2\right)\)
\(=\left(a-b\right)^3\cdot\left(a+b\right)\)
Ta có x + y = a + b
=> (x + y)2 = (a + b)2
=> x2 + y2 + 2xy = a2 + b2 + 2ab
=> xy = ab
Lại có x + y = a + b
=> (x + y)3 = (a + b)3
=> x3 + 3x2y + 3xy2 + y3 = a3 + 3a2b + 3ab2 + b3
=> x3 + y3 + 3xy(x + y) = a3 + b3 + 3ab(a + b)
=> x3 + y3 = a3 + b3 (vì x + y = a + b ; xy = ab)
Bài 2:
a) \(3x^2-7x-10=\left(x+1\right)\left(3x-10\right)\)
b) \(x^2+6x+9-4y^2=\left(x+3\right)^2-\left(2y\right)^2=\left(x+3-2y\right)\left(x+3+2y\right)\)
c) \(x^2-2xy+y^2-5x+5y=\left(x-y\right)^2-5\left(x-y\right)=\left(x-y\right)\left(x-y-5\right)\)
d) \(4x^2-y^2-6x+3y=\left(2x-y\right)\left(2x+y\right)-3\left(2x-y\right)=\left(2x-y\right)\left(2x+y-3\right)\)
e) \(1-2a+2bc+a^2-b^2-c^2=\left(a-1\right)^2-\left(b-c\right)^2=\left(a-1-b+c\right)\left(a-1+b-c\right)\)
f) \(x^3-3x^2-4x+12=\left(x+2\right)\left(x-3\right)\left(x-2\right)\)
g) \(x^4+64=\left(x^2+8\right)^2-16x^2=\left(x^2+8-4x\right)\left(x^2+6+4x\right)\)h) \(x^4-5x^2+4=\left(x+2\right)\left(x+1\right)\left(x-1\right)\left(x-2\right)\)
i) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+16=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+16=\left(x^2+8x+7\right)^2+8\left(x^2+8x+7\right)+16=\left(x^2+8x+11\right)^2\)
a: \(3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(=\left(x+1\right)\left(3x-10\right)\)
b: \(x^2+6x+9-4y^2\)
\(=\left(x+3\right)^2-4y^2\)
\(=\left(x+3-2y\right)\left(x+3+2y\right)\)
c: \(x^2-2xy+y^2-5x+5y\)
\(=\left(x-y\right)^2-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-5\right)\)
Đặt \(P=a^2+b^2+c^2+ab+bc+ca\)
\(P=\dfrac{1}{2}\left(a+b+c\right)^2+\dfrac{1}{2}\left(a^2+b^2+c^2\right)\)
\(P\ge\dfrac{1}{2}\left(a+b+c\right)^2+\dfrac{1}{6}\left(a+b+c\right)^2=6\)
Dấu "=" xảy ra khi \(a=b=c=1\)
Bài 1 :
\(=\left(x^3-x\right)-\left(6x+6\right)\)
\(=x\left(x^2-1\right)-6\left(x+1\right)\)
\(=x\left(x+1\right)\left(x-1\right)-6\left(x+1\right)\)
\(=\left(x^2-x\right)\left(x+1\right)-6\left(x+1\right)\)
\(=\left(x^2-x-6\right)\left(x+1\right)\)
Bài 2 :
a) \(x^2+y^2=\left(x+y\right)^2-2xy=9+56=65\)
b) \(x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)=-3\left(56+56\right)=-336\)
d) \(x^4+y^4=\left(x^2+y^2\right)^2-2x^2y^2=56^2-2.\left(-28\right)^2=1568\)