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10 tháng 6 2016

Từ \(x=\frac{1}{2}a+\frac{1}{2}b+\frac{1}{2}c=\frac{1}{2}.\left(a+b+c\right)\Rightarrow2x=a+b+c\)

\(M=\left(x-a\right)\left(x-b\right)+\left(x-b\right)\left(x-c\right)+\left(x-c\right)\left(x-a\right)+x^2\)

\(=x^2-xb-ax+ab+x^2-xc-bx+bc+x^2-ax-cx+ac+x^2\)

\(=4x^2-2ax-2bx-2cx+ab+bc+ac\)

\(=4x^2-2x\left(a+b+c\right)+ab+bc+ca\)

Thay 2x=a+b+c,ta đc:

\(M=4x^2-2x.2x+ab+bc+ca=4x^2-4x^2+ab+bc+ca=ab+bc+ca\)

21 tháng 2 2020

a,(a+2b+3c)^2-2(a+2b+3c)*(2a+b)+(2a +b) ^2 = (a+2b+3c-2a-b)2

=(-a+b+3c)2

b,(x-1)*(x+1 ) *(x^2+1)*(x^4+1)*(x^8+1)*(x^16+1)=(x2-1)(x2+1)(x4-1)(x8+1)(x16+1)=(x4+1)(x4-1)(x8+1)(x16+1)=(x8-1)(x8+1)(x16+1)

=(x16-1)(x16+1)=x32-1

22 tháng 2 2020

Đúng

Bài 1:

a) \(\left(a-b^2\right)\left(a+b^2\right)=a^2-b^4\)

b) \(\left(a^2+2a-3\right)\left(a^2+2a+3\right)=\left(a^2+2a\right)^2-9\)

c) \(\left(a^2+2a+3\right)\left(a^2-2a-3\right)=a^2-\left(2a+3\right)^2\)

d) \(\left(a^2-2a+3\right)\left(a^2+2a+3\right)=9-\left(a^2-2a\right)^2\)

e) \(\left(-a^2-2a+3\right)\left(-a^2-2a+3\right)=\left(-a^2-2a+3\right)^2\)

g) \(\left(a^2+2a+3\right)\left(a^2-2a+3\right)=\left(a^2+3\right)^2-4a^2\)

f) \(\left(a^2+2a\right)\left(2a-a^2\right)=4a^2-a^4\)

Bài 2 :

a) \(\left(x+1\right)\left(x^2-x+1\right)=x^3+1\)

b) \(\left(x+y+z\right)^2=\left(x+y+z\right)\left(x+y+z\right)=x^2+xy+xz+yx+y^2+yz+zx+zy+z^2=x^2+2xy+2yz+2xz+y^2+z^2\)

c) \(\left(x-y+z\right)^2=\left(x-y+z\right)\left(x-y+z\right)=x^2-xy+xz-xy+y^2-yz+xz-yz+z^2=x^2+y^2+z^2-2xy+2xz-2yz\)d) \(\left(x-2y\right)\left(x^2+2xy+4y^2\right)=\left(x-2y\right)^3\)

e) \(\left(x-y-z\right)^2=\left(x-y-z\right)\left(x-y-z\right)=x^2-xy-xz-xy+y^2+yz-xz+yz+z^2=x^2-2xy-2xz+2yz+y^2+z^2\)

a: \(=y^2-9\)

b: \(=m^3+n^3\)

c: \(=8-a^3\)

d: \(=\left(a-b-c-a+b-c\right)\left(a-b-c+a-b+c\right)\)

\(=-2c\cdot\left(2a-2b\right)\)

\(=-4ac+4bc\)

f: \(=\left(1-x^3\right)\left(1+x^3\right)=1-x^6\)