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24 tháng 10 2016

⇔2(a+b)2+2(c+d)2+2(a−b)2+2(d−c)2=2(2a2+2b2+2d2+2c2=4(∑a2) 

2 tháng 10 2015

 

\(A=\left[\left(a+b\right)+\left(c+d\right)\right]^2+\left[\left(a+b\right)-\left(c+d\right)\right]^2+\left[\left(a-b\right)+\left(c-d\right)\right]^2+\left[\left(a-b\right)-\left(c-d\right)\right]^2\)

Ta có

\(\left[\left(a+b\right)+\left(c+d\right)\right]^2=\left(a+b\right)^2+2\left(a+b\right)\left(c+d\right)+\left(c+d\right)^2\)

\(\left[\left(a+b\right)-\left(c+d\right)\right]^2=\left(a+b\right)^2-2\left(a+b\right)\left(c+d\right)+\left(c+d\right)^2\)

\(\left[\left(a-b\right)+\left(c-d\right)\right]^2=\left(a-b\right)^2+2\left(a-b\right)\left(c-d\right)+\left(c-d\right)^2\)

\(\left[\left(a-b\right)-\left(c-d\right)\right]^2=\left(a-b\right)^2-2\left(a-b\right)\left(c-d\right)+\left(c-d\right)^2\)

\(A=2\left(a+b\right)^2+2\left(a-b\right)^2+2\left(c+d\right)^2+2\left(c-d\right)^2\)

\(A=2\left(a^2+2ab+b^2+a^2-2ab+b^2+c^2+2cd+d^2+c^2-2cd+d^2\right)\)

\(A=4\left(a^2+b^2+c^2+d^2\right)\)

 

18 tháng 7 2019

\(\left(a+b+c+d\right)^2+\left(a+b-c-d\right)^2+\left(a-b+c-d\right)^2+\left(a-b-c+d\right)^2\)(Sửa lại nha bn viết sai để)

Đặt x=a+b , y=c+d , z=a-b , t=c-d

Khi đó biểu thức bằng

\(\left(x+y\right)^2+\left(x-y\right)^2+\left(z+t\right)^2+\left(z-t\right)^2\)

\(=x^2+y^2+2xy+x^2+y^2-2xy+z^2+t^2+2zt+z^2+t^2-2zt\)

\(=2\left(x^2+y^2+z^2+t^2\right)=2\left[\left(a+b\right)^2+\left(a-b\right)^2+\left(c+d\right)^2+\left(c-d\right)^2\right]\)

\(=2(a^2+b^2-2ab+a^2+b^2-2ab+c^2+d^2+2cd+c^2+d^2-2cd)\)

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

6 tháng 9 2021

a. A = (a + b)3 - (a - b)3

A = \(\left[\left(a+b\right)-\left(a-b\right)\right]\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)

A = (a + b - a + b)\(\left[a^2+2ab+b^2+a^2-b^2+a^2-2ab+b^2\right]\)

A = 2b(a2 + a2 + a2 + 2ab - 2ab + b2 - b2 + b2)

A = 2b(3a2 + b2)

A = 6a2b + 2b3

16 tháng 6 2016

a,Ta đặt : 

a-b-c=x ; b-c-a=y ; c-a-b=z

Ta có:

\(\text{x+y+z=a-b-c+b-c-a+c-a-b=-(a+b+c)}\)

\(\Rightarrow\left(x+y+z\right)^2=\left(a+b+c\right)^2\)

\(\Rightarrow\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(b-c-a\right)^2+\left(c-a-b\right)^2=\left(x+y+z\right)^2+x^2+y^2+z^2\)

\(\Rightarrow\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(b-c-a\right)^2+\left(c-a-b\right)^2=\left(x+y\right)^2+\left(y+z\right)^2+\left(x+z\right)^2\)\(\Rightarrow\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(b-c-a\right)^2+\left(c-a-b\right)^2=4\left(a^2+b^2+c^2\right)\)

27 tháng 6 2017

= a^8 nha

27 tháng 6 2017

bài này là làm j vậy bạn

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

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

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

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

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

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

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

\(=4a^2+4b^2+4c^2\)