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a,Ta đặt :
a-b-c=x ; b-c-a=y ; c-a-b=z
Ta có:
\(\text{x+y+z=a-b-c+b-c-a+c-a-b=-(a+b+c)}\)
\(\Rightarrow\left(x+y+z\right)^2=\left(a+b+c\right)^2\)
\(\Rightarrow\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(b-c-a\right)^2+\left(c-a-b\right)^2=\left(x+y+z\right)^2+x^2+y^2+z^2\)
\(\Rightarrow\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(b-c-a\right)^2+\left(c-a-b\right)^2=\left(x+y\right)^2+\left(y+z\right)^2+\left(x+z\right)^2\)\(\Rightarrow\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(b-c-a\right)^2+\left(c-a-b\right)^2=4\left(a^2+b^2+c^2\right)\)
\(\left(a+b+c+d\right)^2+\left(a+b-c-d\right)^2+\left(a-b+c-d\right)^2+\left(a-b-c+d\right)^2\)(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
\(\left(x+y\right)^2+\left(x-y\right)^2+\left(z+t\right)^2+\left(z-t\right)^2\)
\(=x^2+y^2+2xy+x^2+y^2-2xy+z^2+t^2+2zt+z^2+t^2-2zt\)
\(=2\left(x^2+y^2+z^2+t^2\right)=2\left[\left(a+b\right)^2+\left(a-b\right)^2+\left(c+d\right)^2+\left(c-d\right)^2\right]\)
\(=2(a^2+b^2-2ab+a^2+b^2-2ab+c^2+d^2+2cd+c^2+d^2-2cd)\)
\(=2\left(2a^2+2b^2+2c^2+2d^2\right)=4\left(a^2+b^2+c^2+d^2\right)\)
a. \(\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(b-c-a\right)^2+\left(c-a-b\right)^2=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2-2ab+2bc-2ac+c^2+a^2+b^2-2bc+2ac-2ab+a^2+b^2+c^2+2ab-2ac-2bc=4\left(a^2+b^2+c^2\right)\)b. Bạn làm tương tự câu a , đáp số ra : \(4\left(a^2+b^2+c^2+d^2\right)\)
a/ \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(\Rightarrow2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}\)