Question

[I ♥ Jungkook]

Rút gọn :

\(a,A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\\ b,B=-1^2+2^2-3^2+4^2-...-99^2+100^2\\ c,C=-1^2+2^2-3^2+4^2-...+\left(-1\right)^n\cdot n^2\\ d,D=3\cdot\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\\ e,E=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\\ g,G=\left(a+b+c+d\right)^2+\left(a+b-c-d\right)^2+\left(a+c-b-d\right)^2+\left(a+d-b-c\right)^2\\ h,H=\left(a+b+c\right)^3-\left(b+c-a\right)^3-\left(a+c-b\right)^3+\left(a+b-c\right)^3\\ i,I=\left(a+b\right)^3+\left(b+c\right)^3+\left(c+a\right)^3-3\left(a+b\right)\left(c+b\right)\left(c+a\right)\)

Mọi người ơi, giúp mk vs, đc câu nào hay câu ấy ! Help me!!!!!!!!!!!!!!!!!!

Answer 1

[Aki Tsuki]

nhiều thế, đăng ít một thôi bạn

Answer 2

[Aki Tsuki]

a/ \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(\Rightarrow2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}\)