Cho \(a=b+1.\) Chứng minh: \(\left(a+b\right).\left(a^2+b^2\right).\left(a^4+b^4\right).\left(a^8+b^8\right)=a^{16}-b^{16}\)
\(a+b=1\)
\(\Rightarrow a-b=1\)
\(\Rightarrow VT=\left(a+b\right).\left(a^2+b^2\right).\left(a^4+b^4\right).\left(a^8+b^8\right)\)
\(=\left(a-b\right).\left(a+b\right).\left(a^2+b^2\right).\left(a^4+b^4\right).\left(a^8+b^8\right)\)
\(=\left(a^2-b^2\right).\left(a^2+b^2\right).\left(a^4+b^4\right).\left(a^8+b^8\right)\)
\(=\left(a^4-b^4\right).\left(a^4+b^4\right).\left(a^8+b^8\right)\)
\(=\left(a^8-b^8\right).\left(a^8+b^8\right)\)
\(=\left(a^{16}-b^{16}\right)=VP\)
\(\Rightarrow\) Đpcm.
P/s: Lần sau cậu viết đề rõ ràng ra nhé.
