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Từ a = b + 1 ta suy ra \(a-b=1\)
Do đó : \(\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\left(a^8+b^8\right)...\left(a^{32}+b^{32}\right)=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\left(a^8+b^8\right)...\left(a^{32}+b^{32}\right)=\left(a^2-b^2\right)\left(a^2+b^2\right)...\left(a^{32}+b^{32}\right)=\left(a^4-b^4\right)\left(a^4+b^4\right)...\left(a^{32}+b^{32}\right)\)
Tiếp tục thu gọn theo cách trên ta được đpcm.
Có: \(b=a-1\Rightarrow a-b=1\)
\(\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)...\left(a^{32}+b^{32}\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)...\left(a^{32}+b^{32}\right)\)
\(=\left(a^2-b^2\right)\left(a^2+b^2\right)\left(a^4+b^4\right)...\left(a^{32}+b^{32}\right)\)
\(=\left(a^{32}-b^{32}\right)\left(a^{32}+b^{32}\right)=a^{64}-b^{64}\)
Từ đầu bài
=> 1.\(\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\) \(+...+\left(a^{32}+b^{32}\right)\)= \(a^{64}-b^{64}\)
=> \(\left(a-b\right)\left(a+b\right)+...+\left(a^{32}+b^{32}\right)\)= \(a^{64}+b^{64}\)
=> \(\left(a^2-b^2\right)\left(a^2+b^2\right)+...+\left(a^{32}+b^{32}\right)\)= a^64 + b^64
tương tự sẽ ra kết quả cuối là \(\left(a^{32}-b^{32}\right)\left(a^{32}+b^{32}\right)=a^{64}-b^{64}\left(đpcm\right)\)
Có a = b+1
=> a - b =1
=> (a-b)(a+b)(a^2+b^2)(a^4+b^4)...(a^32+b^32) = (a-b)(a^64-b^64)
=> (a^2-b^2)(a^2+b^2)(a^4+b^4)...(a^32+b^32) = 1 . (a^64 - b^64)
=> (a^4-b^4)(a^4+b^4)(a^8+b^8)(a^16+b^16)(a^32+b^32) = a^64 - b^64
=> (a^8-b^8)(a^8+b^8)(a^16+b^16)(a^32+b^32) = a^64 - b^64
=> (a^16-b^16)(a^16+b^16)(a^32+b^32) = a^64 - b^64
=> (a^32-b^32)(a^32+b^32) = a^64 - b^64
=> a^64-b^64 = a^64 - b^64
=> đpcm
ta có \(a^2-b^2=\left(a+b\right)\left(a-b\right)\) => \(\frac{a^2-b^2}{a-b}=a+b\)
\(a^4-b^4=\left(a^2-b^2\right)\left(a^2+b^2\right)\)=> \(\frac{a^4-b^4}{a^2-b^2}=a^2+b^2\)
\(a^8-b^8=\left(a^4-b^4\right)\left(a^4+b^4\right)\) => \(\frac{a^8-b^8}{a^4-b^4}=a^4+b^4\)
...............................................................................................
\(a^{64}-b^{64}=\left(a^{32}-b^{32}\right)\left(a^{32}+b^{32}\right)\) => \(\frac{a^{64}-b^{64}}{a^{32}-b^{32}}=a^{32}+b^{32}\)
thay vào ta được
\(\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)......\left(a^{32}+b^{32}\right)\)
\(=\frac{a^2-b^2}{a-b}.\frac{a^4-b^4}{a^2-b^2}.\frac{a^8-b^8}{a^4-b^4}.............\frac{a ^{64}-b^{64}}{a^{32}-b^{32}}\)
\(=\frac{a^{64}-b^{64}}{a-b}\)
mà a-b= 1 nên \(\frac{a^{64}-b^{64}}{a-b}=a^{64}-b^{64}\)
