Nếu b=a-1 thì (a+b)(a^2+b^2)(a^4+b^4)...(a^32+b^32)=a^64-b^64
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Chứng minh đắng thức: Nếu a=b+1 thì: (a+b)(a^2+b^2)(a^4+b^4)(a^8+b^8)...(a^32+b^32)=a^64-b^64
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.
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Nếu b=a-1 thì (a+b)(a^2+b^2)(a^4+b^4)...(a^32+b^32)=a^64+b^64
Ai làm đk thì giúp mik vs nhé!Thank nhìu
Có: \(b=a-1\Rightarrow a-b=1\)
\(\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)...\left(a^{32}+b^{32}\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)...\left(a^{32}+b^{32}\right)\)
\(=\left(a^2-b^2\right)\left(a^2+b^2\right)\left(a^4+b^4\right)...\left(a^{32}+b^{32}\right)\)
\(=\left(a^{32}-b^{32}\right)\left(a^{32}+b^{32}\right)=a^{64}-b^{64}\)
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CMR nếu a-b=1 thì
\(\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)........\left(a^{32}+b^{32}\right)=a^{64}-b^{64}\)
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 \(\left(a^{32}-b^{32}\right)\left(a^{32}+b^{32}\right)=a^{64}-b^{64}\left(đpcm\right)\)
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CMR nếu a-b=1 thì:\(\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)...\left(a^{32}+a^{32}\right)=a^{64}-b^{64}\)
ta có \(a^2-b^2=\left(a+b\right)\left(a-b\right)\) => \(\frac{a^2-b^2}{a-b}=a+b\)
\(a^4-b^4=\left(a^2-b^2\right)\left(a^2+b^2\right)\)=> \(\frac{a^4-b^4}{a^2-b^2}=a^2+b^2\)
\(a^8-b^8=\left(a^4-b^4\right)\left(a^4+b^4\right)\) => \(\frac{a^8-b^8}{a^4-b^4}=a^4+b^4\)
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\(a^{64}-b^{64}=\left(a^{32}-b^{32}\right)\left(a^{32}+b^{32}\right)\) => \(\frac{a^{64}-b^{64}}{a^{32}-b^{32}}=a^{32}+b^{32}\)
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 \(\frac{a^{64}-b^{64}}{a-b}=a^{64}-b^{64}\)
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Giúp mình nha...
Chứng minh các đẳng thức sau:
a, Nếu a = b + 1 thì (a + b)(a^2 + b^2)(a^4 + b^4)(a^8 + b^8)...(a^32 + b^32) = a^64 - b^64
b, Nếu a = b + c thì (a^3 + b^3)/(a^3 + c^3) = (a + b)/(a + c)
Chứng minh nếu a=b+1 thì (a+b)(a2 + b2)(a4+b4)(a8+b8)(a16+b16)(a32+b32)=(a64-b64)
cho a=b+1 chứng minh (a+b)(a^2+b^2)(a^4+b^4).......(a^32+b^32)=a^64-b^64
CMR Nếu \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\) \(thì (a+b)\)\(\left(a^2+b^2\right)\left(a^4+b^4\right)\left(a^8+b^8\right)...\left(a^{32}+b^{32}\right)=a^{64}-b^{64}\)
chứng minh rằng
b= a-1 thì
S = ( a+ b)(a^2 + b^2)(a^4 + b^4) .....(a^32 + b^32)= a^64 - b^64
