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\(a,\left(2x-3\right)n-2n\left(n+2\right)\)
\(=n\left(2x-3-2n-4\right)\)
\(=-7n\)
Vì \(-7⋮7\Rightarrow-7n⋮7\) => ĐPCM
\(b,n\left(2n-3\right)-2n\left(n+1\right)\)
\(=n\left(2n-3-2n-2\right)\)
\(=-5n⋮5\) (ĐPCM)
Rút gọn
\(a,\left(3x-5\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)
\(=6x^2+33x-10x-55-6x^2-14x-9x-21\)
\(=-76\)
\(b,\left(x+2\right)\left(2x^2-3x+4\right)-\left(x^2-1\right)\left(2x+1\right)\)
\(=2x^3-3x^2+4x+4x^2-6x+8-2x^3-x^2+2x+1\)
\(=9\)
\(c,3x^2\left(x^2+2\right)+4x\left(x^2-1\right)-\left(x^2+2x+3\right)\left(3x^2-2x+1\right)\)
\(=3x^4+6x^2+4x^3-4x-3x^4+2x^3-x^2-6x^3+4x^2-2x-9x^2+6x-3\)
= -3
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\(\Rightarrow A=2^{2n}-1=4^n-1=\left(4-1\right)\left(4^{n-1}+4^{n-2}+...+4+1\right)=3\cdot\left(4^{n-1}+4^{n-2}+...+4+1\right)⋮3\forall n\in N\)
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\(A=n^4+2n^3+2n^2+2n+1\)
\(=\left(n^2\right)^2+2.n^2\left(n+1\right)+\left(n+1\right)^2-n^2\)
\(=\left(n^2+n+1\right)^2-n^2\)
\(=\left(n^2+n+1-n\right)\left(n^2+n+1+n\right)\)
\(=\left(n^2+1\right)\left(n^2+2n+1\right)=\left(n^2+1\right)\left(n+1\right)^2\) không là số chính phương (đpcm)
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BN thử vào câu hỏi tương tự xem có k?
Nếu có thì bn xem nhé!
Nếu k thì xin lỗi đã làm phiền bn
Hội con 🐄 chúc bạn học tốt!!!
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chỉ cần CM \(Q=2^{2^n}+4^n+1⋮3\) là ok
Với n=1 thì \(Q⋮3\)
Giả sử Q vẫn chia hết cho 3 đến n=k, ta có: \(Q=2^{2^k}+4^k+1⋮3\)
Với n=k+1 thì \(Q=2^{2^k.2}+4^{k+1}+1=2^{2^k}.2^{2^k}+4^k.4+1\)
\(=\left(2^{2^k}.2^{2^k}+2^{2^k}.4^k+2^{2^k}\right)-\left(2^{2^k}.4^k+2^{2^k}-4^k.4-4\right)-3\)
\(=2^{2^k}\left(2^{2^k}+4^k+1\right)-\left(4^k+1\right)\left(2^{2^k}-4\right)-3\)
\(=2^{2^k}Q-\left(4^k+1\right)\left(4^{2^{k-1}}-1-3\right)-3⋮3\) do \(\left(4^{2^{k-1}}-1\right)⋮\left(4-1\right)=3\)
\(2^{2n+1}=2\left(4^n\right)=2\left(3+1\right)^n=2\left(BS3+1\right)=BS3+2=3k+2\)
=>\(2^{2^{2n+1}}+3=2^{3k+2}+3=4\left(8\right)^k+3=4\left(7+1\right)^k+3=4\left(BS7+1\right)+3=BS7+7\)
chia hết cho 7
=> \(A\notin P\)
Thiếu
K\(\ge1\)