a) \(16^5+2^{15}=2^{20}+2^{15}=2^{15}\left(2^5+1\right)=2^{14}\cdot2\cdot33⋮66\)

b) \(3^{m+2}-2^{n+4}+3^m+2^n\)

\(=3^m\cdot9+3-2^n\left(2^4-1\right)\)

\(=3^m\cdot10-2^{n-1}\cdot30\)

\(=30\left(3^{m-1}-2^{n-1}\right)⋮30\)

a) \(A=16^5+2^{15}=2^{20}+2^{15}=2^{15}\left(2^5+1\right)=2^{15}\cdot33=2^{14}\cdot66⋮66\)

b) Sửa đề 

\(B=3^{n+2}-2^{n+4}+3^n+2^n=3^n\left(3^2+1\right)-2^n\left(2^4-1\right)=3^n\cdot10-2^n\cdot15\\ =3^{n-1}\cdot30-2^{n-1}\cdot30=30\left(3^{n-1}-2^{n-1}\right)⋮30\)

(với mọi n nguyên dương)

-Với n=1, ta thấy bthức đúng.

-Với n=k, có: \(\frac{1}{4+1^4}+\frac{3}{4+3^4}+...+\frac{2k-1}{4+\left(2k-1\right)^4}=\frac{k^2}{4k^2+1}=\frac{1}{4}-\frac{1}{4}.\frac{1}{4k^2+1}\)

-Giả sử bthức đúng với n=k+1, có:

\(\left(\frac{1}{4}-\frac{1}{4}.\frac{1}{4\left(k+1\right)^2+1}\right)-\left(\frac{1}{4}-\frac{1}{4}.\frac{1}{4k^2+1}\right)\)

\(=\frac{1}{4}\left(\frac{1}{4k^2+1}-\frac{1}{4\left(k+1\right)^2+1}\right)\)

\(=\frac{2k+1}{\left(4k^2+1\right)\left(4\left(k+1\right)^2+1\right)}=\frac{2k+1}{4+\left(2k+1\right)^4}\)

Vậy ta có đpcm.

Câu hỏi của luu thi thao ly - Toán lớp 8 - Học toán với OnlineMath

1+2+3+4+5+6+7+8+9=133456 hi hi

đào xuân anh sao mày gi sai hả

???????????????????
 

Lời giải:

Đặt \(A=n^4+2n^3-n^2-2n\)

\(\Leftrightarrow A=(n+2)(n^3-n)=n(n+2)(n^2-1)\)

Ta cm \(A\vdots 3\)

+) Nếu \(n\equiv 0\pmod 3\Rightarrow A\vdots 3\)

+) Nếu \(n\equiv \pm 1\pmod 3\Rightarrow n^2\equiv 1\pmod 3\Leftrightarrow n^2-1\vdots 3\)

\(\Rightarrow A\vdots 3\)

Từ hai TH trên suy ra \(A\vdots 3(1)\)

Ta cm \(A\vdots 8\)

\(A=n(n+2)(n-1)(n+1)\)

+) Nếu \(n\equiv 0\pmod 4\Rightarrow\left\{\begin{matrix} n+2\equiv 0\pmod 2\\ n\equiv 0\pmod 4\end{matrix}\right.\Rightarrow n(n+2)\vdots 8\Rightarrow A\vdots 8\)

+) Nếu \(n\equiv 1\pmod {4}\Rightarrow \left\{\begin{matrix} n-1\equiv 0\pmod 4\\ n+1\equiv 0\pmod 2\end{matrix}\right.\Rightarrow (n-1)(n+1)\vdots 8\Rightarrow A\vdots 8\)

+) Nếu \(n\equiv 2\pmod 4\Rightarrow\left\{\begin{matrix} n\equiv 0\pmod 2\\ n+2\equiv 2+2\equiv 0\pmod 4\end{matrix}\right.\Rightarrow n(n+2)\vdots 8\Rightarrow A\vdots 8\)

+) Nếu \(n\equiv 3\pmod 4\Rightarrow\left\{\begin{matrix} n-1\equiv 0\pmod 2\\ n+1\equiv 3+1\equiv 0\pmod 4\end{matrix}\right.\Rightarrow (n-1)(n+1)\vdots 8\Rightarrow A\vdots 8\)

Từ các TH trên suy ra \(A\vdots 8(2)\)

Từ \((1),(2),\text{UCLN(8,3)=1}\Rightarrow A\vdots 24\)

Ta có: \(n^4+2n^3-n^2-2n\)

\(=\left(n^4+2n^3\right)-\left(n^2+2n\right)\)

\(=n^3\left(n+2\right)-n\left(n+2\right)\)

\(=\left(n+2\right)\left(n^3-n\right)\)

=> \(n^4+2n^3-n^2-2n⋮24\)

\(n^4+2n^3+2n^2+2n+1=\left(n^4+2n^3+n^2\right)+\left(n^2+2n+1\right)=\left(n^2+1\right)\left(n+1\right)^2\)

Voi n=0 

=>n⁴+2n³+2n²+2n+1=1=1²

Em xin mạn phép sửa đề: Chứng minh với mọi số nguyên n thì A (là cái biểu thức bên trên) luôn không âm.

Ta có: \(A=n^2\left(n+1\right)^2+\left(n+1\right)^2=\left(n+1\right)^2\left(n^2+1\right)\ge0\)

Suy ra đpcm.

Gọi d là ước chung lớn nhất của 2 số. Nhiệm vụ của ta là chứng minh d=1.

a) 2n+3, n+2 \(⋮d\)

\(\Rightarrow\left(2n+3\right)-\left(n+2\right)⋮d\)

\(\Rightarrow1⋮d\)

b) n+1, 3n+4

\(\Rightarrow\left(3n+4\right)-3\left(n+1\right)⋮d\)

\(\Rightarrow1⋮d\)

c) 2n+3, 3n+4

\(\Rightarrow3\left(2n+3\right)-2\left(3n+4\right)⋮d\)

\(\Rightarrow1⋮d\)

𝓪, 𝓖𝓸̣𝓲 𝓤̛𝓒𝓛𝓝\(\left(2n+3,n+2\right)=d\)

\(\Rightarrow2n+3⋮d\)  

\(\Rightarrow n+2⋮d\Rightarrow2.\left(n+2\right)⋮d\Rightarrow2n+4⋮d\)

\(\Rightarrow2n+4-2n+3⋮d\Rightarrow1⋮d\Rightarrow d=1\)

\(\Rightarrow\)𝓤̛𝓒𝓛𝓝\(\left(2n+3,n +2\right)=1\)

𝓥𝓪̣̂𝔂 \(2n+3,n+2\) 𝓵𝓪̀ 𝓱𝓪𝓲 𝓼𝓸̂́ 𝓷𝓰𝓾𝔂𝓮̂𝓷 𝓽𝓸̂́ 𝓬𝓾̀𝓷𝓰 𝓷𝓱𝓪𝓾

𝓫, 𝓖𝓸̣𝓲 𝓤̛𝓒𝓛𝓝\(\left(n+1,3n+4\right)=d\)

\(\Rightarrow3n+4⋮d\)

\(\Rightarrow n+1⋮d\Rightarrow3\left(n+1\right)⋮d\Rightarrow3n+3⋮d\)

\(\Rightarrow3n+4-\left(3n+3\right)⋮d\Rightarrow1⋮d\Rightarrow d=1\)

\(\Rightarrow\)𝓤̛𝓒𝓛𝓝\(\left(n+1,3n+4\right)=1\)

𝓥𝓪̣̂𝔂 \(n+1,3n+4\) 𝓵𝓪̀ 𝓱𝓪𝓲 𝓼𝓸̂́ 𝓷𝓰𝓾𝔂𝓮̂𝓷 𝓽𝓸̂́ 𝓬𝓾̀𝓷𝓰 𝓷𝓱𝓪𝓾

𝓑𝓪̣𝓷 𝓸̛𝓲, 𝓬𝓱𝓸 𝓶𝓲̀𝓷𝓱 𝓼𝓾̛̉𝓪 𝓵𝓪̣𝓲 𝓸̛̉ 𝓬𝓪̂𝓾 𝓪 𝓷𝓱𝓪, 𝓬𝓱𝓸̂̃ 2𝓷+4-(2𝓷+3) 𝓹𝓱𝓪̉𝓲 𝓽𝓱𝓮̂𝓶 𝓷𝓰𝓸𝓪̣̆𝓬 𝓸̛̉ 2𝓷+3 𝓷𝓱𝓪!