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\(b,3^{n+2}-2^{n+2}+3^n-2^n⋮10\)
Ta có: \(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=3^n.9-2^n.4+3^n-2^n\)
\(=3^n.10-2^n.5\)
Với: \(n\ge1\Rightarrow2^n⋮2\Rightarrow2^n.5⋮10\)
\(3^n.10⋮10\)
\(\Rightarrow3^n.10-2^n.5⋮10\)
\(\Rightarrow\)Ta có đpcm (viết ra cái đề ý)
\(d,7^6+7^5-7^4⋮11\)
Ta có: \(7^6+7^5-7^4=7^4\left(7^2+7-1\right)\)
\(=7^4\left(49+7-1\right)\)
\(=7^4.55\)
Trong tích có thừa số \(55⋮11\)
\(\Rightarrow\)Ta có đpcm (viết ra cái đề ý)
Chứng minh rằng :
\(a.\)
\(3^{n+2}-2^{n+2}+3^n-2^n⋮10\)
\(b.\)
\(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}⋮6\)
\(.a.\) \(3^{n+2}-2^{n+2}+3^n-2^n⋮10\)
Ta có : \(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=3^n.\left(3^2+2\right)-2^n.\left(2^2+1\right)\)
\(=3^n.10-2^{n-1}.2.5\)
\(=3^n.10-2^{n-1}.10\)
\(=10.\left(3^n-2^{n-1}\right)⋮10\) \(\left(dpcm\right)\)
Vậy : \(3^{n+2}-2^{n+2}+3^n-2^n⋮10\)
\(.b.\) \(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}⋮6\)
\(=3^n.30+2^n.12\)
\(=6\left(3^n.5+2^{n+1}\right)⋮6\) \(\left(dpcm\right)\)
Vậy : \(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}⋮6\)
a)\(VT=3^{n+2}-2^{n+2}+3^n-2^n\)
\(=\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)
\(=3^n\left(3^2+1\right)-2^n\left(2^2+1\right)\)
\(=3^n\cdot10-2^n\cdot5\)
\(=3^n\cdot10-2^{n-1}\cdot2\cdot5\)
\(=3^n\cdot10-2^{n-1}\cdot10\)
\(=10\cdot\left(3^n-2^{n-1}\right)⋮10\)
b)\(VT=3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}\)
\(=\left(3^{n+3}+3^{n+1}\right)+\left(2^{n+3}+2^{n+2}\right)\)
\(=3^{n+1}\left(3^2+1\right)+2^{n+2}\left(2+1\right)\)
\(=3^{n+1}\cdot10+2^{n+2}\cdot3\)
\(=3^n\cdot3\cdot2\cdot5+2^{n+1}\cdot2\cdot3\)
\(=3^n\cdot5\cdot6+2^{n+1}\cdot6\)
\(=6\cdot\left(3^n\cdot5+2^{n+1}\right)⋮6\)
\(B=\left(3^{n+3}-2^{n+3}+3^{n+1}-2^{n+1}\right)\)
\(=3^{n+1}\left(3^2+1\right)-2^{n+1}\left(2^2+1\right)\)
\(=3^{n+1}.10-2^{n+1}.5\)
\(=3^{n+1}.10+2^n.2.5\)
\(=3^{n+1}.10+2^n.10\)
\(=10\left(3^{n+1}+2^n\right)\)\(⋮\)\(10\)\(\left(đpcm\right)\)
\(Â=3^{n+3}+3^{n+1}+2^{n+3}+2^{n+1}\)
\(=3^n\left(3^3+3\right)+2^{n+1}\left(2^2+1\right)\)
\(=3^n.30+2^{n+1}.\left(2^2+2\right).\frac{1}{2}\)
\(=3^n.30+2^{n+1}.6.\frac{1}{2}\)
Mà \(3^n.30⋮6;2^{n+1}.6.\frac{1}{2}⋮6\)
\(\Rightarrow3^n.30+2^{n+1}.6.\frac{1}{2}⋮6\)
\(\Rightarrow A⋮6\left(đpcm\right)\)
a) \(3^{n+2}-2^{n+2}+3^n-2^n\)
\(\Rightarrow\left(3^n\cdot3^2+3^n\right)-\left(2^n\cdot2^2+2^n\right)\)
\(\Rightarrow3^n\left(3^2+1\right)-2^n\left(2^2+1\right)\)
\(\Rightarrow3^n\cdot10-2^n\cdot5\)
\(\Rightarrow3^n\cdot10-2^{n-1}\cdot\left(2\cdot5\right)\)
\(\Rightarrow10\left(3^n-2^n\right)\) chia hết cho 10
b) \(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}\)
\(\Rightarrow3^n\cdot3^3+3^n\cdot3+2^n\cdot2^3+2^n\cdot2^2\)
\(\Rightarrow3^n\left(3^3+3\right)+2^n\left(2^3+2^2\right)\)
\(\Rightarrow3^n\cdot30+2^n\cdot12\)
\(\Rightarrow3^n\cdot6\cdot5+2^n\cdot2\cdot6\)
\(\Rightarrow6\left(3^n\cdot5+2^n\cdot2\right)\) chia hết cho 6
Bài 2:
1: \(5^n+5^{n+2}=650\)
\(\Leftrightarrow5^n\cdot26=650\)
\(\Leftrightarrow5^n=25\)
hay x=2
2: \(32^{-n}\cdot16^n=1024\)
\(\Leftrightarrow\dfrac{1}{32^n}\cdot16^n=1024\)
\(\Leftrightarrow\left(\dfrac{1}{2}\right)^n=1024\)
hay n=-10
13: \(9\cdot27^n=3^5\)
\(\Leftrightarrow3^{3n}=3^5:3^2=3^3\)
=>3n=3
hay n=1
b: \(=3^n\cdot\left(3^2+1\right)-2^n\cdot\left(2^2+1\right)\)
\(=3^n\cdot10-2^n\cdot5\)
\(=3^n\cdot10-2^{n-1}\cdot10⋮10\)
c: \(=3^n\left(3^2+3\right)+2^n\left(2^3+2^2\right)\)
\(=3^n\cdot12+2^n\cdot12⋮6\)