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b) dễ lắm cậu tự làm nha , tách ra thành 2 vế rồi rút gọn lại
c) \(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=3^n.9-2^n.4+3^n.1-2^n.1\)
\(=3^n.\left(9+1\right)-2^n.\left(4+1\right)\)
\(=3^n.10-2^n.5\)
\(=3^n.10-2^{n-1}.2.5\)
\(=3^n.10-2^{n-1}.10\)
\(=10.\left(3^n.2^{n-1}\right)\)
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
\(A=3^{n+2}-2^{n+2}+3^n-2^n\)
=> \(A=3^n.3^2-2^n.2^2+3^n-2^n\)
=> \(A=3^n\left(3^2+1\right)-2^n.\left(2^2+1\right)\)
=> \(A=3^n.10-2^n.5\)
=> \(A=3^n.10-2^{n-1}.2.5\)
=> \(A=3^n.10-2^{n-1}.10\)
=> \(A=10\left(3^n-2^{n-1}\right)\)
Vì 10 chia hết cho 10 => \(10\left(3^n-2^{n-1}\right)\) chia hết cho 10 => A chia hết cho 10
A=3^n+2-2^n+2+3^n-2^n
A=3^n.3^2-2^n.2^2+3^n-2^n
A=3^n.10-2^n.5
A=10(3^n-2^n-1)
=> A chia hết cho 10.
Ta có:\(3^{n+2}-2^{n+2}+3^n-2^n\)
=\(\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)
=\(3^n\left(3^2+1\right)-2^n\left(2^2+1\right)\)
=\(3^n.10-2^n.5\)
=\(3^n.10-2^{n-1}.2.5\)
=\(3^n.10-2^{n-1}.10\)
=\(\left(3^n-2^{n-1}\right).10⋮10\)
\(\Rightarrow3^{n+2}-2^{n+2}+3^n-2^n⋮10\)
Nhớ tick cho mình nha!
\(3^{n+2}-2^{n+2}+3^n-2^n=3^n \left(3^2+1\right)+2^n\left(2^2+1\right)=3^n.10+2^{n-1}.10=10\left(3^n-2^{n-1}\right)\) chia hết cho 10
ta có
3^n+2=3^n.3^2
2^n-2=2^n.2^2
=>3^n.3^2 - 2^n.2^2 + 3^n-2^n=3^n.(3^2+1) - 2^n.(2^2+1)=3^n.10 - 2^n.5
Mà 3^n.10 luôn chia hết cho 10
2^n.5=10.2^n-1 luôn chia hết cho 10
=>3^n.10-2^n.5 chia hết cho 10
=>3^n+2 - 2^n+2 + 3^2 - 2^n chia hết cho 10