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a) \(10^n+1-6\cdot10^n=\left(1-6\right)10^n+1=-5\cdot10^n+1\)
b) \(90\cdot10^n-10^2-2+10^n+1=\left(90-1+1\right)\cdot10^n-2+1=90\cdot10^n-1\)
c) \(2,5\cdot56^n-3=\frac{5}{2}\cdot56^n-3\)
a) 2^n (2^3 + 2^2 -2^1+1)=2^n(8+4-2+1)
=2^n * 11
b)10^n ( 90 -10^2 + 10 )=10^N * 0
= 0
\(d,2,5.5^{n-3}.2.5+5^n-6.5^{n-1}=5.5.5^{n-3}+5^n-6.5^{n-1}=5^2.5^{n-3}+5^n-6.5^{n-1}\)
\(=5^{n-3+2}+5^n-6.5^{n-1}=5^{n-1}\left(1+5-6\right)=5^{n-1}.0=0\)
a, \(10^{n+1}-6.10^n=10^n\left(10-6\right)=4.10^n\)
b. \(2^{n+3}+2^{n+2}-2^{n+1}+2^n=2^n\left(2^3+2^2-2+1\right)=2^n\left(8+4-2+1\right)=11.2^n\)
a) \(3^n+3^{n+2}=3^n.\left(1+3^2\right)=3^n.\left(1+9\right)=10.3^n\)
b) \(1,5.2^n-2^{n-1}=1,5.2^{1+n-1}-2^{n-1}=1,5.2.2^{n-1}-2^{n-1}\)
\(=3.2^{n-1}-2^{n-1}=2^{n-1}.\left(3-1\right)=2^{n-1}.2=2^n\)
a: \(10^{n+1}=10^n\cdot10\)
b: \(2^{n+3}+2^{n+1}-2^{n+1}+2^n\)
\(=2^n\cdot8+2^n=9\cdot2^n\)
c: \(90\cdot10^k-10^{k+2}+10^{k+1}\)
\(=90\cdot10^k+10^k\cdot10-10^k\cdot100=0\)
a)10^n+1-10^n b)2^n+3+2^n+2+2^n
=(10^n-10^n)+1 =3.(2^n)+(3+2+0)
=0+1=1 =6^3n+5
vớ vẩn,đây là bài toán lớp 6 chứ lớp 7 đâu ra