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Ta có:\(5^n.2,5-30.5^n-6.5^n-1=5^n.\left(25-30-6\right)-1=5^n.\left(-11\right)-1\)-1
a) \(10^{n+1}-6.10^n\)
\(=10^n.10-6.19^n\)
\(=10^n.\left(10-6\right)\)
\(=10^n.4\)
b) \(2^{n+3}+2^{n+2}-2^{n+1}+2^n\)
\(=2^n.2^3+2^n.2^2-2^n.2+2^n.1\)
\(=2^n.\left(2^3+2^2-2+1\right)\)
\(=2^n.11\)
c) \(90.10^k-10^{k+2}+10^{k+1}\)
\(=90.10^k-10^k.10^2+10^k.10\)
\(=10^k.\left(90-10^2+10\right)\)
\(=0\)
d) \(2,5.5^{n-3}.10+5^n-6.5^{n-1}\)
\(=\dfrac{2,5.5^n.10}{5^3}+5^n-\dfrac{6.5^n}{5}\)
\(=\dfrac{5^n}{5}+5^n-\dfrac{6.5^n}{5}\)
\(=\dfrac{5^n+5^{n+1}-6.5^n}{5}=\dfrac{5^n+5^n.5-6.5^n}{5}=\dfrac{5^n\left(1+5-6\right)}{5}=\dfrac{0}{5}=0\)
\(^{10^{n+1}-6.10^n}\)=\(^{10^n-10^n-6.10^n}\)
=\(^{4.10^n}\)
=>rút gọn thành 4.10^n
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
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
\(d,2,5.5^{n-3}.2.5+5^n-6.5^{n-1}=5.5.5^{n-3}+5^n-6.5^{n-1}=5^2.5^{n-3}+5^n-6.5^{n-1}\)
\(=5^{n-3+2}+5^n-6.5^{n-1}=5^{n-1}\left(1+5-6\right)=5^{n-1}.0=0\)
a, \(10^{n+1}-6.10^n=10^n\left(10-6\right)=4.10^n\)
b. \(2^{n+3}+2^{n+2}-2^{n+1}+2^n=2^n\left(2^3+2^2-2+1\right)=2^n\left(8+4-2+1\right)=11.2^n\)
\(2,5.5^{n-3}.10+5^n-6.5^{n-1}\)
=25.\(5^n\):3+\(5^n\)\(-\)6.\(5^n\):5
=\(\dfrac{25}{3}\).\(5^n\)+\(5^n\)\(-\)\(\dfrac{6}{5}\).\(5^n\)
=\(5^n\).\(\left(\dfrac{25}{3}+1-\dfrac{6}{5}\right)\)
=\(5^n\).\(\dfrac{158}{15}\)
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) \(2,5\cdot56^n-3=\frac{5}{2}\cdot56^n-3\)
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