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(1981 x 1982 - 990) : (1980 x 1982 + 992)
=(1980 x 1982+1982 -990) : (1980 x 1982 +992)
=(1980 x 1982 + 992) : ( 1980 x 1982 + 992)
=1
B=[(45.79+45.21)]:90-5^2]:5+2^3 B=[(45.79+45.21):90-25]:5+8 B=[(45.(79+21):65]:13 B=[(45.100):65]:13 B=[4500:65]:13 B=4500:65:13
Bài 1: a) \(M=1+5+5^2+...+5^{100}\)
\(5M=5+5^2+5^3+...+5^{101}\)
\(5M-M=\left(5+5^2+5^3+...+5^{101}\right)-\left(1+5+5^2+...+5^{100}\right)\)
\(4M=5^{101}-1\)
\(M=\frac{5^{101}-1}{4}\)
b) \(N=2+2^2+...+2^{100}\)
\(2N=2^2+2^3+...+2^{101}\)
\(2N-N=\left(2^2+2^3+...+2^{101}\right)-\left(2+2^2+...+2^{100}\right)\)
\(N=2^{101}-2\)
Bài 2:
a) \(16^{32}=\left(2^4\right)^{32}=2^{128}\)
\(32^{16}=\left(2^5\right)^{16}=2^{80}\)
Vì \(2^{128}>2^{80}\Rightarrow16^{32}>32^{16}\)
B=1/3+1/3^2+1/3^3+...+1/3^8
=1/3+1/3.1/3+1/3.1/3.1.3+...+1/3
=1/3(1+2.1/3+3.1/3+4.1/3+5.1/3+6.1/3+7.1/3+8.1/3)
=1/3[1+1/3(2+3+4+5+6+7+8)]
=1/3[1+1/3.35]
=1/3[1+35/3]
=1/3.38/3
=38/9
\(B=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\)
\(\Rightarrow3B=1+\frac{1}{3}+...+\frac{1}{3^7}\)
\(\Rightarrow3B-B=\left(1+\frac{1}{3}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)\)
\(\Rightarrow2B=1-\frac{1}{3^8}\)
\(\Rightarrow B=\frac{1-\frac{1}{3^8}}{2}\)
\(\Rightarrow B=\frac{3280}{6561}\)