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1/2=1/2
1/3+1/4>1/4+1/4=1/2
1/5+…+1/8>4*1/8=1/2
1/9+…+1/16>8*1/16=1/2
1/2+1/3+1/4+…+1/16>4*1/2=2
1/2+1/3+1/4+…+1/63>1/2+1/3+1/4+…+1/16
suy ra: 1/2+1/3+1/4+…+1/63>2
Giải
1/2 + 1/3 + 1/4 + ... + 1/63 > 1/31 + 1/31 + ... + 1/31 ( 62 số hạng 1/31 )
hay 1/2 + 1/3 + 1/4 + ... + 1/63 > 62 x 1/31
nên 1/2 +1/3 + 1/4 + ... + 1/63 > 1 (dpcm)
ko biết đúng ko nữa
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Ta có:
\(A=\frac{1}{3}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{195}\)
\(\Rightarrow2A=\frac{2}{3}+\frac{2}{15}+\frac{2}{21}+...+\frac{2}{195}\)
\(=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\)
\(=\frac{1}{1}-\frac{1}{15}\)
\(=\frac{14}{15}\)
\(\Rightarrow2A=\frac{14}{15}\Rightarrow A=\frac{14}{15}\div2=\frac{7}{15}\)
Vậy A = 7/15
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a) S=1 + 2 + 2^2 + 2^3 +...+ 2^63
2S=2 + 2^2 + 2^3 + 2^4 +...+ 2^64
S=2S-S=(2 + 2^2 + 2^3 + 6^4 +...+ 2^64)-(1 + 2 + 2^2 + 2^3 +...+ 2^63)
S=2 + 2^2 + 2^3 + 2^4 +...+ 2^64 - 1 - 2 - 2^2 - 2^3 -...- 2^63
S=2^64 - 1
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a, \(5-\left(\frac{a}{b}+\frac{1}{2}\right)=2\frac{1}{3}\) => \(\frac{a}{b}+\frac{1}{2}=5-2\frac{1}{3}\) => \(\frac{a}{b}+\frac{1}{2}=\frac{8}{3}\) => \(\frac{a}{b}=\frac{8}{3}-\frac{1}{2}\) => \(\frac{a}{b}=\frac{13}{6}\)
b, \((\frac{3}{4}+2\frac{1}{2}):\frac{3}{5-3}=\left(\frac{3}{4}+\frac{5}{4}\right):\frac{3}{5}-1=\frac{9}{4}:\frac{-2}{5}=\frac{-45}{8}\)
a, 5-(\(\frac{a}{b}\)+\(\frac{1}{2}\))=2\(\frac{1}{3}\)
<=>5-\(\frac{a}{b}-\frac{1}{2}\)=\(\frac{7}{3}\)
<=>\(\frac{a}{b}=5-\frac{1}{2}-\frac{7}{3}\)
<=>\(\frac{a}{b}=\frac{13}{6}\)
b,(\(\frac{3}{4}\)+2\(\frac{1}{2}\)):\(\frac{3}{5}\)-3
=(\(\frac{3}{4}\)+\(\frac{5}{2}\)).\(\frac{5}{3}\)-3
=\(\frac{23}{4}\).\(\frac{5}{3}\)-3
=\(\frac{115}{12}\)-3
=\(\frac{115-36}{12}\)
=\(\frac{79}{12}\)
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câu nào dạng cũng giống nhau, ko biết 1 câu là ko giải đc toàn bộ
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1)\(2^3\cdot37-2^3\cdot63-10=2^3\left(37-63\right)-10=8\cdot-26-10\)=-218
2)\(2^3+2^2+2^4=2^2\left(1+2+4\right)=4\cdot7=28\)
3)\(5^3-5=5\left(5^2-1\right)=5\cdot24=120\)
4)\(3+3^2+3^4=3\left(1+3+3^3\right)=3\cdot13=39\)
5)\(x^{n+1}-x^n=x^n\left(x-1\right)\)