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Bài 1 mik học xong quên hết òi (mấy bài kia là hok biết luôn :V)
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)
=>\(2S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\)
=>\(2S-S=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)\)
=>\(S=1-\frac{1}{2^9}=\frac{511}{512}\)
Vậy \(S=\frac{511}{512}\)
Ta có : \(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+......+\frac{1}{2^9}\)
\(\Rightarrow2S=1+\frac{1}{2}+\frac{1}{2^3}+....+\frac{1}{2^8}\)
\(\Rightarrow2S-S=1-\frac{1}{2^9}\)
\(\Leftrightarrow S=1-\frac{1}{2^9}\)
S=\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+\(\frac{1}{5^2}\)+...+\(\frac{1}{18^2}\)+\(\frac{1}{19^2}\)
S<\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+\(\frac{1}{4.5}\)+...+\(\frac{1}{17.18}\)+\(\frac{1}{18.19}\)
S<1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+\(\frac{1}{4}\)-\(\frac{1}{5}\)+...+\(\frac{1}{17}\)-\(\frac{1}{18}\)+\(\frac{1}{18}\)-\(\frac{1}{19}\)
S<1-\(\frac{1}{19}\)
\(\Rightarrow\)S<\(\frac{18}{18}\)
bài khó nhất nhé
2. Ta có :
\(P=\frac{1}{49}+\frac{2}{48}+\frac{3}{47}+...+\frac{48}{2}+\frac{49}{1}\)
cộng vào 48 phân số đầu với 1, trừ phân số cuối đi 48 ta được :
\(P=\left(\frac{1}{49}+1\right)+\left(\frac{2}{48}+1\right)+\left(\frac{3}{47}+1\right)+...+\left(\frac{48}{2}+1\right)+\left(\frac{49}{1}-48\right)\)
\(P=\frac{50}{49}+\frac{50}{48}+\frac{50}{47}+...+\frac{50}{2}+\frac{50}{50}\)
\(P=\frac{50}{50}+\frac{50}{49}+\frac{50}{48}+...+\frac{50}{2}\)
\(P=50.\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+...+\frac{1}{2}\right)\)
\(\Rightarrow\frac{S}{P}=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{48}+\frac{1}{49}+\frac{1}{50}}{50.\left(\frac{1}{50}+\frac{1}{49}+\frac{1}{48}+...+\frac{1}{2}\right)}=\frac{1}{50}\)
a, \(B=\frac{19^{31}+5}{19^{32}+5}< \frac{19^{31}+5+90}{19^{32}+5+90}=\frac{19^{31}+95}{19^{32}+95}=\frac{19\left(19^{30}+5\right)}{19\left(19^{31}+5\right)}=\frac{19^{30}+5}{19^{31}+5}=A\)
b, Ta có: \(\frac{1}{A}=\frac{2^{20}-3}{2^{18}-3}=\frac{2^2.\left(2^{18}-3\right)+9}{2^{18}-3}=4+\frac{9}{2^{18}-3}\)
\(\frac{1}{B}=\frac{2^{22}-3}{2^{20}-3}=\frac{2^2\left(2^{20}-3\right)+9}{2^{20}-3}=4+\frac{9}{2^{20}-3}\)
Vì \(\frac{9}{2^{18}-3}>\frac{9}{2^{20}-3}\)\(\Rightarrow\frac{1}{A}>\frac{1}{B}\Rightarrow A< B\)
c, Câu hỏi của truong nguyen kim
S = (1/2 + 3/2 + 5/2 + ... + 19/2) + (1+2+3+...+9)
= A + 45
A = \(\frac{1+3+5+...+19}{2}\)
Tử số có số số hạng là: (19-1)/2 + 1 = 10 số
=> A = \(\frac{\left(19+1\right)x10:2}{2}=20x5:2=50\)
=> S = 50 + 45 = 95
S= (1+2+3+4+...+9) . \(\left(\frac{1}{2}+\frac{3}{2}+\frac{5}{2}+...+\frac{19}{2}\right)\)
S=(1+9).9:2.\(\frac{1+3+5+7+...+19}{2}\)
S=45.\(\frac{\left[\left(19-1\right):2+1\right].\left(19+1\right):2}{2}\)
S=45.50=2250