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\(S=\frac{1}{3}+\frac{1}{9}+\frac{1}{18}+\frac{1}{30}+\frac{1}{45}+...+\frac{1}{14850}\)
\(\Rightarrow\frac{3}{2}S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(S=\frac{1}{3}+\frac{1}{9}+\frac{1}{30}+\frac{1}{45}+...+\frac{1}{14850}\)
\(\Rightarrow\frac{3}{2}S=\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{9900}\)
\(=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)
Vậy S = \(\frac{99}{100}:\frac{3}{2}\) = \(\frac{33}{50}\)
\(F=\left(\frac{3}{1.8}+\frac{3}{8.15}+\frac{3}{15.22}+...+\frac{3}{106.113}\right)\)\(-\)\(\left(\frac{25}{50.55}+\frac{25}{55.60}+\frac{25}{60.65}+...+\frac{25}{95.100}\right)\)
\(=\frac{3}{7}\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{15}+...+\frac{1}{106}-\frac{1}{113}\right)\) - \(5\left(\frac{1}{50}-\frac{1}{55}+\frac{1}{55}-\frac{1}{60}+...+\frac{1}{95}-\frac{1}{100}\right)\)
\(=\frac{3}{7}\left(\frac{1}{3}-\frac{1}{113}\right)-5\left(\frac{1}{50}-\frac{1}{100}\right)\)
\(=\frac{3}{7}.\frac{110}{339}-5.\frac{1}{100}\)
\(=\frac{1}{7}-\frac{1}{20}=\frac{13}{140}\)
= \(\frac{3}{7}\left(\frac{7}{1.8}+\frac{7}{8.15}+...+\frac{7}{106.103}\right)-5\left(\frac{5}{50.55}+\frac{5}{55.60}+...+\frac{5}{95.100}\right)\)
=\(\frac{3}{7}\left(1-\frac{1}{8}+\frac{1}{8}-\frac{1}{15}+...+\frac{1}{106}-\frac{1}{113}\right)-5\left(\frac{1}{50}-\frac{1}{55}+\frac{1}{55}-\frac{1}{60}+...+\frac{1}{95}-\frac{1}{100}\right)\)
=\(\frac{3}{7}\left(1-\frac{1}{113}\right)-5\left(\frac{1}{50}-\frac{1}{100}\right)\)
=\(\frac{3}{7}\cdot\frac{112}{113}-5\cdot\frac{1}{100}\)
=\(\frac{48}{113}-\frac{1}{20}\)
=\(\frac{847}{2260}\)
ta có 1/3=10/30
1/21+1/22+...+1/30 có 10 p/số
mà 1/21>1/30
1/22>1/30
....
1/29>1/30
1/30=1/30
=>1/21+..1/30>1/30+....1/30 có 10 phân số
=>1/21+...1/30>1/3
Ta có :
\(S=\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^9}\)
\(\Leftrightarrow\)\(3S=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\)
\(\Leftrightarrow\)\(3S-S=\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^8}\right)-\left(\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^9}\right)\)
\(\Leftrightarrow\)\(2S=\frac{1}{3}-\frac{1}{3^9}\)
\(\Leftrightarrow\)\(2S=\frac{3^8-1}{3^9}\)
\(\Leftrightarrow\)\(S=\frac{3^8-1}{2.3^9}\)
Ở đây mk chỉ ghi \(...\) cho nhanh nếu bạn làm vào vở thì ghi đầy đủ ra nhé
S = 1/21 + 1/22 + ... + 1/30
Số lượng số của S là :
( 30 - 21 ) : 1 + 1 = 10 ( số )
Ta có : 1/21 > 1/30 , 1/22 > 1/30 , ... 1/29 > 1/30 , 1/30 = 1/30
=> 1/21 + 1/22 + ...+ 1/30 ( 10 số ) > 1/30 + 1/30 + ...+ 1/30 ( 10 số )
=> S > 1/30 . 10
=> S > 1/3
Chúc bạn học giỏi !!!!
Ta có :
1/21 > 1/30
1/22 > 1/30
.........
1/29 > 1/30
=> S > 1/30 + 1/30 + ...... + 1/30 ( có 10 phân số 1/30 )
= 10/30 = 1/3
=>S > 1/3
Tk mk nha
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<\(\frac{18}{18}\)
b) Đặt B = A : C ta có:
\(A=\frac{5^3}{6}+\frac{5^3}{12}+\frac{5^3}{20}+\frac{5^3}{42}+\frac{5^3}{56}+\frac{5^3}{72}+\frac{5^3}{90}\)
\(A=5^3.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}\right)\)
\(A=5^3.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(A=5^3.\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(A=\frac{5^3.2}{5}\)
\(A=5^2.2\)
\(\Rightarrow A=50\)
\(C=\frac{1124.2247-1123}{1124+1123.2247}\)
\(C=\frac{\left(1123+1\right).2274-1123}{1123.2247+1124}\)
\(C=\frac{1123.2247-2247-1123}{1123.2247+1124}\)
\(C=\frac{1123.2247+1124}{1123.2247+1124}=1\)
\(\Rightarrow B=50:1=50\)
Vậy B = 50
\dfrac{1}{9}+\dfrac{1}{10}< \dfrac{1}{8}+\dfrac{1}{8}=\dfrac{1}{4}
\(\frac{1}{5}+\frac{1}{9}+\frac{1}{10}+\frac{1}{41}+\frac{1}{42}\)
\(< \frac{1}{5}+\frac{1}{8}+\frac{1}{8}+\frac{1}{40}+\frac{1}{40}\)
\(=\frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{1}{2}\)
S=\(\frac{1}{3}.\left(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+\frac{1}{21}+...+\frac{1}{4950}\right)\)
S=\(\frac{1}{3}.2\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}\right)\)
S=\(\frac{2}{3}.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
S=\(\frac{2}{3}.\left(1-\frac{1}{100}\right)=\frac{2}{3}.\frac{99}{100}=\frac{33}{50}\)
\(\frac{33}{50}>\frac{30}{50}=\frac{3}{5}->S>\frac{3}{5}\)