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Ta có:
\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{50}\right)\)
\(=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)(1)
Lại có:
\(B\)\(=\dfrac{2013}{51}+\dfrac{2013}{52}+...+\dfrac{2013}{100}\)
\(=2013\left(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\right)\)(2)
Từ (1),(2)\(\Rightarrow\dfrac{B}{A}=2013\)
\(\Rightarrow\dfrac{B}{A}\) là số nguyên
Ta có:
A\(=\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot6}+....+\dfrac{1}{99\cdot100}\)
=\(1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...\dfrac{1}{99}-\dfrac{1}{100}\)
=\(\left(1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}...\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}...\dfrac{1}{100}\right)\)
=\(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)-2\cdot\left(\dfrac{1}{2}+\dfrac{1}{4}...+\dfrac{1}{100}\right)\)
=\(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{50}\right)\)
=\(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)
Và:
B=\(\dfrac{2013}{51}+\dfrac{2013}{52}+...+\dfrac{2013}{100}\)
=\(2013\cdot\left(\dfrac{1}{51}+\dfrac{1}{52}+...\dfrac{1}{100}\right)\)
\(\Rightarrow\dfrac{B}{A}=2013\)
Vậy\(\dfrac{B}{A}\)là một số nguyên
Lời giải:
$x=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{98.99}+\frac{1}{99.100}+\frac{1}{100}$
$=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{99-98}{98.99}+\frac{100-99}{99.100}+\frac{1}{100}$
$=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}$
$=1$
`# \text {DNamNgV}`
\(x-\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}-...-\dfrac{1}{98\cdot99}=\dfrac{1}{100}+\dfrac{1}{99\cdot100}\)
\(\Rightarrow x-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}\right)=\dfrac{1}{100}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow x-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{98}-\dfrac{1}{99}\right)=\dfrac{1}{99}\)
\(\Rightarrow x-\left(1-\dfrac{1}{99}\right)=\dfrac{1}{99}\)
\(\Rightarrow x-\dfrac{98}{99}=\dfrac{1}{99}\)
\(\Rightarrow x=\dfrac{1}{99}+\dfrac{98}{99}\)
\(\Rightarrow x=\dfrac{99}{99}\)
\(\Rightarrow x=1\)
Vậy, `x = 1.`
a, \(\dfrac{1}{2!}+\dfrac{2}{3!}+...+\dfrac{99}{100!}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}< 1\)
\(\Rightarrowđpcm\)
d, \(D=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow3D=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)
\(\Rightarrow3D-D=\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)\)
\(\Rightarrow2D=1-\dfrac{1}{3^{99}}\)
\(\Rightarrow D=\dfrac{1}{2}-\dfrac{1}{3^{99}.2}< \dfrac{1}{2}\)
\(\Rightarrowđpcm\)
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}-1-\dfrac{1}{2}-...-\dfrac{1}{25}\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
\(\Rightarrowđpcm\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=1-\dfrac{1}{100}=\dfrac{99}{100}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ A=1-\dfrac{1}{100}=\dfrac{99}{100}\)
Ta có : \(B\text{=}\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+....+\dfrac{1}{99.100}\)
\(B\text{=}\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(B\text{=}\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{100}\)
\(B\text{=}\dfrac{247}{300}\)
Ta có : \(\dfrac{7}{12}\text{=}\dfrac{175}{300};\dfrac{5}{6}\text{=}\dfrac{250}{300}\)
Vì : \(\dfrac{175}{300}< \dfrac{247}{300}< \dfrac{250}{300}\)
\(\Rightarrowđpcm\)
1.
\(\dfrac{1}{2!}+\dfrac{2}{3!}+\dfrac{3}{4!}+...+\dfrac{99}{100!}\)
\(=\dfrac{2-1}{2!}+\dfrac{3-1}{3!}+\dfrac{4-1}{4!}+...+\dfrac{100-1}{100!}\)
\(=\dfrac{1}{1!}-\dfrac{1}{2!}+\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{3!}-\dfrac{1}{4!}+...+\)\(\dfrac{1}{99!}-\dfrac{1}{100!}\)
\(=1-\dfrac{1}{100!}< 1\)
2.
\(\dfrac{1.2-1}{2!}+\dfrac{2.3-1}{3!}+\dfrac{3.4-1}{4!}+...+\)\(\dfrac{1}{100!}\)
Ta có:
\(=\dfrac{1.2}{2!}-\dfrac{1}{2!}+\dfrac{2.3}{3!}-\dfrac{1}{3!}+\dfrac{3.4}{4!}-\dfrac{1}{4!}+...+\)\(\dfrac{99.100}{100!}-\dfrac{1}{100}\)
\(=\left(\dfrac{1.2}{2!}+\dfrac{2.3}{3!}+\dfrac{3.4}{4!}+...+\dfrac{99.100}{100!}\right)\)\(-\left(\dfrac{1}{2!}+\dfrac{1}{3!}+...+\dfrac{1}{100!}\right)\)
\(=\left(1+1+\dfrac{1}{2!}+...+\dfrac{1}{98!}\right)\)\(-\left(\dfrac{1}{2!}+\dfrac{1}{3!}+...+\dfrac{1}{100!}\right)\)
\(=2-\dfrac{1}{99!}-\dfrac{1}{100!}< 2\)
Ta có:
\(\dfrac{1.2-1}{2!}+\dfrac{2.3-1}{3!}+...+\dfrac{99.100-1}{100!}\)
\(=\dfrac{1.2}{2!}-\dfrac{1}{2!}+\dfrac{2.3}{3!}-\dfrac{1}{3!}+...+\dfrac{99.100}{100!}-\dfrac{1}{100!}\)
\(=\left(\dfrac{1.2}{2!}+\dfrac{2.3}{3!}+...+\dfrac{99.100}{100!}\right)-\left(\dfrac{1}{2!}+\dfrac{1}{3!}+...+\dfrac{1}{100!}\right)\)
\(=\left(1+1+\dfrac{1}{2!}+...+\dfrac{1}{98!}\right)-\left(\dfrac{1}{2!}+\dfrac{1}{3!}+...+\dfrac{1}{100!}\right)\)
\(=1+1-\dfrac{1}{99!}-\dfrac{1}{100!}\)
\(=2-\dfrac{1}{99!}-\dfrac{1}{100!}< 2\)
Vậy \(\dfrac{1.2-1}{2!}+\dfrac{2.3-1}{3!}+...+\dfrac{99.100-1}{100!}< 2\) (Đpcm)
`#3107`
`a)`
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{1999\cdot2000}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{1999}-\dfrac{1}{2000}\)
\(=1-\dfrac{1}{2000}\)
\(=\dfrac{1999}{2000}\)
`b)`
\(\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+...+\dfrac{1}{100\cdot103}?\)
\(=\dfrac{1}{3}\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+\dfrac{3}{7\cdot10}+...+\dfrac{3}{100\cdot103}\right)\)
\(=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}\cdot\left(1-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}\cdot\dfrac{102}{103}\)
\(=\dfrac{34}{103}\)
`c)`
\(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-....-\dfrac{1}{6}-\dfrac{1}{2}\)
\(=\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)\)
\(=\dfrac{8}{9}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\right)\)
\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
\(=\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)\)
\(=\dfrac{8}{9}-\dfrac{8}{9}\\ =0\)
b) Sửa đề:
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{100.103}\)
\(=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}.\left(1-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}.\left(\dfrac{103}{103}-\dfrac{1}{103}\right)\)
\(=\dfrac{1}{3}.\dfrac{102}{103}\)
\(=\dfrac{34}{103}\)
A= \(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{99\cdot100}\)
= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
= \(1-\dfrac{1}{100}\)
= \(\dfrac{99}{100}\)
Ta có :
\(\dfrac{7}{12}=\dfrac{7\cdot50}{12\cdot50}=\dfrac{350}{600}\)
\(\dfrac{99}{100}=\dfrac{99\cdot6}{100\cdot6}=\dfrac{594}{600}\)
\(\dfrac{5}{6}=\dfrac{5\cdot100}{6\cdot100}=\dfrac{500}{600}\)
Chỗ này hình như bn viết sai đề nha
QĐMS lên phải là \(\dfrac{7}{12}< \dfrac{5}{6}< A\) chứ
Bn xem lại đề ik nha ☺
THANKSSSSSSSSSSSS
\(P=\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}{\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}}\\ \Rightarrow P=\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}{\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}}\\ \)
\(\Rightarrow P=\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}{\left(1+\dfrac{1}{3}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ \Rightarrow P=\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}{\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\)
\(\Rightarrow P=\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}{\left(1+\dfrac{1}{2}+...+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{50}\right)}\\ \Rightarrow P=\dfrac{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}{\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}}\\ \Rightarrow P=1\)