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2, a-b=ab => a=ab+b => a=b(a+1)
thay a=b(a+1) vào a:b ta có: => b:b(a+1)=a+1
Theo bài ra ta có: a:b=a-b
=> a+1=a-b
=>-b=1
=> b=-1
Thay b=-1 vào a-b=ab ta có : a-(-1)=-a
=> a +1=-a
=>a=-1/2
Vậy a=-1/2. b=-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, \(D=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)
\(\Rightarrow3D=1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)
\(\Rightarrow3D-D=\left(1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)\)
\(\Rightarrow2D=1-\dfrac{1}{3^{99}}\)
\(\Rightarrow D=\dfrac{1}{2}-\dfrac{1}{3^{99}.2}< \dfrac{1}{2}\)
\(\Rightarrowđpcm\)
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}-1-\dfrac{1}{2}-...-\dfrac{1}{25}\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+...+\dfrac{1}{50}\)
\(\Rightarrowđpcm\)
Ta thấy:
\(1\cdot2^2=2^2;2\cdot3^2>3^2;3\cdot4^2>4^2;...;49\cdot50^2>50^2\)
\(\Rightarrow\dfrac{1}{1.2^2}=\dfrac{1}{2^2};\dfrac{1}{2\cdot3^2}< \dfrac{1}{3^2};\dfrac{1}{3\cdot4^2}< \dfrac{1}{4^2};...;\dfrac{1}{49\cdot50^2}< \dfrac{1}{50^2}\)
\(\Rightarrow\dfrac{1}{1\cdot2^2}+\dfrac{1}{2\cdot3^2}+\dfrac{1}{3\cdot4^2}+...+\dfrac{1}{49\cdot50^2}< \dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{50^2}\)
hay A<B
Vậy A<B
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
1. Tính:
a. \(\dfrac{\text{−1 }}{\text{4 }}+\dfrac{\text{5 }}{\text{6 }}=\dfrac{-3}{12}+\dfrac{10}{12}=\dfrac{7}{12}\)
b. \(\dfrac{\text{5 }}{\text{12 }}+\dfrac{\text{-7 }}{8}=\dfrac{10}{24}+\dfrac{-21}{24}=\dfrac{-11}{24}\)
c. \(\dfrac{-7}{6}+\dfrac{-3}{10}=\dfrac{-35}{30}+\dfrac{-9}{30}=\dfrac{-44}{30}=\dfrac{-22}{15}\)
d.\(\dfrac{-3}{7}+\dfrac{5}{6}=\dfrac{-18}{42}+\dfrac{35}{42}=\dfrac{17}{42}\)
2. Tính :
a. \(\dfrac{2}{14}-\dfrac{5}{2}=\dfrac{2}{14}-\dfrac{35}{14}=\dfrac{-33}{14}\)
b.\(\dfrac{-13}{12}-\dfrac{5}{18}=\dfrac{-39}{36}-\dfrac{10}{36}=\dfrac{49}{36}\)
c.\(\dfrac{-2}{5}-\dfrac{-3}{11}=\dfrac{-2}{5}+\dfrac{3}{11}=\dfrac{-22}{55}+\dfrac{15}{55}=\dfrac{-7}{55}\)
d. \(0,6--1\dfrac{2}{3}=\dfrac{6}{10}--\dfrac{5}{3}=\dfrac{3}{5}+\dfrac{5}{3}=\dfrac{9}{15}+\dfrac{25}{15}=\dfrac{34}{15}\)
3. Tính :
a.\(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4}{156}+\dfrac{-3}{156}=\dfrac{-7}{156}\)
b.\(\dfrac{-6}{9}-\dfrac{12}{16}=\dfrac{2}{3}-\dfrac{3}{4}=\dfrac{8}{12}-\dfrac{9}{12}=\dfrac{-17}{12}\)
c. \(\dfrac{-3}{7}-\dfrac{-2}{11}=\dfrac{-3}{7}+\dfrac{2}{11}=\dfrac{-33}{77}+\dfrac{14}{77}=\dfrac{-19}{77}\)
d.\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{1}+\dfrac{1}{10}\)
\(=\dfrac{10}{10}-\dfrac{1}{10}\)
= \(\dfrac{9}{10}\)
Chế Kazuto Kirikaya thử tham khảo thử đi !!!
Mấy câu trên kia dễ rồi mình chữa mình câu \(c\) bài \(3\) thôi nhé Kazuto Kirikaya
d) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
Lời giải:
Ta có:
\(\frac{1}{1.2^2}=\frac{1}{2^2}\)
\(2.3^2>3^2\Rightarrow \frac{1}{2.3^2}< \frac{1}{3^2}\)
\(3.4^2> 4^2\Rightarrow \frac{1}{3.4^2}< \frac{1}{4^2}\)
...........
\(49.50^2> 50^2\Rightarrow \frac{1}{49.50^2}< \frac{1}{50^2}\)
Cộng theo từng vế các BĐT:
\(\Rightarrow \frac{1}{1.2^2}+\frac{1}{2.3^2}+\frac{1}{3.4^2}+....+\frac{1}{49.50^2}< \frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{50^2}\)
\(\Leftrightarrow A< B\)
Vậy ta có đpcm.
khó nhìn lắm ạ