\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)

\(A=\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{30}+..+\dfrac{1}{9900}\)

\(A=\left(\dfrac{1}{2}+\dfrac{1}{12}\right)+\left(\dfrac{1}{30}+...+\dfrac{1}{9900}\right)\)

\(A>\dfrac{1}{2}+\dfrac{1}{12}\Rightarrow A>\dfrac{7}{12}\left(1\right)\)

\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A=\left(1-\dfrac{1}{2}+\dfrac{1}{3}\right)-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A=\dfrac{5}{6}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A< \dfrac{5}{6}\left(2\right)\)

\(\Rightarrow\dfrac{7}{12}< A< \dfrac{5}{6}\rightarrowđpcm\)

Ta có :

\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+..........+\dfrac{1}{99.100}\)

\(\Leftrightarrow A=\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{30}+............+\dfrac{1}{99.100}>\dfrac{1}{2}+\dfrac{1}{12}=\dfrac{7}{12}\)

\(\Leftrightarrow A>\dfrac{1}{12}\)\(\left(1\right)\)

Lại có :

\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...........+\dfrac{1}{99.100}\)

\(\Leftrightarrow A=\left(1-\dfrac{1}{2}+\dfrac{1}{3}\right)-\left(\dfrac{1}{4}-\dfrac{1}{5}\right)-.........-\left(\dfrac{1}{98}-\dfrac{1}{99}\right)-\dfrac{1}{100}\)

\(\Leftrightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)

\(\Leftrightarrow A< \dfrac{5}{6}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{7}{12}< A< \dfrac{5}{6}\rightarrowđpcm\)

Ta có: \(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)

\(\Rightarrow A>\dfrac{1}{1.2}+\dfrac{1}{3.4}=\dfrac{1}{2}+\dfrac{1}{12}=\dfrac{14}{24}=\dfrac{7}{12}\)\(\left(1\right)\)

Lại có: \(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\)

\(A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(A=\left(1-\dfrac{1}{2}+\dfrac{1}{3}\right)-\left(\dfrac{1}{4}-\dfrac{1}{5}\right)-\left(\dfrac{1}{6}-\dfrac{1}{7}\right)-...-\left(\dfrac{1}{98}-\dfrac{1}{99}\right)-\dfrac{1}{100}\)

\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)\(\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) \(\Rightarrow\dfrac{7}{12}< A< \dfrac{5}{6}\)

Vậy \(\dfrac{7}{12}< A< \dfrac{5}{6}\) ( Điều phải chứng minh ).

Ta có:

\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\\ A=\left(\dfrac{1}{1.2}+\dfrac{1}{3.4}\right)+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\\ A=\dfrac{7}{12}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}>\dfrac{7}{12}\left(1\right)\\ \Rightarrow A>\dfrac{7}{12}\left(1\right)\)

Ta lại có:

\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}\\ A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ A=\left(1-\dfrac{1}{2}+\dfrac{1}{3}\right)-\left(\dfrac{1}{4}-\dfrac{1}{5}\right)-...\left(\dfrac{1}{98}-\dfrac{1}{99}\right)-\dfrac{1}{100}\\ A=\dfrac{5}{6}-\left(\dfrac{1}{4}-\dfrac{1}{5}\right)-...\left(\dfrac{1}{98}-\dfrac{1}{99}\right)-\dfrac{1}{100}< \dfrac{5}{6}\\ \Rightarrow A=< \dfrac{5}{6}\left(2\right)\)

Từ (1) và (2) suy ra: \(\dfrac{7}{12}< A< \dfrac{5}{6}\left(dpcm\right)\)

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

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 : \(\dfrac{175}{300}< \dfrac{247}{300}< \dfrac{250}{300}\)

\(\Rightarrowđpcm\)

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

a) $A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}$

$=>A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}$

$=>A=(1+\dfrac{1}{3}+...+\dfrac{1}{99})-(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100})$

$=>A=(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{99}+\dfrac{1}{100})-(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{100}.2)$

$=>A=(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100})-(1+\dfrac{1}{2}+...+\dfrac{1}{50})$

$=>A=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}$

b) Ta có : $A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}$

$=>A=(1-\dfrac{1}{2}+\dfrac{1}{3})-(\dfrac{1}{4}-\dfrac{1}{5})-...-(\dfrac{1}{98}-\dfrac{1}{99})-\dfrac{1}{100}$

$=>A<1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}$

d, `3,15+2,4=5,55`

e, \(\dfrac{5}{7}.\dfrac{2}{11}+\dfrac{5}{7}.\dfrac{9}{11}=\dfrac{5}{7}\left(\dfrac{2}{11}+\dfrac{9}{11}\right)=\dfrac{5}{7}.\dfrac{11}{11}=\dfrac{5}{7}.1=\dfrac{5}{7}\)

f, `1,25.3,6+3,6.8,75=3,6(1,25+8,75)=3,6.10=36`

\(h,\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =1-\dfrac{1}{100}\\ =\dfrac{99}{100}\)

\(e\dfrac{5}{7}\times\left(\dfrac{2}{11}+\dfrac{9}{11}\right)=\dfrac{5}{7}\times1=\dfrac{5}{7}\)

\(f3.6\times\left(1.25+8.75\right)=3.6\times10=36\)

2) $\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}$

$=>\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1$

$=>\dfrac{x+4}{2000}+\dfrac{2000}{2000}+\dfrac{x+3}{2001}+\dfrac{2001}{2001}=\dfrac{x+2}{2002}+\dfrac{2002}{2002}+\dfrac{x+1}{2003}+\dfrac{2003}{2003}$

$=>\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}$

$=>\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0$

$=>(x+2004)(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}=0$

$=>x+2004=0$

$=>x=-2004$

3) Ta có : $A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{99.100}$

$=>A=\dfrac{1}{2}+\dfrac{1}{12}+...+\dfrac{1}{99.100}>\dfrac{1}{2}+\dfrac{1}{12}=\dfrac{7}{12}$

$=>A>\dfrac{7}{12}(1)$

Ta lại có : $A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}$

$=>A=(1-\dfrac{1}{2}+\dfrac{1}{3})-(\dfrac{1}{4}-\dfrac{1}{5})-...-(\dfrac{1}{98}-\dfrac{1}{99})-\dfrac{1}{100}<(1-\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}$

$=>A<\dfrac{5}{6}(2)$

Từ (1)(2) => đpcm.

Bài 4 :

\(a=0,5\)

\(b=-1\)

p /s : Không biết cách làm