bài 2:

\(A=9.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)

\(A=9.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(A=9.\left(1-\dfrac{1}{100}\right)=9.\left(\dfrac{100}{100}-\dfrac{1}{100}\right)=\dfrac{891}{100}\)

bài 3:

\(=>\dfrac{x}{3}=\dfrac{5}{8}+\dfrac{1}{8}=\dfrac{8}{8}=1=\dfrac{3}{3}\)

\(=>x=3\)

A = \(\dfrac{9}{1.2}\)+ \(\dfrac{9}{2.3}\)+\(\dfrac{9}{3.4}\)+......+\(\dfrac{99}{99.100}\)

A = 9( \(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+.......+\(\dfrac{1}{99.100}\))

A = 9( 1-\(\dfrac{1}{2}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+........+\(\dfrac{1}{99}\)-\(\dfrac{1}{100}\))

A = 9 ( 1 - \(\dfrac{1}{100}\))

A = 9 . \(\dfrac{99}{100}\)

A = \(\dfrac{891}{100}\)

\(A=\dfrac{9}{1\cdot2}+\dfrac{9}{2\cdot3}+\dfrac{9}{3\cdot4}+...+\dfrac{9}{98\cdot99}+\dfrac{9}{99\cdot100}\)

\(=9\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{98\cdot99}+\dfrac{1}{99\cdot100}\right)\)

\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=9\left(1-\dfrac{1}{100}\right)\)

\(=9\left(\dfrac{100}{100}-\dfrac{1}{100}\right)\)

\(=9\cdot\dfrac{99}{100}\)

\(=\dfrac{891}{100}\)

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.`

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`

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

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

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

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)

b, \(\left(1-\dfrac{1}{100}\right)\left(1-\dfrac{1}{99}\right)...\left(1-\dfrac{1}{2}\right)=\dfrac{99.98...1}{100.99...2}=\dfrac{1}{100}\)

A = \(\dfrac{3}{4}\).\(\dfrac{8}{9}\).\(\dfrac{15}{16}.\)\(\dfrac{24}{25}\)...\(\dfrac{9800}{9801}\)

A = \(\dfrac{1.3}{2.2}\).\(\dfrac{2.4}{3.3}\).\(\dfrac{3.5}{4.4}\)...\(\dfrac{98.100}{99.99}\)

A = \(\dfrac{1}{2}.\dfrac{100}{99}\)

A = \(\dfrac{50}{99}\) 

B = \(\dfrac{1.2+2.3+3.4+...+98.99}{98.99.100}\)

Đặt tử số là C Thì 

C = 1.2 + 2.3 + 3.4 +...+ 98.99

C = \(\dfrac{1}{3}\).(1.2.3 + 2.3.3 + 3.4.3 + ...+ 98.99.3)

C = \(\dfrac{1}{3}\).[1.2.3 + 2.3.(4-1) + 3.4.(5-2) +...+ 98.99.(100-97)]

C = \(\dfrac{1}{3}\).[1.2.3 -1.2.3+2.3.4- 2.3.4 + 2.4.5 - .... - 97.98.99 + 98.99.100]

C = \(\dfrac{1}{3}\).98.99.100

B = \(\dfrac{\dfrac{1}{3}.98.99.100}{98.99.100}\) 

B = \(\dfrac{1}{3}\) = \(\dfrac{33}{99}\) < \(\dfrac{50}{99}\) = A

Vậy B < A

B1: Tính nhanh:

\(E=\dfrac{-9}{10}\cdot\dfrac{5}{14}+\dfrac{1}{10}\cdot\dfrac{-9}{2}+\dfrac{1}{7}\cdot\dfrac{-9}{10}\)

\(E=\dfrac{-9}{10}\cdot\dfrac{5}{14}+\dfrac{-9}{10}\cdot\dfrac{1}{2}+\dfrac{1}{7}\cdot\dfrac{-9}{10}\)

\(E=\dfrac{-9}{10}\cdot\left(\dfrac{5}{14}+\dfrac{1}{2}+\dfrac{1}{7}\right)\)

\(E=\dfrac{-9}{10}\cdot\left(\dfrac{5}{14}+\dfrac{7}{14}+\dfrac{2}{14}\right)\)

\(E=\dfrac{-9}{10}\cdot1=\dfrac{-9}{10}\)

B2: Chứng tỏ rằng:

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}< 1\)

Ta có: \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}\)

\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Leftrightarrow1-\dfrac{1}{100}=\dfrac{99}{100}\)

Mà \(\dfrac{99}{100}< 1\)

\(\Rightarrow\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{99\cdot100}< 1\)

Tick mình nha!

a) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{99.100}\)

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

=\(\dfrac{1}{1}+0+0+...+0-\dfrac{1}{100}\)

=\(1-\dfrac{1}{100}\)

= \(\dfrac{99}{100}\)

a)

=

=

=

=

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

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

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

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

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

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

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

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

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

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

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