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24 tháng 4 2023

      (92.3+93.4+...+9199.200).�=25

⇒[9.(12-13+13-14+...+1199-1200)]�=25

⇒[9.(12-1200)]�=25

⇒(9.99200)�=25

⇒  891200�=25

⇒        �=25:891200

⇒        �=80891

Vậy, 

5 tháng 5 2022

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:

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

\(=>x=3\)

13 tháng 2 2023

\(A=\dfrac{9}{1.2}+\dfrac{9}{2.3}+\dfrac{9}{3.4}+...+\dfrac{9}{98.99}+\dfrac{9}{99.100}\)
\(=9\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\right)\)
\(=9\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=9\left(1-\dfrac{1}{100}\right)\)
\(=9.\dfrac{99}{100}\)
\(=\dfrac{891}{100}\)

13 tháng 2 2023

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

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

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

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

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

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

10 tháng 6 2017

1)Tính

a)\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+..........+\dfrac{1}{9.10}\)

=\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=1-\dfrac{1}{10}\)

\(=\dfrac{9}{10}\)

b)\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.........+\dfrac{1}{99.100}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+..............+\dfrac{1}{99}-\dfrac{1}{100}\)

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

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

2) tìm x

\(a\)) \(\dfrac{2}{5}+\dfrac{4}{5}x-\dfrac{7}{5}\)\(=\dfrac{9}{5}\)

\(\dfrac{4}{5}x+\dfrac{7}{5}=\dfrac{9}{5}-\dfrac{2}{5}\)

\(\dfrac{4}{5}x+\dfrac{7}{5}=\dfrac{7}{5}\)

\(\dfrac{4}{5}x=\dfrac{7}{5}-\dfrac{7}{5}\)

\(\dfrac{4}{5}x=0\)

\(x=0:\dfrac{4}{5}\)

\(x=0\)

b)\(\dfrac{2}{5}x-\dfrac{6}{4}=\dfrac{8}{5}\)

\(\dfrac{2}{5}x=\dfrac{8}{5}+\dfrac{6}{4}\)

\(\dfrac{2}{5}x=\dfrac{31}{10}\)

\(x=\dfrac{31}{10}:\dfrac{2}{5}\)

\(x=\dfrac{31}{4}\)

10 tháng 6 2017

1. Tính:

a. \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{9.10}\)

= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

= \(\dfrac{1}{1}-\dfrac{1}{10}\)

= \(\dfrac{10}{10}-\dfrac{1}{10}\)

= \(\dfrac{9}{10}\)

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

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

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

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

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

2. Tìm x, biết:

a. \(\dfrac{2}{5}+\dfrac{4}{5}x-\dfrac{7}{5}=\dfrac{9}{5}\)

\(\dfrac{4}{5}x-\dfrac{7}{5}=\dfrac{9}{5}-\dfrac{2}{5}\)

\(\dfrac{4}{5}x-\dfrac{7}{5}=\dfrac{7}{5}\)

\(\dfrac{4}{5}x=\dfrac{7}{5}+\dfrac{7}{5}\)

\(\dfrac{4}{5}x=\dfrac{14}{5}\)

\(x=\dfrac{14}{5}:\dfrac{4}{5}\)

\(x=\dfrac{14}{5}.\dfrac{5}{4}\)

\(x=14.\dfrac{1}{4}\)

\(x=\dfrac{14}{4}\)

Vậy \(x=\dfrac{14}{4}\)

b. \(\dfrac{2}{5}x-\dfrac{6}{4}=\dfrac{8}{5}\)

\(\dfrac{2}{5}x=\dfrac{8}{5}+\dfrac{6}{4}\)

\(\dfrac{2}{5}x=\dfrac{32}{20}+\dfrac{30}{20}\)

\(\dfrac{2}{5}x=\dfrac{62}{20}\)

\(\dfrac{2}{5}x=\dfrac{31}{10}\)

\(x=\dfrac{31}{10}:\dfrac{2}{5}\)

\(x=\dfrac{31}{10}.\dfrac{5}{2}\)

\(x=\dfrac{31}{2}.\dfrac{2}{2}\)

\(x=\dfrac{31}{2}.1\)

\(x=\dfrac{31}{2}\)

Vậy \(x=\dfrac{31}{2}\)

bài này mk tự làm ko sao chép trên mạnghihi

nếu thấy đúng thì tick đúng cho mk nhavui

d) Ta có: \(x+\dfrac{4}{5\cdot9}+\dfrac{4}{9\cdot13}+...+\dfrac{4}{41\cdot45}=\dfrac{-37}{45}\)

\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+...+\dfrac{1}{41}-\dfrac{1}{45}=\dfrac{-37}{45}\)

\(\Leftrightarrow x+\dfrac{1}{5}-\dfrac{1}{45}=\dfrac{-37}{45}\)

\(\Leftrightarrow x=\dfrac{-37}{45}+\dfrac{1}{45}-\dfrac{1}{5}=\dfrac{-36}{45}-\dfrac{1}{5}=\dfrac{-4}{5}-\dfrac{1}{5}=-1\)

Vậy: x=-1

26 tháng 4 2021

Câu 1: 

\(\Rightarrow \left[\begin{array}{} x+\frac{1}{2}=0\\ \frac{2}{3}-2x=0 \end{array} \right.\)

\(\Leftrightarrow \left[\begin{array}{} x=\frac{-1}{2}\\ x=\frac{1}{3} \end{array} \right.\)

Vậy phương trình có tập nghiệm S={\(\frac{-1}{2};\frac{1}{3}\)}

Câu 2: 

\(\Rightarrow \left[\begin{array}{} 3x-10=0\\ 5-\frac{1}{2}x=0 \end{array} \right.\)

\(\Leftrightarrow \left[\begin{array}{} x-=\frac{10}{3}\\ x=10 \end{array} \right.\)

Vậy phương trình có tập nghiệm S={\(10;\frac{10}{3}\)}

Câu 3: 

\(\Leftrightarrow \frac{1}{3}x=\frac{65}{4}-\frac{53}{4}\)

\( \Leftrightarrow \frac{1}{3}x=\frac{12}{4}\)

\(\Leftrightarrow x=9\)

Vậy phương trình có tập nghiệm S={9}

Câu 4: 

\(\Leftrightarrow \frac{2}{3}x=\frac{2}{3}\)

\(\Leftrightarrow x=1\)

Vậy phương trình có tập nghiệm S={1}

Câu 5: 

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x(x+1)}=\frac{2010}{2011}\)

\(\Leftrightarrow 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2010}{2011}\)

\(\Leftrightarrow 1-\frac{1}{x+1}=\frac{2010}{2011}\)

\(\Leftrightarrow \frac{x}{x+1}=\frac{2010}{2011}\)

\(\Rightarrow 2010x+2010=2011x\)

\(\Leftrightarrow x=2010\)

Vậy phương trình có tập nghiệm S={2010}

 

 

 

26 tháng 4 2021

cảm ơn bạn Hoàng Bình Bảo nha nhưng mà đây là toán lớp 6 mà bạn 

a: \(=\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{18\cdot19}-\dfrac{1}{19\cdot20}\)

=1/2-1/380

=179/380

b: \(=\dfrac{1}{1\cdot3}-\dfrac{1}{3\cdot5}+\dfrac{1}{3\cdot5}-\dfrac{1}{5\cdot7}+...+\dfrac{1}{21\cdot23}-\dfrac{1}{23\cdot25}\)

\(=\dfrac{1}{3}-\dfrac{1}{575}=\dfrac{572}{1725}\)

c: \(=1+\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}-\dfrac{1}{20}-\dfrac{1}{21}\)

=1-1/21

=20/21

d: \(=\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)\cdot...\cdot\left(1-\dfrac{1}{121}\right)\)

\(=\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{10}{11}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{12}{11}\)

\(=\dfrac{2}{11}\cdot\dfrac{12}{2}=\dfrac{12}{11}\)

15 tháng 4 2023

câu a và b thì sau 1 lúc thì mk hiểu

còn câu c thì mk chưa, câu d thì bị sai đề

20 tháng 3 2022

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`

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

 

20 tháng 3 2022

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

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

 

23 tháng 8 2023

Trả lời cho bạn đỗ manh tiến

20 tháng 3 2022

\(x\cdot\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)=1\\ x\cdot\left(1-\dfrac{1}{50}\right)=1\\ \dfrac{49}{50}x=1\\ x=1:\dfrac{49}{50}\\ x=\dfrac{50}{49}\)