\(\dfrac{x+1}{2022}\)+\(\dfrac{x+2}{2021}\)+.....+
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AH
Akai Haruma
Giáo viên
8 tháng 5 2023

Lời giải:
PT $\Leftrightarrow (\frac{x+1}{2022}+1)+(\frac{x+2}{2021}+1)+...+(\frac{x+23}{2000}+1)=0$

$\Leftrightarrow \frac{x+2023}{2022}+\frac{x+2023}{2021}+...+\frac{x+2023}{2000}=0$

$\Leftrightarrow (x+2023)(\frac{1}{2022}+\frac{1}{2021}+...+\frac{1}{2000})=0$
Dễ thấy tổng trong () luôn dương 

$\Rightarrow x+2023=0$

$\Leftrightarrow x=-2023$

20 tháng 3 2017

a)\(\dfrac{5}{23}.\dfrac{17}{26}+\dfrac{5}{23}.\dfrac{10}{26}-\dfrac{5}{23}\)

\(=\dfrac{5}{23}\left(\dfrac{17}{26}+\dfrac{10}{26}-1\right)\)

\(=\dfrac{5}{23}.\left(\dfrac{27}{26}-1\right)\)

\(=\dfrac{5}{23}.\dfrac{1}{26}\)

\(=\dfrac{5}{598}\)

b)\(\dfrac{1}{7}.\dfrac{5}{9}+\dfrac{5}{9}.\dfrac{2}{7}+\dfrac{5}{9}.\dfrac{1}{7}+\dfrac{5}{9}.\dfrac{3}{7}\)

\(=\dfrac{5}{9}.\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)

\(=\dfrac{5}{9}.1=\dfrac{5}{9}\)

20 tháng 3 2017

a)\(\dfrac{5}{23}.\dfrac{17}{26}+\dfrac{5}{23}.\dfrac{10}{26}-\dfrac{5}{23}\)

\(=\dfrac{5}{23}.\left(\dfrac{17}{26}+\dfrac{10}{26}-1\right)\)

\(=\dfrac{5}{23}.\left(\dfrac{27}{26}-\dfrac{26}{26}\right)\)

=\(\dfrac{5}{23}.\dfrac{1}{26}\)

\(=\dfrac{5}{598}\)

b)\(\dfrac{1}{7}.\dfrac{5}{9}+\dfrac{5}{9}.\dfrac{2}{7}+\dfrac{5}{9}.\dfrac{1}{7}+\dfrac{5}{9}.\dfrac{3}{7}\)

\(=\dfrac{5}{9}.\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)

\(=\dfrac{5}{9}.\left(\dfrac{7}{7}\right)\)

=\(\dfrac{5}{9}.1\)

\(=\dfrac{5}{9}\)

22 tháng 1 2018

làm như ép người ta quá vậy, lẽ ko ai trả lời cho cậu đâu

23 tháng 8 2017

\(a.\)

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

\(\Rightarrow\dfrac{1}{6}-2+5x=x-\dfrac{3}{5}\)

\(\Rightarrow\dfrac{1}{6}-2+\dfrac{3}{5}=-5x+x\)

\(\Rightarrow-4x=-\dfrac{37}{30}\)

\(\Rightarrow4x=\dfrac{37}{30}\)

\(\Rightarrow x=\dfrac{37}{120}\)

\(b.\)

\(\dfrac{3}{2}x-1\dfrac{1}{2}=x-\dfrac{3}{4}\)

\(\Rightarrow\dfrac{3}{2}x-\dfrac{3}{2}=x-\dfrac{3}{4}\)

\(\Rightarrow\dfrac{3}{2}x-x=\dfrac{3}{2}-\dfrac{3}{4}\)

\(\Rightarrow\dfrac{1}{2}x=\dfrac{3}{4}\)

\(\Rightarrow x=\dfrac{3}{2}\)

\(c.\)

\(x-\dfrac{5}{4}=\dfrac{1}{3}-\dfrac{3}{4}x\)

\(\Rightarrow x+\dfrac{3}{4}x=\dfrac{1}{3}+\dfrac{5}{4}\)

\(\Rightarrow\dfrac{7}{4}x=\dfrac{19}{12}\)

\(\Rightarrow x=\dfrac{19}{21}\)

\(d.\)

\(\dfrac{3}{2}\left(x-\dfrac{5}{3}\right)-\dfrac{7}{5}=x+1\)

\(\Rightarrow\dfrac{3}{2}x-\dfrac{5}{2}-\dfrac{7}{5}=x+1\)

\(\Rightarrow\dfrac{3}{2}x-\dfrac{39}{10}=x+1\)

\(\Rightarrow\dfrac{3}{2}x-x=\dfrac{39}{10}+1\)

\(\Rightarrow\dfrac{1}{2}x=\dfrac{49}{10}\)

\(\Rightarrow x=\dfrac{49}{5}\)

\(e.\)

\(\dfrac{2}{3}\left|4x+3\right|=\dfrac{6}{15}\)

\(\Rightarrow\left|4x+3\right|=\dfrac{3}{5}\)

\(\Rightarrow\left[{}\begin{matrix}4x+3=\dfrac{3}{5}\\4x+3=-\dfrac{3}{5}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}4x=-\dfrac{12}{5}\\4x=-\dfrac{18}{5}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\x=-\dfrac{9}{10}\end{matrix}\right.\)

23 tháng 8 2017

a) \(\dfrac{1}{3}.\left(\dfrac{1}{2}-6\right)+5x=x-\dfrac{3}{5}\Leftrightarrow\dfrac{1}{6}-2+5x=x-\dfrac{3}{5}\)

\(\Leftrightarrow5x-x=-\dfrac{3}{5}-\dfrac{1}{6}+2\Leftrightarrow4x=\dfrac{37}{30}\Leftrightarrow x=\dfrac{\dfrac{37}{30}}{4}=\dfrac{37}{120}\)

vậy \(x=\dfrac{37}{120}\)

b) \(\dfrac{3}{2}x-1\dfrac{1}{2}=x-\dfrac{3}{4}\Leftrightarrow\dfrac{3}{2}x-x=\dfrac{-3}{4}+1\dfrac{1}{2}\Leftrightarrow\dfrac{1}{2}x=\dfrac{-3}{4}+\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{1}{2}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}.2=\dfrac{6}{4}=\dfrac{3}{2}\) vậy \(x=\dfrac{3}{2}\)

c) \(x-\dfrac{5}{4}=\dfrac{1}{3}-\dfrac{3}{4}x\Leftrightarrow x+\dfrac{3}{4}x=\dfrac{1}{3}+\dfrac{5}{4}\Leftrightarrow\dfrac{7}{4}x=\dfrac{19}{12}\)

\(\Leftrightarrow x=\dfrac{\dfrac{19}{12}}{\dfrac{7}{4}}=\dfrac{19}{21}\) vậy \(x=\dfrac{19}{21}\)

d) \(\dfrac{3}{2}\left(x-\dfrac{5}{3}\right)-\dfrac{7}{5}=x+1\Leftrightarrow\dfrac{3}{2}x-\dfrac{5}{2}-\dfrac{7}{5}=x+1\)

\(\Leftrightarrow\dfrac{3}{2}x-x=1+\dfrac{5}{2}+\dfrac{7}{5}\Leftrightarrow\dfrac{1}{2}x=\dfrac{49}{10}\Leftrightarrow x=\dfrac{49}{10}.2=\dfrac{49}{5}\)

vậy \(x=\dfrac{49}{5}\)

e) \(\dfrac{2}{3}\left|4x+3\right|=\dfrac{6}{15}\Leftrightarrow\left|4x+3\right|=\dfrac{\dfrac{6}{15}}{\dfrac{2}{3}}=\dfrac{3}{5}\)

th1 : \(4x+3\ge0\Leftrightarrow4x\ge-3\Leftrightarrow x\ge\dfrac{-3}{4}\)

\(\Rightarrow\left|4x+3\right|=\dfrac{3}{5}\Leftrightarrow4x+3=\dfrac{3}{5}\Leftrightarrow4x=\dfrac{3}{5}-3=\dfrac{-12}{5}\)

\(\Leftrightarrow x=\dfrac{\dfrac{-12}{5}}{4}=\dfrac{-3}{5}\left(tmđk\right)\)

th2: \(4x+3< 0\Leftrightarrow4x< -3\Leftrightarrow x< \dfrac{-3}{4}\)

\(\Rightarrow\left|4x+3\right|=\dfrac{3}{5}\Leftrightarrow-\left(4x+3\right)=\dfrac{3}{5}\Leftrightarrow-4x-3=\dfrac{3}{5}\)

\(\Leftrightarrow4x=-3-\dfrac{3}{5}=\dfrac{-18}{5}\Leftrightarrow x=\dfrac{\dfrac{-18}{5}}{4}=\dfrac{-9}{10}\left(tmđk\right)\)

vậy \(x=\dfrac{-3}{5};x=\dfrac{-9}{10}\)

4 tháng 3 2017

a) \(\dfrac{7}{13}\)\(\times\)\(\dfrac{7}{15}\)-\(\dfrac{5}{12}\)\(\times\)\(\dfrac{21}{39}+\dfrac{49}{91}\)\(\times\)\(\dfrac{8}{15}\)

= \(\dfrac{7}{13}\)\(\times\)\(\dfrac{7}{15}\)-\(\dfrac{5}{12}\times\dfrac{7}{13}+\dfrac{7}{13}\times\dfrac{8}{15}\)

= \(\dfrac{7}{13}\left(\dfrac{7}{15}-\dfrac{5}{12}+\dfrac{8}{15}\right)\)

= \(\dfrac{7}{13}\times\dfrac{7}{12}\)

= \(\dfrac{49}{156}\)

b) \(\left(\dfrac{12}{199}+\dfrac{23}{200}-\dfrac{34}{201}\right)\times\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)

= \(\left(\dfrac{12}{199}+\dfrac{23}{200}-\dfrac{34}{201}\right)\times0\)

= 0

Bài 1: 

a: \(A=\dfrac{1\left(\dfrac{1}{13}-\dfrac{1}{17}-\dfrac{1}{23}\right)}{2\left(\dfrac{1}{13}-\dfrac{1}{17}-\dfrac{1}{23}\right)}\cdot\dfrac{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}{\dfrac{7}{6}-\dfrac{7}{8}+\dfrac{7}{10}}+\dfrac{6}{7}\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{7}+\dfrac{6}{7}=\dfrac{1}{7}+\dfrac{6}{7}=1\)

b: \(B=2000:\left[\dfrac{\dfrac{2}{5}-\dfrac{2}{9}+\dfrac{2}{11}}{\dfrac{7}{5}-\dfrac{7}{9}+\dfrac{7}{11}}\cdot\dfrac{-\dfrac{7}{6}+\dfrac{7}{8}-\dfrac{7}{10}}{\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}}\right]\)

\(=2000:\left[\dfrac{2}{7}\cdot\dfrac{-7}{2}\right]=-2000\)

c: \(C=10101\cdot\left(\dfrac{5}{111111}+\dfrac{1}{111111}-\dfrac{4}{111111}\right)\)

\(=10101\cdot\dfrac{2}{111111}=\dfrac{2}{11}\)

31 tháng 3 2019
https://i.imgur.com/eWLEslg.jpg
31 tháng 3 2019

Mk làm cho bn câu b

31 tháng 3 2017

b) \(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{499}{1000}\)

\(\dfrac{2}{6}+\dfrac{2}{12}+\dfrac{2}{20}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{499}{1000}\)

\(\dfrac{2}{2.3}+\dfrac{2}{3.4}+\dfrac{2}{4.5}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{499}{1000}\)

31 tháng 3 2017

mik lỡ bấm nhầm rồi, phần sau bn tự nghĩ nhé, sorry

20 tháng 7 2017

\(\dfrac{3}{-2}=\dfrac{-9}{6};\dfrac{-1}{-7}=\dfrac{1}{7}\)

Sắp xếp:

\(\dfrac{-1}{-7};\dfrac{0}{8};\dfrac{-7}{6};\dfrac{3}{-2}\)

20 tháng 7 2017

thanks Mới vô

13 tháng 8 2017

\(4\dfrac{3}{4}+\left(-0,37\right)+\dfrac{1}{8}+\left(-1,28\right)+\left(-2,5\right)+3\dfrac{1}{12}\)

\(=\left(4\dfrac{3}{4}+\dfrac{1}{8}+3\dfrac{1}{12}\right)+\left[\left(-0,37\right)+\left(-1,28\right)+\left(-2,5\right)\right]\)

\(=\left(\dfrac{19}{4}+\dfrac{1}{8}+\dfrac{37}{12}\right)+\left(-4,15\right)\)

\(=\dfrac{191}{24}-4,15\)

\(=\dfrac{457}{120}=3\dfrac{97}{120}\)

13 tháng 8 2017

\(4\dfrac{3}{4}+\left(-0,37\right)+\dfrac{1}{8}+\left(-1,28\right)+\left(-2,5\right)+3\dfrac{1}{12}\)

\(=\dfrac{19}{4}+\left(-\dfrac{37}{100}\right)+\dfrac{1}{8}+\left(-\dfrac{32}{25}\right)+\left(-\dfrac{5}{2}\right)+\dfrac{37}{12}\)

=\(\dfrac{475}{100}+\left(-\dfrac{37}{100}\right)+\dfrac{1}{8}+\left(-\dfrac{32}{25}\right)+\left(-\dfrac{5}{2}\right)+\dfrac{37}{12}\)

=\(\dfrac{219}{50}+\dfrac{25}{200}+\left(-\dfrac{256}{200}\right)+\left(-\dfrac{5}{2}\right)+\dfrac{37}{12}\)

=\(\dfrac{219}{50}+\left(-\dfrac{231}{200}\right)+\left(-\dfrac{30}{12}+\dfrac{37}{12}\right)\)

=