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\(\dfrac{x+23}{2021}+\dfrac{x+22}{2022}+\dfrac{x+21}{2023}+\dfrac{x+20}{2024}=-4\)
Vì \(\dfrac{x+23}{2021}+\dfrac{x+22}{2022}+\dfrac{x+21}{2023}+\dfrac{x+20}{2024}=-4\)
\(\Rightarrow\dfrac{x+23}{2021}+\dfrac{x+22}{2022}+\dfrac{x+21}{2023}+\dfrac{x+20}{2024}+4=0\)
\(\Rightarrow\left(\dfrac{x+23}{2021}+1\right)+\left(\dfrac{x+22}{2022}+1\right)+\left(\dfrac{x+21}{2023}+1\right)+\left(\dfrac{x+20}{2024}+1\right)=0\)
\(\Rightarrow\dfrac{x+2044}{2021}+\dfrac{x+2044}{2022}+\dfrac{x+2044}{2023}+\dfrac{x+2044}{2024}=0\)
\(\Rightarrow\left(x+2044\right)\left(\dfrac{1}{2021}+\dfrac{1}{2022}+\dfrac{1}{2023}+\dfrac{1}{2024}\right)=0\)
\(\Rightarrow x+2044=0\left(\dfrac{1}{2021}+\dfrac{1}{2022}+\dfrac{1}{2023}+\dfrac{1}{2024}\ne0\right)\)
\(\Rightarrow x=-2024\)
\(\dfrac{x+2017}{x+2018}=\dfrac{2022}{2023}\)
\(\Leftrightarrow2023x+4080391=2022x+4080396\)
=>x=5
tìm giá trị lớn nhất của P = \(\dfrac{|x-2022|-|x-2023|+|x-2024|+2022}{|x-2022|+|x-2023|+|x-2024|}\)
Giải:
a) \(\dfrac{1}{3}x+\dfrac{1}{5}-\dfrac{1}{2}x=1\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{6}x=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{1}{6}x=\dfrac{-21}{20}\)
\(\Leftrightarrow x=\dfrac{-63}{10}\)
Vậy ...
b) \(\dfrac{3}{2}\left(x+\dfrac{1}{2}\right)-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{3}{2}x+\dfrac{3}{4}-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{11}{8}x=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{-4}{11}\)
Vậy ...
Các câu sau làm tương tự câu b)
2, \(\Rightarrow\left\{{}\begin{matrix}\\\dfrac{5}{4}x-\dfrac{7}{2}=0\\\dfrac{5}{8}x+\dfrac{3}{5}=0\\\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{14}{5}\\\\x=\dfrac{-24}{25}\\\end{matrix}\right.\)
Các câu dễ tự làm :v
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
a: \(\Leftrightarrow\dfrac{7}{2}x-\dfrac{3}{4}=\dfrac{1}{2}x+\dfrac{5}{2}\)
\(\Leftrightarrow3x=\dfrac{5}{2}+\dfrac{3}{4}=\dfrac{10}{4}+\dfrac{3}{4}=\dfrac{13}{4}\)
=>x=13/12
b: \(\Leftrightarrow x\cdot\left(\dfrac{2}{3}-\dfrac{1}{2}\right)=-\dfrac{1}{3}+\dfrac{2}{5}\)
\(\Leftrightarrow x\cdot\dfrac{1}{6}=\dfrac{-5+6}{15}=\dfrac{1}{15}\)
\(\Leftrightarrow x=\dfrac{1}{15}:\dfrac{1}{6}=\dfrac{2}{5}\)
c: \(\Leftrightarrow x\cdot\dfrac{1}{3}+x\cdot\dfrac{2}{5}+\dfrac{2}{5}=0\)
\(\Leftrightarrow x\cdot\dfrac{11}{15}=-\dfrac{2}{5}\)
\(\Leftrightarrow x=-\dfrac{2}{5}:\dfrac{11}{15}=\dfrac{-2}{5}\cdot\dfrac{15}{11}=\dfrac{-30}{55}=\dfrac{-6}{11}\)
d: \(\Leftrightarrow-\dfrac{1}{3}x+\dfrac{1}{2}+\dfrac{2}{3}-x-\dfrac{1}{2}=5\)
\(\Leftrightarrow-\dfrac{4}{3}x+\dfrac{2}{3}=5\)
\(\Leftrightarrow-\dfrac{4}{3}x=5-\dfrac{2}{3}=\dfrac{13}{3}\)
\(\Leftrightarrow x=\dfrac{13}{3}:\dfrac{-4}{3}=\dfrac{-13}{4}\)
e: \(\Leftrightarrow\left(\dfrac{x+2015}{5}+1\right)+\left(\dfrac{x+2016}{4}+1\right)=\left(\dfrac{x+2017}{3}+1\right)+\left(\dfrac{x+2018}{2}+1\right)\)
=>x+2020=0
hay x=-2020
a: =>||12x-1/2|-2|=-2/3x3/4=-6/12=-1/2(loại)
b: =>2/3-1/3x-1/2+2/3x=2x+2/3
=>-5/3x=1/2
=>x=-1/2:5/3=-1/2x3/5=-3/10
c: =>|3/2x+1/4|=2+3/4=11/4
=>3/2x+1/4=11/4 hoặc 3/2x+1/4=-11/4
=>3/2x=5/2 hoặc 3/2x=-3
=>x=3/5 hoặc x=-3:3/2=-2
a. \(\dfrac{-39}{7}:x=26\)
x = \(\dfrac{-39}{7}:26\)
x = \(\dfrac{-3}{14}\)
b. \(x:\dfrac{13}{5}=\dfrac{7}{4}\)
x = \(\dfrac{7}{4}.\dfrac{13}{5}\)
x = \(\dfrac{91}{20}\)
c. x = \(\dfrac{-3}{5}-\dfrac{1}{2}\)
x = \(\dfrac{-11}{10}\)
d. \(x-\dfrac{3}{4}=\dfrac{9}{4}\)
x = \(\dfrac{9}{4}+\dfrac{3}{4}\)
x = 3
e. \(\dfrac{7}{8}:x=\dfrac{14}{3}\)
x = \(\dfrac{7}{8}:\dfrac{14}{3}\)
x = \(\dfrac{3}{16}\)
f. \(x:\dfrac{8}{3}=\dfrac{13}{3}\)
x = \(\dfrac{13}{3}.\dfrac{8}{3}\)
x = \(\dfrac{104}{9}\)
g. x = \(\dfrac{4}{10}-\dfrac{2}{5}\)
x = 0
chúc bạn học tốt 
Ta có: \(\dfrac{x+4}{2021}+\dfrac{x+3}{2022}=\dfrac{x+2}{2023}+\dfrac{x+1}{2024}\)
=>\(\left(\dfrac{x+4}{2021}+1\right)+\left(\dfrac{x+3}{2022}+1\right)-\left(\dfrac{x+2}{2023}+1\right)-\left(\dfrac{x+1}{2024}+1\right)=0\)
=>\(\dfrac{x+2025}{2021}+\dfrac{x+2025}{2022}-\dfrac{x+2025}{2023}-\dfrac{x+2025}{2024}=0\)
=>x+2025=0
=>x=-2025