\(\dfrac{x-25-124}{2015}\) + \(\dfrac{x-124-2015}{25}\) + \(\dfrac{x-2015-25}{124}\) = 3

\(\dfrac{x-15-124}{2015}\) - 1 + \(\dfrac{x-124-2015}{25}\) - 1 + \(\dfrac{x-2015-25}{124}\) - 1 = 0

\(\dfrac{x-15-124-2015}{2015}\)+\(\dfrac{x-124-2015-25}{25}\)+\(\dfrac{x-2015-25-124}{124}\) = 0

\(\dfrac{x-\left(15+124+2015\right)}{2015}\)+\(\dfrac{x-\left(124+2015+25\right)}{25}\)+\(\dfrac{x-\left(2015+25+124\right)}{124}\) = 0

(\(x\) - 2164).(\(\dfrac{1}{2015}\)+\(\dfrac{1}{25}\)+\(\dfrac{1}{124}\)) = 0

\(x-2164\) = 0

\(x\)              = 2164

trừ mỗi vế cho 2 rồi tách -2 thành -1và -1

X=1 nhé

     \(\frac{x+2014}{2015}+\frac{x+2015}{2016}=\frac{x+2016}{2017}+\frac{x+2017}{2018}\)

\(\Leftrightarrow\)\(\frac{x+2014}{2015}-1+\frac{x+2015}{2016}-1=\frac{x+2016}{2017}-1+\frac{x+2017}{2018}-1\)

\(\Leftrightarrow\)\(\frac{x-1}{2015}+\frac{x-1}{2016}=\frac{x-1}{2017}+\frac{x-1}{2018}\)

\(\Leftrightarrow\)\(\left(x-1\right)\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\right)=0\)

\(\Leftrightarrow\)\(x-1=0\)   ( do  1/2015 + 1/2016 - 1/2017 - 1/2018 # 0 )

\(\Leftrightarrow\) \(x=1\)

\(\frac{x+2015}{x-2015}=\frac{y+2017}{y-2017}\)

\(\frac{x+2015}{y+2017}=\frac{x-2015}{y-2017}\)

Áp dụng tính chất của dãy tỉ số bằng nhau,ta có :

\(\frac{x+2015}{y+2017}=\frac{x-2015}{y-2017}=\frac{\left(x+2015\right)-\left(x-2015\right)}{\left(y+2017\right)-\left(y-2017\right)}=\frac{2015}{2017}\)( 1 )

\(\frac{x+2015}{y+2017}=\frac{x-2015}{y-2017}=\frac{\left(x+2015\right)+\left(x-2015\right)}{\left(y+2017\right)+\left(y-2017\right)}=\frac{x}{y}\)( 2 )

Từ ( 1 ) và ( 2 ) \(\Rightarrow\frac{x}{y}=\frac{2015}{2017}\)

1.
\(\left(\frac{3}{1\times3}+\frac{3}{3\times5}+\frac{3}{5\times7}+...+\frac{3}{97\times99}\right)-x:\frac{3}{2}=\frac{7}{3}\\ \left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{97\times99}\right):\frac{3}{2}-x:\frac{3}{2}=\frac{7}{3}\\\left[\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)-x\right]:\frac{3}{2}=\frac{7}{3}\\ \left(1-\frac{1}{99}\right)-x=\frac{7}{3}\times\frac{3}{2}\\ \frac{98}{99}-x=\frac{7}{2}\\ x=\frac{98}{99}-\frac{7}{2}=\frac{-497}{198}\)

2.\(\frac{x}{y}=\frac{4}{3}\Rightarrow\hept{\begin{cases}x=4a\\y=3a\\x-y=4a-3a=a\end{cases}}\\ \left(x-y\right)^{2015}=5^{2015}\Rightarrow x-y=5\\ \Rightarrow a=5\Rightarrow\hept{\begin{cases}x=4\times5=20\\y=3\times5=15\end{cases}}\)

1.

\(\Rightarrow1225+x:5=2015\\ \Rightarrow x:5=790\\ \Rightarrow x=3950\)

x=3950

\(\frac{x-4}{2015}-\frac{1}{2015}=\frac{10-2x}{2015}\)

\(\Rightarrow\frac{x-4}{2015}-\frac{10-2x}{2015}=\frac{1}{2015}\)

\(\Rightarrow\frac{x-4-\left(10-2x\right)}{2015}=\frac{1}{2015}\)

\(\Rightarrow\frac{\left(x+2x\right)-\left(4+10\right)}{2015}=\frac{1}{2015}\)

\(\Rightarrow\frac{3x-14}{2015}=\frac{1}{2015}\)

\(\Rightarrow\left(3x-14\right).2015=2015\)

\(\Rightarrow3x-14=1\) ( bớt cả 2 vế đi 2015 lần )

\(\Rightarrow3x=15\)

\(\Rightarrow x=5\)

Vậy \(x=5\)

X=5

a) \(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)

\(\Leftrightarrow\frac{x+2015}{5}+\frac{5}{5}+\frac{x+2016}{4}+\frac{4}{4}=\frac{x+2017}{3}+\frac{3}{3}+\frac{x+2018}{2}+\frac{2}{2}\)

\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2002}{2}\)

\(\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)

\(\Leftrightarrow\left(x+2020\right).\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)

\(\Leftrightarrow x+2020=0\)

\(\Leftrightarrow x=-2020\)

Vậy : \(x=-2020\)

Chúc bạn học tốt !!

a) \(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\\ \left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\\ \frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2020}{2}\\ \frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\\ \left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\\ \Rightarrow x+2020=0\\ \Rightarrow x=-2020\)

Vậy x = -2020

b) \(\frac{x+2015}{5}+\frac{x+2016}{6}=\frac{x+2017}{7}+\frac{x+2018}{8}\\ \left(\frac{x+2015}{5}-1\right)+\left(\frac{x+2016}{6}-1\right)=\left(\frac{x+2017}{7}-1\right)+\left(\frac{x+2018}{8}-1\right)\\ \frac{x+2010}{5}+\frac{x+2010}{6}=\frac{x+2010}{7}+\frac{x+2010}{8}\\ \frac{x+2010}{5}+\frac{x+2010}{6}-\frac{x+2010}{7}-\frac{x+2010}{8}=0\\ \left(x+2010\right)\left(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\right)=0\\ \Rightarrow x+2010=0\\ \Rightarrow x=-2010\)

Vậy x = -2010