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

\(\Leftrightarrow\left(3x+2\right)\left(5x+1\right)=\left(3x-1\right)\left(5x+7\right)\)

\(\Leftrightarrow15x^2+3x+10x+2=15x^2+21x-5x-7\)

=>16x-7=13x+2

=>3x=9

hay x=3

b: \(\dfrac{x+1}{2016}+\dfrac{x}{2017}=\dfrac{x+2}{2015}+\dfrac{x+3}{2014}\)

\(\Leftrightarrow\left(\dfrac{x+1}{2016}+1\right)+\left(\dfrac{x}{2017}+1\right)=\left(\dfrac{x+2}{2015}+1\right)+\left(\dfrac{x+3}{2014}+1\right)\)

=>x+2017=0

hay x=-2017

e: \(\left(2x-3\right)^2=144\)

=>2x-3=12 hoặc 2x-3=-12

=>2x=15 hoặc 2x=-9

=>x=15/2 hoặc x=-9/2

a: =>\(\dfrac{2x-4}{2014}+\dfrac{2x-2}{2016}< \dfrac{2x-1}{2017}+\dfrac{2x-3}{2015}\)

=>\(\dfrac{2x-2018}{2014}+\dfrac{2x-2018}{2016}< \dfrac{2x-2018}{2017}+\dfrac{2x-2018}{2015}\)

=>2x-2018<0

=>x<2019

b: \(\Leftrightarrow\left(\dfrac{3-x}{100}+\dfrac{4-x}{101}\right)>\dfrac{5-x}{102}+\dfrac{6-x}{103}\)

=>\(\dfrac{x-3}{100}+\dfrac{x-4}{101}-\dfrac{x-5}{102}-\dfrac{x-6}{103}< 0\)

=>\(x+97< 0\)

=>x<-97

\(\dfrac{x+4}{2014}+\dfrac{x+3}{2015}=\dfrac{x+2}{2016}+\dfrac{x+1}{2017}\)

\(\dfrac{x+4}{2014}+1+\dfrac{x+3}{2015}+1=\dfrac{x+2}{2016}+1+\dfrac{x+1}{2017}+1\)

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

\(\left(x+2018\right)\left(\dfrac{1}{2014}+\dfrac{1}{2015}-\dfrac{1}{2016}-\dfrac{1}{2017}\right)=0\\ x+2018=0\\ x=-2018\)

\(\dfrac{1-x}{2017}+\dfrac{2-x}{2016}=\dfrac{3-x}{2015}+\dfrac{4-x}{2014}\)

\(\left(\dfrac{1-x}{2017}+1\right)+\left(\dfrac{2-x}{2016}+1\right)=\left(\dfrac{3-x}{2015}+1\right)+\left(\dfrac{4-x}{2014}+1\right)\)

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

\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}-\dfrac{2018-x}{2015}-\dfrac{2018-x}{2014}=0\)

\(\Leftrightarrow\left(2018-x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)

Mà \(\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)\ne0\)

\(\Leftrightarrow2018-x=0\Leftrightarrow x=2018\)

Vậy ...

\(\dfrac{1-x}{2017}+\dfrac{2-x}{2016}=\dfrac{3-x}{2015}+\dfrac{4-x}{2014}\)

\(\Leftrightarrow\left(\dfrac{1-x}{2017}+1\right)+\left(\dfrac{2-x}{2016}+1\right)=\left(\dfrac{3-x}{2015}+1\right)+\left(\dfrac{4-x}{2014}+1\right)\)

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

\(\Leftrightarrow\dfrac{2018-x}{2017}+\dfrac{2018-x}{2016}-\dfrac{2018-x}{2015}-\dfrac{2018-x}{2014}=0\)

\(\Leftrightarrow\left(2018-x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)

Mà \(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\ne0\)

\(\Leftrightarrow2018-x=0\Leftrightarrow x=2018\)

Vậy ....

\(\dfrac{1-18x}{2017}+\dfrac{2-18x}{2016}=\dfrac{3-18x}{2015}+\dfrac{4-18x}{2014}\)

\(\Rightarrow\left(\dfrac{1-18x}{2017}+1\right)+\left(\dfrac{2-18x}{2016}+1\right)=\left(\dfrac{3-18x}{2015}+1\right)+\left(\dfrac{4-18x}{2014}+1\right)\)

\(\Rightarrow\dfrac{2018-18x}{2017}+\dfrac{2018-18x}{2016}-\dfrac{2018-18x}{2015}-\dfrac{2018-18x}{2014}=0\)

\(\Rightarrow\left(2018-18x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)

\(\Rightarrow2018-18x=0\Rightarrow x=\dfrac{1009}{9}\)

Vậy.............

Chúc bạn học tốt!!!

\(\dfrac{1-18x}{2017}+\dfrac{2-18x}{2016}=\dfrac{3-18x}{2015}+\dfrac{4-18x}{2014}\)

\(\Rightarrow\left(\dfrac{1-18x}{2017}+1\right)+\left(\dfrac{2-18x}{2016}+1\right)=\left(\dfrac{3-18x}{2015}+1\right)+\left(\dfrac{4-18x}{2014}+1\right)\)

\(\Rightarrow\dfrac{2018-18x}{2017}+\dfrac{2018-18x}{2016}-\dfrac{2018-18x}{2015}-\dfrac{2018-18x}{2014}=0\)

\(\Rightarrow\left(2018-18x\right)\left(\dfrac{1}{2017}+\dfrac{1}{2016}-\dfrac{1}{2015}-\dfrac{1}{2014}\right)=0\)

\(\Rightarrow2018-18x=0\)

\(\Rightarrow18x=2018-0\)

\(\Rightarrow18x=2018\)

\(\Rightarrow x=2018:18\)

\(\Rightarrow x=\dfrac{1009}{9}\)

1,

đặt A= \(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+....+\(\dfrac{1}{2016}\)+\(\dfrac{1}{2017}\)

2A=1+\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+....+\(\dfrac{1}{2015}\)+\(\dfrac{1}{2016}\)

2A-A=(1+\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+....+\(\dfrac{1}{2015}\)+\(\dfrac{1}{2016}\))-(\(\dfrac{1}{2}\)+\(\dfrac{1}{3}\)+\(\dfrac{1}{4}\)+....+\(\dfrac{1}{2016}\)+\(\dfrac{1}{2017}\))

A=1-\(\dfrac{1}{2017}\)

A=\(\dfrac{2016}{2017}\)

vậy A=\(\dfrac{2016}{2017}\)

Bạn ơi hnhf như đề bài phải là tính \(^{\dfrac{a}{b}}\)chứ k thì làm sao mak tính đc phần b

Ta có: \(\dfrac{2x-5}{2015}+\dfrac{2x-6}{2016}=\dfrac{2x+3}{2007}+\dfrac{2x+4}{2006}\)

\(\Leftrightarrow\dfrac{2x-5}{2015}+1+\dfrac{2x-6}{2016}+1=\dfrac{2x+3}{2007}+1+\dfrac{2x+4}{2006}+1\)

\(\Leftrightarrow\dfrac{2x+2010}{2015}+\dfrac{2x+2010}{2016}-\dfrac{2x+2010}{2007}-\dfrac{2x+2010}{2006}=0\)

\(\Leftrightarrow\left(2x+2010\right)\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)=0\)

mà \(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2007}-\dfrac{1}{2006}\ne0\)

nên 2x+2010=0

\(\Leftrightarrow2x=-2010\)

hay x=-1005

Vậy: S={-1005}

\(\dfrac{2x-5}{2015}+\dfrac{2x-6}{2016}=\dfrac{2x+3}{2007}+\dfrac{2x+4}{2006}\)
= \(\left(\dfrac{2x-5}{2015}-1\right)+\left(\dfrac{2x-6}{2016}-1\right)=\left(\dfrac{2x+3}{2007}+1\right)+\left(\dfrac{2x+4}{2006}+1\right)\)
=\(\dfrac{2x-2010}{2015}+\dfrac{2x-2010}{2016}=\dfrac{2x+2010}{2007}+\dfrac{2x+2010}{2006}\)
=\(\dfrac{2x-2010}{2015}+\dfrac{2x-2010}{2016}-\dfrac{2x-2010}{2007}-\dfrac{2x-2010}{2006}\)=0

=(2x-2010)\(\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)\)=0
vì \(\left(\dfrac{1}{2015}+\dfrac{1}{2016}-\dfrac{1}{2007}-\dfrac{1}{2006}\right)\)≠0
⇒ 2x-2010=0
⇔2x=2010
⇔x=1005