Cộng 1 vào từng phân số ta sẽ đc

\(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{101}+\frac{x+100}{102}+\frac{x+100}{103}\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}\right)=0\)

\(\Rightarrow x=-100\)

\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=\frac{x-1}{101}+\frac{x-2}{102}+\frac{x-3}{103}\)

<=> \(\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1=\frac{x-1}{101}+1+\frac{x-2}{102}+1+\frac{x-3}{103}+1\)

<=> \(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{101}+\frac{x+100}{102}+\frac{x+100}{103}\)

<=> \(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}\right)=0\)

<=> x + 100 = 0 (vì \(\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}\right)\ne0\))

<=> x = -100

\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=\frac{x-1}{101}+\frac{x-2}{102}+\frac{x-3}{103}\\ \Leftrightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1=\frac{x-1}{101}+1+\frac{x-2}{102}+1+\frac{x-3}{103}+1\\ \Leftrightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{101}-\frac{x+100}{102}-\frac{x+100}{103}=0\\ \Leftrightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}\right)=0\\ \Leftrightarrow x+100=0\\ \Leftrightarrow x=-100\)

Vậy tập nghiệm của phương trình trên là \(S=\left\{-100\right\}\)

x=100

Ta sẽ có: 1-1+1+1-1+1-1+1=0

\(\frac{x-1}{99}-\frac{x+1}{101}+\frac{x-2}{98}-\frac{x+2}{102}+\frac{x-3}{97}-\frac{x+3}{103}+\frac{x-4}{96}-\frac{x+4}{104}=0\)

\(\Rightarrow\frac{x-1}{99}-1-\frac{x+1}{101}+1+\frac{x-2}{98}-1-\frac{x+2}{102}+1+\frac{x-3}{97}-1-\frac{x+3}{103}+1+\frac{x-4}{96}-1-\frac{x+4}{104}+1=0\)

\(\Rightarrow\frac{x-100}{99}-\frac{x-100}{101}+\frac{x-100}{98}-\frac{x-100}{102}+\frac{x-100}{97}-\frac{x-100}{103}+\frac{x-100}{96}-\frac{x-100}{104}=0\)

\(\Rightarrow\left(x-100\right).\left(\frac{1}{99}-\frac{1}{101}+\frac{1}{98}-\frac{1}{102}+\frac{1}{97}-\frac{1}{103}+\frac{1}{96}-\frac{1}{104}\right)=0\)

Vì \(\frac{1}{99}>\frac{1}{101};\frac{1}{98}>\frac{1}{102};\frac{1}{97}>\frac{1}{103};\frac{1}{96}>\frac{1}{104}\)

\(\Rightarrow\frac{1}{99}-\frac{1}{101}+\frac{1}{98}-\frac{1}{102}+\frac{1}{97}-\frac{1}{103}+\frac{1}{96}-\frac{1}{104}\ne0\)

\(\Rightarrow x-100=0\)

\(\Rightarrow x=100\)

Vậy \(x=100\)

x thuoc R

\(\dfrac{x-1}{99}-\dfrac{x+1}{101}+\dfrac{x-2}{98}-\dfrac{x+2}{102}+\dfrac{x-3}{97}-\dfrac{x+3}{103}+\dfrac{x-4}{96}-\dfrac{x+4}{104}=0\)
<=> \(\dfrac{x-1}{99}-1-\dfrac{x+1}{101}-1+\dfrac{x-2}{98}-1-\dfrac{x-2}{102}-1+\dfrac{x-3}{97}-1-\dfrac{x+3}{103}-1+\dfrac{x-4}{96}-1-\dfrac{x+4}{104}=0\)

\(a,\frac{x+1}{65}+\frac{x+2}{64}=\frac{x+3}{63}+\frac{x+4}{62}\)

\(\Rightarrow\left[\frac{x+1}{65}+1\right]+\left[\frac{x+2}{64}+1\right]=\left[\frac{x+3}{63}+1\right]+\left[\frac{x+4}{62}+1\right]\)

\(\Rightarrow\frac{x+1+65}{65}+\frac{x+2+64}{64}=\frac{x+3+63}{63}+\frac{x+4+62}{62}\)

\(\Rightarrow\frac{x+66}{65}+\frac{x+66}{64}=\frac{x+66}{63}+\frac{x+66}{62}\)

\(\Rightarrow\frac{x+66}{65}+\frac{x+66}{64}=\frac{x+66}{63}+\frac{x+66}{62}=0\)

\(\Rightarrow\left[x+66\right]\left[\frac{1}{65}+\frac{1}{64}-\frac{1}{63}+\frac{1}{62}\right]=0\)

Mà \(\frac{1}{65}+\frac{1}{64}-\frac{1}{63}+\frac{1}{62}\ne0\)

\(\Rightarrow x+66=0\)

\(\Rightarrow x=0-66=-66\)

Auto làm nốt câu b

a,  Cộng cả 2 vế với 2 

Ta có \(\frac{x+1}{64}+\frac{x+2}{63}+2=\frac{x+3}{62}+\frac{x+4}{61}+2\)

\(\left(\frac{x+1}{64}+\frac{64}{64}\right)+\left(\frac{x+2}{63}+\frac{63}{63}\right)=\left(\frac{x+3}{62}+\frac{62}{62}\right)+\left(\frac{x+4}{61}+\frac{61}{61}\right)\)

=>  \(\frac{x+65}{64}+\frac{x+65}{63}=\frac{x+65}{62}+\frac{x+65}{61}\)\(\)

=> \(\frac{x+65}{64}+\frac{x+65}{63}-\frac{x+65}{62}-\frac{x+65}{61}=0\)

=> \(\left(x+65\right)\left(\frac{1}{64}+\frac{1}{63}-\frac{1}{62}-\frac{1}{61}\right)=0\)

Do \(\frac{1}{64}+\frac{1}{63}-\frac{1}{62}-\frac{1}{61}\ne0\)=> \(x+65=0\)

=> \(x=-65\)

b ,  Lm tương tự như Câu a

Chúc bn hok tốt

a) \(\frac{x+1}{65}+\frac{x+2}{64}=\frac{x+3}{63}+\frac{x+4}{62}\)

\(\Leftrightarrow\frac{x+1}{65}+\frac{x+2}{64}+2=\frac{x+3}{63}+\frac{x+4}{62}+2\)

\(\Leftrightarrow\left(\frac{x+1}{65}+1\right)+\left(\frac{x+2}{64}+1\right)=\left(\frac{x+3}{63}+1\right)+\left(\frac{x+4}{62}+1\right)\)

\(\Leftrightarrow\frac{x+1+65}{65}+\frac{x+2+64}{64}=\frac{x+3+63}{63}+\frac{x+4+62}{62}\)

\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{64}=\frac{x+66}{63}+\frac{x+66}{62}\)

\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{64}-\frac{x+66}{63}-\frac{x+66}{62}=0\)

\(\Leftrightarrow\left(x+66\right)\left(\frac{1}{65}+\frac{1}{64}-\frac{1}{63}-\frac{1}{62}\right)=0\)

\(\Leftrightarrow x+66=0\)

\(\Leftrightarrow x=-66\)

CÂu b) làm tương tự:

- Trừ 3 cho hai vế ( câu a) mk cộng 2 cho hai vế)

- Tách -3 = -1-1-1 rồi kết hợp với mỗi hạng tử

CỐ LÊN NHÉ

NẾU bạn KHÔNG HIỂU thì câu b) mik sẽ làm kĩ càng và rõ ràng hơn cho bạn hiểu

Ta có:  \(\frac{x-18}{2018}=\frac{x-17}{2017}\)

\(\Rightarrow\left(x-18\right).2017=\left(x-17\right).2018\)( tính chất của 2 tỉ số bằng nhau )

\(2017x-2017.18=2018x-2018.17\)

\(2018.17-2017.18=2018x-2017x\)

\(\left(2017+1\right).17-2017.\left(17+1\right)=x\)

\(2017.17+17-2017.17-2017=x\)

\(x=-2000\)

Vậy \(x=-2000\)

\(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x-1}{101}+\frac{x-2}{102}\)

\(\Rightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)=\left(\frac{x-1}{101}+1\right)+\left(\frac{x-2}{102}+1\right)\) ( cộng cả 2 vế thêm 2 )

\(\frac{x+100}{99}+\frac{x+100}{98}=\frac{x+100}{101}+\frac{x+100}{102}\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}-\frac{x+100}{101}-\frac{x+100}{102}=0\)

\(\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{101}-\frac{1}{100}\right)=0\)

Ta có: \(\frac{1}{99}+\frac{1}{98}-\frac{1}{101}-\frac{1}{100}\ne0\)

\(\Rightarrow x+100=0\)

​​​\(x=-100\)​

Vậy \(x=-100\)

a, \(\frac{x-18}{2018}=\frac{x-17}{2017}\)

=>\(\frac{x-18}{2018}+1=\frac{x-17}{2017}+1\)

=>\(\frac{x-18+2018}{2018}=\frac{x-17+2017}{2017}\)

=>\(\frac{x+2000}{2018}=\frac{x+2000}{2017}\)

=>\(\frac{x+2000}{2018}-\frac{x+2000}{2017}=0\)

=>\(\left(x+2000\right)\left(\frac{1}{2018}-\frac{1}{2017}\right)=0\)

Mà \(\frac{1}{2018}-\frac{1}{2017}\ne0\)

=>x+2000=0 => x=-2000

b, 

=>\(\frac{x+1}{99}+1+\frac{x+2}{98}+1=\frac{x-1}{101}+1+\frac{x-2}{102}+1\)

=>\(\frac{x+1+99}{99}+\frac{x+2+98}{98}=\frac{x-1+101}{101}+\frac{x-2+102}{102}\)

=>\(\frac{x+100}{99}+\frac{x+100}{98}=\frac{x+100}{101}+\frac{x+100}{102}\)

=>\(\frac{x+100}{99}+\frac{x+100}{98}-\frac{x+100}{101}-\frac{x+100}{102}=0\)

=>\(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{101}-\frac{1}{102}\right)=0\)

Mà \(\frac{1}{99}+\frac{1}{98}-\frac{1}{101}-\frac{1}{102}\ne0\)

=>x+100=0 => x=-100

\(\frac{x-1}{2018}=\frac{x-17}{2017}\)

\(\Rightarrow\frac{x-18}{2018}=\frac{x-17}{2017}=\frac{\left(x-18\right)-\left(x-17\right)}{2018-2017}=-\frac{1}{1}=-1\)

\(\Rightarrow x-18=-2018\)

\(\Rightarrow x=-2000\)

Uyên ơi m đừng mất dậy đừng chửi con gái h đú vc

c, Trừ hai vế cho 6 

Vế trái thì lấy từng số hạng trừ 1 là được

thế tức là phải như nào hả bạn

Không biết

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