\(\frac{x+1}{99}+\frac{x+2}{99}+\frac{x+3}{99}+\frac{x+4}{99}=-4\)

=>\(\frac{\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+\left(x+4\right)}{99}=-4\)

=> (x+1)+(x+2)+(x+3)+(x+4)=-4.99=-396

=>4x+10=-396

4x=-406

x=-406:4=-101,5

a) ĐK: \(x\ne0;x\ne-1\)

\(\left(\frac{1}{x^2+x}-\frac{2-x}{x+1}\right):\left(\frac{1}{2}+x-2\right)\)

\(=\left(\frac{x+1-2+x}{\left(x^2+x\right)\left(x+1\right)}\right):\left(\frac{1+2x+4}{2}\right)\)

\(=\frac{2x-1}{\left(x^2+x\right)\left(x+1\right)}:\frac{2x+5}{2}\)\(=\frac{2\left(2x-1\right)}{\left(x^2+x\right)\left(x+1\right)\left(2x+5\right)}\)?? hình như hết tính tiếp được rồi :v

P/s: Có phải đề là tính giá trị biểu thức không?

a,\(\frac{17}{12}\)

b,\(\frac{-134}{81}\)

c,x=2

d,x=7

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

\(\frac{1}{3}\) + \(\frac{5}{6}\): \(\left(x-2\frac{1}{5}\right)\)= \(\frac{3}{4}\)

<=> \(\frac{5}{6}\):\(\left(x-2\frac{1}{5}\right)\)= \(\frac{3}{4}\)- \(\frac{1}{3}\)

<=> \(\frac{5}{6}\) : \(\left(x-2\frac{1}{5}\right)\) = \(\frac{5}{12}\)

<=> \(\left(x-2\frac{1}{5}\right)\) =    \(\frac{5}{6}\) : \(\frac{5}{12}\)

,<=> \(\left(x-2\frac{1}{5}\right)\)=   2 

<=. x = 2 + \(\frac{11}{5}\)

<=> x = \(\frac{21}{5}\)

a) \(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x+3}{97}+\frac{x+4}{96}\)

\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1=\frac{x+3}{97}+1+\frac{x+4}{96}+1\)

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

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

Vì 1/99 + 1/98 - 1/97 - 1/96 khác 0

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

b) \(\frac{x-3}{47}+\frac{x-2}{48}=\frac{x-1}{49}+1\)

\(\Rightarrow\frac{x-3}{47}-1+\frac{x-2}{48}-1=\frac{x-1}{49}+1-2\)

\(\Rightarrow\frac{x-50}{47}+\frac{x-50}{48}-\frac{x-50}{49}=0\)

\(\Rightarrow\left(x-50\right)\left(\frac{1}{47}+\frac{1}{48}-\frac{1}{49}\right)=0\)

Vì 1/47 + 1/48 - 1/49 khác 0

Nên x -50 = 0 => x = 50

\(\frac{-11}{9}\le x+\frac{11}{18}\Leftrightarrow x\ge\frac{-11}{6}\)

\(-4\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{6}\right)\le x-\frac{2}{3}.\left(\frac{1}{3}-\frac{1}{2}-\frac{3}{4}\right)\)

\(\Rightarrow\frac{-13}{9}\le x-\frac{-11}{18}\)

\(\Leftrightarrow\frac{-13}{9}\le x+\frac{11}{18}\)

\(\Rightarrow x\ge\frac{-37}{18}\)

Học tốt nhé !! ^-^

cảm ơn Lâm Việt Phúc nhé