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

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

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

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

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

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

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

Dễ thấy \(\left(\frac{1}{99}< \frac{1}{98}< \frac{1}{97}< \frac{1}{96}\right)\)nên \(\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)\ne0\)

\(\Rightarrow x+100=0\Rightarrow x=-100\)

Vậy x = -100

\(\frac{109-x}{91}+\frac{107-x}{93}+\frac{105-x}{95}+\frac{103-x}{97}+4=0\)

\(\Rightarrow\frac{109-x}{91}+1+\frac{107-x}{93}+1+\frac{105-x}{95}+1+\frac{103-x}{97}+1=0\)

\(\Rightarrow\frac{200-x}{91}+\frac{200-x}{93}+\frac{200-x}{95}+\frac{200-x}{97}=0\)

\(\Rightarrow\left(200-x\right)\left(\frac{1}{91}+\frac{1}{93}-\frac{1}{95}-\frac{1}{97}\right)=0\)

Dễ thấy \(\left(\frac{1}{91}>\frac{1}{93}>\frac{1}{95}>\frac{1}{97}\right)\)nên \(\left(\frac{1}{91}+\frac{1}{93}-\frac{1}{95}-\frac{1}{97}\right)\ne0\)

\(\Rightarrow200-x=0\Rightarrow x=200\)

Vậy x = 200

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

     \(\frac{x+200}{99}+\frac{x+200}{98}=\frac{x+200}{97}+\frac{x+200}{96}\)

   \(\frac{x+200}{99}+\frac{x+200}{98}-\frac{x+200}{97}-\frac{x+200}{96}=0\)

\(\left(x+200\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)

mà\(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\ne0\)

==> x+200=0

<=>x=-200

Vậy nghiệm của phương trình là x=-200

c,  \(\frac{109-x}{91}+1+\frac{107-x}{93}+1+\frac{105-x}{95}+1+\frac{103-x}{97}+1=0\)

      \(\frac{200-x}{91}+\frac{200-x}{93}+\frac{200-x}{95}+\frac{200-x}{97}=0\)

\(\left(200-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=0\)

mà  \(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\ne0\)

==>200-x=0

<=>x=200

vậy nghiệm của pt là x=200

a, \(2\left(\frac{11x}{12}+\frac{1}{3}\right)=2-\frac{x}{6}\)

\(2\left(\frac{11x+4}{12}\right)-2+\frac{x}{6}=0\)

\(\frac{44x+8}{12}-2+\frac{x}{6}=0\)

\(\frac{44x+8}{12}-\frac{24}{12}+\frac{2x}{12}=0\)

\(\frac{44x+8-24+2x}{12}=\frac{46x-16}{12}=0\)

\(\Leftrightarrow46x-16=0\)

\(\Leftrightarrow46x=16\Rightarrow x=\frac{8}{23}\)

Vậy nghiệm của pt là x=8/23

k mk

\(\frac{X+1}{99}+1+\frac{X+2}{98}+1+\frac{x+3}{97}+1+\frac{X+4}{96}+1=0\)

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

(x+1)/99+(x+2)/98+(x+3)/97=(x+4)/96+(x+5)/95+(x+6)/94

[(x+1)/99 +1]+[(x+2)/98 +1]+[(x+3)/97 +1]-3=[(x+4)/96 +1]+[(x+5)/95 +1]+[(x+6)/94 +1]-3

[(x+1+99)/99+(x+2+98)/98+(x+3+97)/97]-3=[(x+4+96)/96+(x+5+95)/95+(x+6+94)/94]-3

(x+100)/99+(x+100)/98+(x+100)/97=(x+100)/96+(x+100)/95+(x+100)/94

(x+100)(1/99+1/98+1/97)=(x+100)(1/96+1/95+1/94)

(x+100)(1/99+1/98+1/97)-(x+100)(1/96+1/95+1/94)=0

(x+100)(1/99+1/98+1/97-1/96-1/95-1/94)=0

Ma : 1/99+1/98+1/97-1/96-1/95-1/94 \(\ne\)0

=>x+100=0

=>x=-100

k mk nha khong hieu noi mk nha.

1/3x-1/2=(3/5-4x)15/7

1/3x-1/2=9/7-60/7x

1/3x+60/7x=1/2+9/7

187/21x=25/14

x=75/374

k mk nha ban.

Ghi rõ hơn được không? Tớ không hiểu chỗ xuống dòng ấy...

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