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

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

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

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

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

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

(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...

a)

\(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)

\(\Leftrightarrow (x-23)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)

Dễ thấy: \(\frac{1}{24}>\frac{1}{26}; \frac{1}{25}>\frac{1}{27}\Rightarrow \frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}>0\)

$\Rightarrow \frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\neq 0$

Do đó $x-23=0\Rightarrow x=23$

b)

PT \(\Leftrightarrow \frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}\)

\(\Leftrightarrow (x+100)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)

Dễ thấy: $\frac{1}{98}< \frac{1}{96}; \frac{1}{97}< \frac{1}{95}$

$\Rightarrow \frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}< 0$ hay khác $0$

$\Rightarrow x+100=0\Rightarrow x=-100$

c)

PT \(\Leftrightarrow \frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)

\(\Leftrightarrow \frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)

\(\Leftrightarrow (x+2005)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)

Dễ thấy $\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}<0$ hay khác $0$

Do đó $x+2005=0\Rightarrow x=-2005$

d)

PT \(\Leftrightarrow \frac{201-x}{99}+1+\frac{203-x}{97}+1+\frac{205-x}{96}+1=0\)

\(\Leftrightarrow \frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{96}=0\)

\(\Leftrightarrow (300-x)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{96}\right)=0\)

Dễ thấy \(\frac{1}{99}+\frac{1}{97}+\frac{1}{96}>0\) hay khác $0$

Do đó $300-x=0\Rightarrow x=300$

(109-x)/91+(107-x)/93+(105-x)/95+(103-x)/97=-4

[(109-x)/91 +1]+[(107-x)/93 +1]+[(105-x)/95 +1]+[(103-x)/97 +1]-4=-4

(109+91-x)/91+(107+93-x)/93+(105+95-x)/95+(103+97-x)/97=-4+4

(200-x)/91+(200-x)/93+(200-x)/95+(200-x)/97=0

(200-x)(1/91+1/93+1/95+1/97)=0

Ma : 1/91+1/93+1/95+1/97\(\ne\)0

=>200-x=0

=>x=200

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

\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=\frac{x+4}{96}+\frac{x+5}{95}+\frac{x+6}{94}\)

\(\Leftrightarrow\)\(\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1=\frac{x+4}{96}+1+\frac{x+5}{95}+\frac{x+6}{94}+1\)

\(\Leftrightarrow\)\(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}+\frac{x+100}{94}\)

\(\Leftrightarrow\)(x+100)(\(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}-\frac{1}{94}\))=0

\(\Leftrightarrow\)x+100=0(vì \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}-\frac{1}{94}\ne0\))

\(\Leftrightarrow\)x=-100

\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=\frac{x+4}{96}+\frac{x+5}{95}+\frac{x+6}{94}\)

\(\Rightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)+\left(\frac{x+5}{95}+1\right)+\left(\frac{x+6}{94}+1\right)\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}+\frac{x+100}{94}\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{96}-\frac{x+100}{95}-\frac{x+100}{94}=0\)

\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}-\frac{1}{94}\right)=0\)

Mà \(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}-\frac{1}{94}\ne0\)

\(\Rightarrow x+100=0\)

\(\Rightarrow x=-100\)

Vậy \(x=-100\)

\(\frac{109-x}{91}+\frac{107-x}{93}+\frac{105-x}{95}+\frac{103-x}{97}=-2\)

\(\Leftrightarrow\frac{109-x}{91}+1+\frac{107-x}{93}+1+\frac{105-x}{95}+1+\frac{103-x}{97}+1=2\)

\(\Leftrightarrow\left(200-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=2\)

\(\Rightarrow x=200-\frac{2}{\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}}\)

Bạn tự bấm máy, kết quả chẳng đẹp gì

\(A=2B\) thì còn có lý