Ta có: \(\dfrac{x+1}{99}+\dfrac{x+2}{98}+...+\dfrac{x+50}{50}+50=0\)

\(\Leftrightarrow\dfrac{x+1}{99}+1+\dfrac{x+2}{98}+1+...+\dfrac{x+50}{50}+1=0\)

\(\Leftrightarrow\dfrac{x+100}{99}+\dfrac{x+100}{98}+...+\dfrac{x+100}{50}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{99}+\dfrac{1}{98}+...+\dfrac{1}{50}\right)=0\)

mà \(\dfrac{1}{99}+\dfrac{1}{98}+...+\dfrac{1}{50}>0\)

nên x+100=0

hay x=-100

Vậy: S={-100}

\(\dfrac{x+1}{99}+\dfrac{x+2}{98}+...+\dfrac{x+50}{50}+50=0\)

\(\Leftrightarrow\left(\dfrac{x+1}{99}+1\right)+\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)+...+\left(\dfrac{x+50}{50}+1\right)=0\)

\(\Leftrightarrow\dfrac{x+100}{99}+\dfrac{x+100}{98}+...+\dfrac{x+100}{50}=0\)

\(\Leftrightarrow\left(x+100\right).\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+...+\dfrac{1}{50}\right)=0\)

\(\Leftrightarrow x+100=0\) (vì \(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+...+\dfrac{1}{50}>0\) )

\(\Leftrightarrow x=-100\)

Do VT là tổng của các giá trị tuyệt đối nên \(\ge0\Rightarrow100x\ge0\Rightarrow x\ge0\)

\(PT\Leftrightarrow\left(x+x+x+...+x\right)+\left(1+2+3+...+99\right)=100x\) (có 99x số x)

\(\Leftrightarrow99x+4950=100x\Leftrightarrow100x-99x=x=4950\)

Vậy \(x=4950\)

Dễ thấy \(x\ge0\)

\(\Rightarrow x+1+x+2+x+3+...+x+99=100x\)

giúp mik đi ạ đang gấp

x+1<x+2<x+3<x+4 ( với mọi x)

\(\dfrac{1}{100}\) < \(\dfrac{1}{99}\)<\(\dfrac{1}{3}\) <\(\dfrac{1}{2}\)

=>\(\dfrac{x+1}{100}\)+\(\dfrac{x+2}{99}\) <\(\dfrac{x+3}{3}\)+\(\dfrac{x+4}{2}\) là đúng

​\(|x-99|^{100}+|x-100|^{101}=1\)​

* Nếu \(x=99\)\(\Rightarrow\) \(|99-99|^{100}+|99-100|^{101}=0+1=1\)(  đúng )

\(\Rightarrow x=99\)là một nghiệm của phương trình

* Nếu \(x=100\)\(\Rightarrow|100-99|^{100}+|100-100|^{101}=1+0=1\)( đúng )

\(\Rightarrow x=100\)là một nghiệm của phương trình

* Nếu \(x< 99\)\(\Rightarrow x-100< 99-100\)\(\Rightarrow x-100< -1\)

\(\Rightarrow|x-100|^{101}>1\)\(\Leftrightarrow|x-99|^{100}+|x-100|^{101}>1\)\(\Rightarrow\)Phương trình vô nghiệm

* Nếu \(x>100\)\(\Rightarrow x-99>100-99\)\(\Rightarrow x-99>1\)

\(\Rightarrow|x-99|^{100}>1\)\(\Rightarrow|x-99|^{100}+|x-100|^{101}>1\)\(\Rightarrow\)Phương trình vô nghiệm

* Nếu \(99< x< 100\)\(\Rightarrow99-99< x-99< 100-99\)\(\Rightarrow0< x-99< 1\)

\(\Rightarrow|x-99|=x-99\)\(\left(1\right)\)

Cũng có : \(99< x< 100\)\(\Rightarrow99-100< x-100< 100-100\)\(\Rightarrow-1< x-100< 0\)

\(\Rightarrow|x-100|=-x+100\)\(\left(2\right)\)

Từ \(\left(1\right)\)và \(\left(2\right)\)\(\Rightarrow|x-99|+|x-100|=x-99-x+100\)

\(\Rightarrow|x-99|+|x-100|=1\)

Ta lại có : \(|x-99|^{100}< |x-99|\)Do(  \(0< |x-99|< 1\))

\(|x-100|^{101}< |x-100|\)Do ( \(0< |x-100|< 1\)

\(\Rightarrow|x-99|^{100}+|x-100|^{101}< |x-99|+|x-100|\)

\(\Rightarrow|x-99|^{100}+|x-100|^{101}< 1\)

\(\Leftrightarrow\)Phương trình vô nghiệm

Vậy phương trình có hai nghiệm duy nhất là \(x\in\left\{99;100\right\}\)

Bạn ơi bạn chia trường hợp kiểu gì vậy , với cả trường hợp cuối mình không hiểu gì đâu bạn ơi

a) \(\dfrac{x+1}{2004}+\dfrac{x+2}{2003}=\dfrac{x+3}{2002}+\dfrac{x+4}{2001}\) 

⇔ \(\dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1=\dfrac{x+3}{2002}+1+\dfrac{x+4}{2001}+1\)

⇔ \(\dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}=\dfrac{x+2005}{2002}+\dfrac{x+2005}{2001}\)

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

Vì\(\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\right)\)<0 nên phương trinh đã cho tương đương:

x+2005=0 ⇔x=-2005

b) \(\dfrac{201-x}{99}+\dfrac{203-x}{97}+\dfrac{205-x}{95}+3=0\) 

⇔ \(\dfrac{201-x}{99}+1+\dfrac{203-x}{97}+1+\dfrac{205-x}{95}+1=0\)

⇔ \(\dfrac{300-x}{99}+\dfrac{300-x}{97}+\dfrac{300-x}{95}=0\)

⇔ \(\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\)

Vì \(\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)>0\) nên phương trình đã cho tương đương:

300-x=0 ⇔ x=300

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

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

\(\frac{x+5}{95}+\frac{x+3}{97}+\frac{x+1}{99}=\frac{x+15}{85}+\frac{x+20}{80}+\frac{x+25}{75}.\)

\(\frac{x+5}{95}+1+\frac{x+3}{97}+1+\frac{x+1}{99}+1-\frac{x+15}{85}-1-\frac{x+20}{80}-1-\frac{x+25}{75}-1=0\)

\(\frac{x+100}{95}+\frac{x+100}{97}+\frac{x+100}{99}-\frac{x+100}{85}-\frac{x+100}{80}-\frac{x+100}{75}=0\)

\(\left(x+100\right).\left(\frac{1}{95}+\frac{1}{97}+\frac{1}{99}-\frac{1}{85}-\frac{1}{80}-\frac{1}{75}\right)=0\)

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

\(\frac{1}{95}+\frac{1}{97}+\frac{1}{99}-\frac{1}{85}-\frac{1}{80}-\frac{1}{75}\ne0\)

minh biet

ta có : 

\(\left|x+1\right|+\left|x-1\right|=1+\left|\left(x-1\right)\left(x+1\right)\right|\)

\(\Leftrightarrow\left|x-1\right|\left|x+1\right|-\left|x-1\right|-\left|x+1\right|+1=0\)

\(\Leftrightarrow\left(\left|x-1\right|-1\right)\left(\left|x+1\right|-1\right)=0\Leftrightarrow\orbr{\begin{cases}\left|x-1\right|=1\\\left|x+1\right|=1\end{cases}}\)

\(\Leftrightarrow x\in\left\{-2,0,2\right\}\)