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\(\Leftrightarrow\left(\dfrac{x-1}{99}-1\right)+\left(\dfrac{x-99}{1}-1\right)+\left(\dfrac{x-3}{97}-1\right)+\left(\dfrac{x-7}{93}-1\right)+\left(\dfrac{x-5}{95}-1\right)+\left(\dfrac{x-95}{5}-1\right)=0\)=>x-100=0

hay x=100

11 tháng 2 2020

các bạn giúp mình với a. Mình cảm ơn trước

11 tháng 2 2020

\(\frac{x+1}{99}+\frac{x+3}{97}+\frac{x+5}{95}=\frac{x+7}{93}+\frac{x+9}{91}+\frac{x+11}{89}\)

\(\Rightarrow\frac{x+1}{99}+1+\frac{x+3}{97}+1+\frac{x+5}{95}+1\)\(=\frac{x+7}{93}+1+\frac{x+9}{91}+1+\frac{x+11}{89}+1\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}\)\(=\frac{x+100}{93}+\frac{x+100}{91}+\frac{x+100}{89}\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}\)\(-\frac{x+100}{93}-\frac{x+100}{91}-\frac{x+100}{89}=0\)

\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{93}-\frac{1}{91}-\frac{1}{89}\right)=0\)

Mà \(\left(\frac{1}{99}< \frac{1}{97}< \frac{1}{95}< \frac{1}{93}< \frac{1}{91}< \frac{1}{89}\right)\)nên \(\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{93}-\frac{1}{91}-\frac{1}{89}\right)< 0\)

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

Vậy x = -100

30 tháng 12 2022

\(\Leftrightarrow\left(\dfrac{x-1}{99}-1\right)+\left(\dfrac{x-99}{1}-1\right)+\left(\dfrac{x-3}{97}-1\right)+\left(\dfrac{x-97}{3}-1\right)+\left(\dfrac{x-5}{95}-1\right)+\left(\dfrac{x-95}{5}-1\right)=0\)

=>x-100=0

=>x=100

Sửa đề: \(\frac{x-1}{99}+\frac{x-3}{97}+\frac{x-5}{95}+\frac{x-7}{93}+\frac{x-95}{5}+x=105\)

Ta có: \(\frac{x-1}{99}+\frac{x-3}{97}+\frac{x-5}{95}+\frac{x-7}{93}+\frac{x-95}{5}+x=105\)

\(\Leftrightarrow\frac{x-1}{99}+\frac{x-3}{97}+\frac{x-5}{95}+\frac{x-7}{93}+\frac{x-95}{5}+x-105=0\)

\(\Leftrightarrow\frac{x-1}{99}-1+\frac{x-3}{97}-1+\frac{x-5}{95}-1+\frac{x-7}{93}-1+\frac{x-95}{5}-1+x-100=0\)

\(\Leftrightarrow\frac{x-100}{99}+\frac{x-100}{97}+\frac{x-100}{95}+\frac{x-100}{93}+\frac{x-100}{5}+\frac{x-100}{1}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}+\frac{1}{93}+\frac{1}{5}+1\right)=0\)

\(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}+\frac{1}{93}+\frac{1}{5}+1\ne0\)

nên x-100=0

hay x=100

Vậy: x=100

4 tháng 3 2020

\(\frac{x+2}{2002}+\frac{x+5}{1999}+\frac{x+201}{1803}=-3\)

\(\Rightarrow\frac{x+2}{2002}+1+\frac{x+5}{1999}+1+\frac{x+201}{1803}+1=0\)

\(\Rightarrow\frac{x+2004}{2002}+\frac{x+2004}{1999}+\frac{x+2004}{1803}=0\)

\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)=0\)

Dễ thấy \(\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)>0\)nên x + 2004 = 0

Vậy x = -2004

26 tháng 1 2018

c, Trừ hai vế cho 6 

Vế trái thì lấy từng số hạng trừ 1 là được

8 tháng 2 2018

thế tức là phải như nào hả bạn

4 tháng 3 2020

\(\frac{x+2}{2002}+\frac{x+5}{1999}+\frac{x+201}{1803}=-3\)

\(\Leftrightarrow\frac{x+2}{2002}+1+\frac{x+5}{1999}+1+\frac{x+201}{1803}+1=-3+1+1+1\)

\(\Leftrightarrow\frac{x+2004}{2002}+\frac{x+2004}{1999}+\frac{x+2004}{1803}=0\)

\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)=0\)

\(\Leftrightarrow x+2004=0\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\ne0\right)\)

<=> x=-2004

4 tháng 3 2020

a,\(\frac{x+2}{2002}+\frac{x+5}{1999}+\frac{x+201}{1803}=-3\)

\(< =>\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+5}{1999}+1\right)+\left(\frac{x+201}{1803}+1\right)=0\)

\(< =>\frac{x+2004}{2002}+\frac{x+2004}{1999}+\frac{x+2004}{1803}=0\)

\(< =>\left(x+2004\right).\left(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\right)=0\)

Do \(\frac{1}{2002}+\frac{1}{1999}+\frac{1}{1803}\ne0\)

\(=>x+2004=0\)

\(=>x=-2004\)