Ta có : \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+2}{98}\)

\(\Rightarrow\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+2}{98}+1\)

\(\Rightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)

=> \(\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}\right)=\left(x+100\right).\left(\frac{1}{94}+\frac{1}{92}\right)\)

=> \(\left(x+100\right).\left(\frac{1}{98}+\frac{1}{96}\right)-\left(x+100\right)\left(\frac{1}{94}+\frac{1}{92}\right)=0\)

=> \(\left(x+100\right).\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\Rightarrow x+100=0\left(\text{vì }\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\right)\)

=> x = - 100 

Vậy x = - 100

\(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)

\(\Leftrightarrow\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+8}{92}+1\)

\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)

\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)

Vì \(\frac{1}{92}>\frac{1}{94}>\frac{1}{96}>\frac{1}{98}\)nên \(\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)< 0\)

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

\(\Leftrightarrow\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0.Ma:\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}< 0\Rightarrow x=-100\)

\(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)

\(\Leftrightarrow\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)

Vì : \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\)

Do đó : \(x+100=0\Leftrightarrow x=-100\)

Vậy phương trình có nghiệm : \(x=-100\)

Chúc bạn học tốt !!!

a)\(\dfrac{1}{2}\)(x+1)+\(\dfrac{1}{4}\)(x+3)=3-\(\dfrac{1}{3}\)(x+2)

\(\Leftrightarrow\)\(\dfrac{1}{2}\)x+\(\dfrac{1}{2}\)+\(\dfrac{1}{4}\)x+\(\dfrac{3}{4}\)=3-\(\dfrac{1}{3}\)x-\(\dfrac{2}{3}\)

\(\Leftrightarrow\)\(\dfrac{1}{2}\)x+\(\dfrac{1}{4}\)x+\(\dfrac{1}{3}\)x=-\(\dfrac{1}{2}\)-\(\dfrac{3}{4}\)+3-\(\dfrac{2}{3}\)

\(\Leftrightarrow\)\(\dfrac{13}{12}\)x=\(\dfrac{13}{12}\)

\(\Leftrightarrow\)x=1

Vậy nghiệm của pt là x=1

b)\(\dfrac{x+2}{98}\)+\(\dfrac{x+4}{96}\)=\(\dfrac{x+6}{94}\)+\(\dfrac{x+8}{92}\)

\(\Leftrightarrow\)\(\dfrac{x+2}{98}\)+\(\dfrac{x+4}{96}\)-\(\dfrac{x+6}{94}\)-\(\dfrac{x+8}{92}\)=0

\(\Leftrightarrow\)(\(\dfrac{x+2}{98}\)+1)+(\(\dfrac{x+4}{96}\)+1)-(\(\dfrac{x+6}{94}\)+1)-(\(\dfrac{x+8}{92}\)+1)=0

\(\Leftrightarrow\)\(\dfrac{x+2+98}{98}\)+\(\dfrac{x+4+96}{96}\)-\(\dfrac{x+6+94}{94}\)-\(\dfrac{x+8+92}{92}\)=0

\(\Leftrightarrow\)\(\dfrac{x+100}{98}\)+\(\dfrac{x+100}{96}\)-\(\dfrac{x+100}{94}\)-\(\dfrac{x+100}{92}\)=0

\(\Leftrightarrow\)(x+100)(\(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\))=0

\(\Leftrightarrow\)x+100=0(vì\(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\)\(\ne\)0)

\(\Leftrightarrow\)x=-100

Vậy nghiệm của pt là x=-100

a) \(\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)\)

\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)

\(\Leftrightarrow x+100=0\) ( do \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\) )

\(\Leftrightarrow x=-100\)

b) \(3x^2+2x-1=0\)

\(\Leftrightarrow3x^2+3x-x-1=0\)

\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=-1\end{matrix}\right.\)

\(\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+8}{92}+1\)

\(\Rightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)

\(\Rightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)

\(\Rightarrow\left(x+100\right).\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)

\(\Rightarrow x+100=0\)

\(\Rightarrow x=-100\)

Vậy : \(x=-100\)

\(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)

\(\Leftrightarrow\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)\)

\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}+\frac{x+100}{92}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)

Vì : \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\Leftrightarrow x+100=0\Leftrightarrow x=-100\)

Vậy \(x=-100\)

Chúc bạn học tốt !!!

\(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)

\(\Leftrightarrow\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}+-\frac{1}{94}-\frac{1}{92}\right)=0\)

Vì \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\)

Do đó : \(x+100=0\Leftrightarrow x=-100\)

Vậy phương trình có nghiệm : \(x=-100\)

a. \(\Leftrightarrow\dfrac{x+2}{98}+1+\dfrac{x+4}{96}+1=\dfrac{x+6}{94}+1+\dfrac{x+8}{92}+1\)

\(\Leftrightarrow\dfrac{x+100}{98}+\dfrac{x+100}{96}=\dfrac{x+100}{94}+\dfrac{x+100}{92}\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\right)=0\)

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

c. \(\Leftrightarrow3x^2+3x-x-1=0\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\Leftrightarrow\left[\begin{matrix}x+1=0\\3x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\frac{x+8}{92}+1+\frac{x+7}{93}+1+\frac{x+6}{94}+1\ge\frac{x+2}{98}+1+\frac{x+3}{97}+1+\frac{x+4}{96}+1\)

\(\Leftrightarrow\frac{x+100}{92}+\frac{x+100}{93}+\frac{x+100}{94}\ge\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{92}-\frac{1}{98}+\frac{1}{93}-\frac{1}{97}+\frac{1}{94}-\frac{1}{96}\right)\ge0\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{6}{92.98}+\frac{4}{93.97}+\frac{2}{94.96}\right)\ge0\)

\(\Leftrightarrow x+100\ge0\Rightarrow x\ge-100\)

a) Ta có: 3x(x-1)=(x-1)(x+2)

⇔3x(x-1)-(x-1)(x+2)=0

⇔(x-1)(3x-x-2)=0

⇔(x-1)(2x-1)=0

⇔2(x-1)²=0

mà 2≠0

nên (x-1)²=0

⇔x-1=0

hay x=1

Vậy: x=1

b) Ta có: \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)

\(\Leftrightarrow\frac{21\left(4x+3\right)}{105}-\frac{15\left(6x-2\right)}{105}-\frac{35\left(5x+4\right)}{105}-\frac{315}{105}=0\)

\(\Leftrightarrow84x+63-90x+30-175x-140-315=0\)

\(\Leftrightarrow-181x-362=0\)

\(\Leftrightarrow-181x=362\)

hay x=-2

Vậy: x=-2

c) Ta có: \(\frac{1}{2}\left(x+1\right)+\frac{1}{4}\left(x+3\right)=3-\frac{1}{2}\left(x+2\right)\)

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

\(\Leftrightarrow\frac{x}{2}+\frac{x}{4}+\frac{5}{4}-\frac{-x}{2}-2=0\)

\(\Leftrightarrow\frac{x}{2}+\frac{x}{4}+\frac{x}{2}-\frac{3}{4}=0\)

\(\Leftrightarrow\frac{x}{4}+x-\frac{3}{4}=0\)

\(\Leftrightarrow\frac{x}{4}+\frac{4x}{4}-\frac{3}{4}=0\)

\(\Leftrightarrow5x-3=0\)

\(\Leftrightarrow5x=3\)

hay \(x=\frac{3}{5}\)

Vậy: \(x=\frac{3}{5}\)

d) Ta có: \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)

\(\Leftrightarrow\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+8}{92}+1\)

\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)

mà \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\)

nên x+100=0

hay x=-100

Vậy: x=-100

Cộng hai vào mỗi vễ của phương trình ta có

\(\Leftrightarrow\frac{x+2}{98}+\frac{x+4}{96}+2=\frac{x+6}{94}+\frac{x+8}{92}+2\)

\(\Leftrightarrow\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{X+8}{92}+1\)

\(\Leftrightarrow\frac{x+2+98}{98}+\frac{x+4+96}{96}=\frac{x+6+94}{94}+\frac{x+8+92}{92}\)

\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)

\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)

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

Vậy S={-100}

cho mình hỏi tại sao lại phải công 2 vào 

toan lop 7 ma khong biet con gia vo toan 8