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ML
1
XO
27 tháng 1 2020
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
C
2
BA
31 tháng 10 2019
\(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
\(\Leftrightarrow\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)\)
\(\Leftrightarrow\frac{100+x}{98}+\frac{100+x}{96}-\frac{100+x}{94}-\frac{100+x}{92}=0\)
\(\Rightarrow\left(100+x\right)\left(\frac{1}{98}+\frac{1}{96}+\frac{1}{94}+\frac{1}{92}\right)=0\)
Vì \(\frac{1}{98}+\frac{1}{96}+\frac{1}{94}+\frac{1}{92}\ne0\)
\(\Rightarrow100+x=0\)
\(\Rightarrow x=-100\)
C
31 tháng 10 2019
\(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
\(\Leftrightarrow\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+8}{92}+1\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
Vì \(\frac{1}{98}< \frac{1}{96}< \frac{1}{94}< \frac{1}{92}\)nên \(\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)< 0\)
Vậy \(x+100=0\Leftrightarrow x=-100\)
TD
3
15 tháng 8 2019
\(\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+8}{92}+1\)
\(\Rightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)
\(\Rightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)
\(\Rightarrow\left(x+100\right).\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
\(\Rightarrow x+100=0\)
\(\Rightarrow x=-100\)
Vậy : \(x=-100\)
15 tháng 8 2019
\(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
\(\Leftrightarrow\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}+\frac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
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 !!!
T
1
MA
3 tháng 2 2018
\(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
\(\Leftrightarrow\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)-2=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)-2\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right)\times\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
\(\Leftrightarrow x+100=0\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0\right)\)
\(\Leftrightarrow x=-100\)
Vậy......
NK
6
15 tháng 5 2015
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}
SG
2
24 tháng 4 2017
\(\dfrac{x+2}{98}+\dfrac{x+4}{96}=\dfrac{x+6}{94}+\dfrac{x+8}{92}\)
\(\Leftrightarrow\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+4}{96}+1\right)=\left(\dfrac{x+6}{94}+1\right)+\left(\dfrac{x+8}{92}+1\right)\)
\(\Leftrightarrow\dfrac{x+100}{98}+\dfrac{x+100}{96}-\dfrac{x+100}{94}+\dfrac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\right)=0\)
Vì : \(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\ne0\)
\(\Leftrightarrow x+100=0\Leftrightarrow x=-100\)
Vậy ...............
24 tháng 4 2017
Võ Đông Anh Tuấn Thế nhỡ \(\left(\dfrac{1}{98}+\dfrac{1}{96}-\dfrac{1}{94}-\dfrac{1}{92}\right)=0\) thì x=-100 có phải nghiệm không
31 tháng 7 2022
1: \(\Leftrightarrow\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+4}{96}+1\right)=\left(\dfrac{x+6}{94}+1\right)+\left(\dfrac{x+8}{92}+1\right)\)
=>x+100=0
hay x=-100
2: \(\Leftrightarrow x\cdot\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{4}x+\dfrac{3}{4}=3-\dfrac{1}{3}x-\dfrac{2}{3}\)
=>3/4x+5/4=-1/3x+7/3
=>13/12x=13/12
hay x=1
DT
25 tháng 4 2019
a) \(\Leftrightarrow\frac{x+2}{98}+\frac{x+4}{96}+2=\frac{x+6}{94}+\frac{x+8}{92}+2\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}+\frac{1}{94}+\frac{1}{92}\right)=0\)
\(\Leftrightarrow x+100=0\Leftrightarrow x=-100\)
b)\(\Leftrightarrow\frac{x-12}{77}+\frac{x-11}{78}-2=\frac{x-74}{15}+\frac{x-73}{16}-2\)
\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}-\frac{x-89}{15}-\frac{x-89}{16}=0\)
\(\Leftrightarrow\left(x-89\right)\left(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\right)=0\)
\(\Leftrightarrow x-89=0\Leftrightarrow x=89\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 4 2020
\(\Leftrightarrow\frac{x+8}{92}+1+\frac{x+7}{93}+1+\frac{x+6}{94}+1\ge\frac{x+2}{98}+1+\frac{x+3}{97}+1+\frac{x+4}{96}+1\)
\(\Leftrightarrow\frac{x+100}{92}+\frac{x+100}{93}+\frac{x+100}{94}\ge\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{92}-\frac{1}{98}+\frac{1}{93}-\frac{1}{97}+\frac{1}{94}-\frac{1}{96}\right)\ge0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{6}{92.98}+\frac{4}{93.97}+\frac{2}{94.96}\right)\ge0\)
\(\Leftrightarrow x+100\ge0\Rightarrow x\ge-100\)
\(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
\(\Leftrightarrow\frac{x+2}{98}+1+\frac{x+4}{96}+1=\frac{x+6}{94}+1+\frac{x+8}{92}+1\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
Vì \(\frac{1}{92}>\frac{1}{94}>\frac{1}{96}>\frac{1}{98}\)nên \(\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)< 0\)
\(\Rightarrow x+100=0\Leftrightarrow x=-100\)
\(\Leftrightarrow\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0.Ma:\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}< 0\Rightarrow x=-100\)