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HT
1
10 tháng 4 2018
\(\frac{x+2}{98}+\frac{x+3}{97}=\frac{x+4}{96}+\frac{x+5}{95}\)
\(\Leftrightarrow\frac{x+2}{98}+1+\frac{x+3}{97}+1=\frac{x+4}{96}+1+\frac{x+5}{95}+1\)
\(\Leftrightarrow\frac{x+2+98}{98}+\frac{x+3+97}{97}=\frac{x+4+96}{96}+\frac{x+5+95}{95}\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{96}-\frac{x+100}{95}=0\)
\(\Leftrightarrow\left(x+100\right).\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)
Vì \(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\)
=> x + 100 = 0
=> x = -100
Vậy x = -100
6 tháng 3 2020
\(\frac{x+2}{98}+1+\frac{x+3}{97}+1=\frac{x+4}{96}+1+\frac{x+5}{95}+1\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{96}-\frac{x+100}{95}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)
\(\Leftrightarrow x+100=0\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\right)\)
<=> x=-100
6 tháng 3 2020
ko chép đề nhé
\(\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95} \)
=> \((x+100)(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95})=0\)
vì \((\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}) khác 0\)
=>\(x+100=0\)
=>x=-100
PV
1
9 tháng 3 2022
sửa đề đến đây thôi bạn nhé, do nếu thêm vào thì mình cũng ko biết có quy luật gì nữa :<
\(\dfrac{x-1}{99}-1+\dfrac{x-3}{97}-1+\dfrac{x-5}{95}-1=\dfrac{x-2}{98}-1+\dfrac{x-4}{96}-1\)
\(\Leftrightarrow\dfrac{x-100}{99}+\dfrac{x-100}{97}+\dfrac{x-100}{95}=\dfrac{x-100}{98}+\dfrac{x-100}{96}\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}-\dfrac{1}{98}-\dfrac{1}{96}\ne0\right)=0\Leftrightarrow x=100\)
5 tháng 7 2021
Pt\(\Leftrightarrow\dfrac{x+98}{2}+1+\dfrac{x+96}{4}+1+\dfrac{x+65}{35}+1=\dfrac{x+3}{97}+1+\dfrac{x+5}{95}+1+\dfrac{x+49}{51}+1\)
\(\Leftrightarrow\dfrac{x+100}{2}+\dfrac{x+100}{4}+\dfrac{x+100}{35}-\dfrac{x+100}{97}-\dfrac{x+100}{95}-\dfrac{x+100}{51}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{35}-\dfrac{1}{97}-\dfrac{1}{35}-\dfrac{1}{51}\right)=0\)
\(\Leftrightarrow x+100=0\Leftrightarrow x=-100\)
Vậy...
25 tháng 5 2021
\(\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{x+3}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)
\(\Leftrightarrow\dfrac{x+2+98}{98}+\dfrac{x+3+97}{97}-\dfrac{x+4+96}{96}-\dfrac{x+5+95}{95}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{98}+\dfrac{1}{97}-\dfrac{1}{96}-\dfrac{1}{95}\right)=0\)
\(\Rightarrow x+100=0\)
\(\Leftrightarrow x=-100\)
Vậy...
NV
2
6 tháng 1 2018
Ta có x+1/99 + x+2/98 + x+3/97 = x+4/96 + x+5/95 + x+10/90
=> x+1/99 + x+2/98 + x+3/97 - x+4/96 - x+5/95 - x+10/90=0
=> (x+1/99 + 1) + (x+2/98 + 1) + (x+3/97 +1) - (x+4/96 + 1) - (x+5/95 + 1) - (x+10/90 + 1) = 0
=> x+100/99 + x+100/98 + x+100/97 - x+100/96 - x+100/95 - x+100/90 =0
=> (x+100)(1/99+1/98+1/97-1/96-1/95-1/90) = 0
Mà 1/99+1/98+1/97-1/96-1/95-1/90 khác 0
=> x+100=0 => x=-100
Vậy phương trình có nghiệm là x=-100
6 tháng 1 2018
\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=\frac{x+4}{96}+\frac{x+5}{95}+\frac{x+10}{99}\)
\(\Leftrightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1-\left(\frac{x+4}{96}+1+\frac{x+5}{95}+1+\frac{x+10}{99}+1\right)=0\)
\(\Leftrightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}-\left(\frac{x+100}{96}+\frac{x+100}{95}+\frac{x+100}{90}\right)=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}-\frac{1}{90}\right)=0\)
Mà\(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}-\frac{1}{90}\ne0\)
\(\Rightarrow x+100=0\Rightarrow x=-100\)
2 tháng 2 2017
\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=\frac{x+4}{96}+\frac{x+5}{95}+\frac{x+6}{94}\)
\(\Leftrightarrow\)\(\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1=\frac{x+4}{96}+1+\frac{x+5}{95}+\frac{x+6}{94}+1\)
\(\Leftrightarrow\)\(\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}+\frac{x+100}{94}\)
\(\Leftrightarrow\)(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
2 tháng 2 2017
\(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=\frac{x+4}{96}+\frac{x+5}{95}+\frac{x+6}{94}\)
\(\Rightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)+\left(\frac{x+5}{95}+1\right)+\left(\frac{x+6}{94}+1\right)\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}+\frac{x+100}{94}\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{96}-\frac{x+100}{95}-\frac{x+100}{94}=0\)
\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}-\frac{1}{94}\right)=0\)
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 \(x=-100\)
DT
10 tháng 7 2017
\(\left(\dfrac{x+2}{98}+1\right)+\left(\dfrac{\left(x+3\right)}{97}+1\right)=\left(\dfrac{x+4}{96}+1\right)+\left(\dfrac{x+5}{95}+1\right)\)
\(\Rightarrow\dfrac{x+100}{98}+\dfrac{x+100}{97}=\dfrac{x+100}{96}+\dfrac{x+100}{95}\)
\(\Rightarrow\left(x+100\right).\left(\dfrac{1}{98}+\dfrac{1}{97}\right)=\left(x+100\right).\left(\dfrac{1}{96}+\dfrac{1}{95}\right)\)
\(\Rightarrow\dfrac{1}{98}+\dfrac{1}{97}=\dfrac{1}{96}+\dfrac{1}{95}\) (Vô lí)
Vậy PTVN
(=)(x+2)/98+1+(x+3)/97+1=(x+4)/96+1+(x+5)/95+1
(=)(x+100)/98+(x+100)/97=(x+100)/96+(x+100)/95
(=)(x+100)(1/98+1/97-1/96-1/95)=0
=)x+100=0
=)x=-100