Giải phương trình: x − 1 99 + x − 99 + x − 3 97 + x − 7 93 + x − 5 95 + x − 95 5 = 6 .
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CM
Cao Minh Tâm
11 tháng 12 2018
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NH
0
KN
11 tháng 2 2020
\(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x+3}{97}+\frac{x+4}{96}\)
\(\Rightarrow\frac{x+1}{99}+1+\frac{x+2}{98}+1=\frac{x+3}{97}+1+\frac{x+4}{96}+1\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}=\frac{x+100}{97}+\frac{x+100}{96}\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}-\frac{x+100}{97}-\frac{x+100}{96}=0\)
\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)
Dễ thấy \(\left(\frac{1}{99}< \frac{1}{98}< \frac{1}{97}< \frac{1}{96}\right)\)nên \(\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)\ne0\)
\(\Rightarrow x+100=0\Rightarrow x=-100\)
Vậy x = -100
\(\frac{109-x}{91}+\frac{107-x}{93}+\frac{105-x}{95}+\frac{103-x}{97}+4=0\)
\(\Rightarrow\frac{109-x}{91}+1+\frac{107-x}{93}+1+\frac{105-x}{95}+1+\frac{103-x}{97}+1=0\)
\(\Rightarrow\frac{200-x}{91}+\frac{200-x}{93}+\frac{200-x}{95}+\frac{200-x}{97}=0\)
\(\Rightarrow\left(200-x\right)\left(\frac{1}{91}+\frac{1}{93}-\frac{1}{95}-\frac{1}{97}\right)=0\)
Dễ thấy \(\left(\frac{1}{91}>\frac{1}{93}>\frac{1}{95}>\frac{1}{97}\right)\)nên \(\left(\frac{1}{91}+\frac{1}{93}-\frac{1}{95}-\frac{1}{97}\right)\ne0\)
\(\Rightarrow200-x=0\Rightarrow x=200\)
Vậy x = 200
NH
0
18 tháng 6 2022
\(\Leftrightarrow\left(\dfrac{x-1}{99}-1\right)+\left(\dfrac{x-99}{1}-1\right)+\left(\dfrac{x-3}{97}-1\right)+\left(\dfrac{x-7}{93}-1\right)+\left(\dfrac{x-5}{95}-1\right)+\left(\dfrac{x-95}{5}-1\right)=0\)=>x-100=0
hay x=100
PN
2
KN
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
NH
1
KN
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
D
22 tháng 1 2022
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
11 tháng 2 2020
b, \(\frac{x+1}{99}+1+\frac{x+2}{98}+1=\frac{x+3}{97}+1+\frac{x+4}{96}+1\)
\(\frac{x+200}{99}+\frac{x+200}{98}=\frac{x+200}{97}+\frac{x+200}{96}\)
\(\frac{x+200}{99}+\frac{x+200}{98}-\frac{x+200}{97}-\frac{x+200}{96}=0\)
\(\left(x+200\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\right)=0\)
mà\(\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}\ne0\)
==> x+200=0
<=>x=-200
Vậy nghiệm của phương trình là x=-200
11 tháng 2 2020
c, \(\frac{109-x}{91}+1+\frac{107-x}{93}+1+\frac{105-x}{95}+1+\frac{103-x}{97}+1=0\)
\(\frac{200-x}{91}+\frac{200-x}{93}+\frac{200-x}{95}+\frac{200-x}{97}=0\)
\(\left(200-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=0\)
mà \(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\ne0\)
==>200-x=0
<=>x=200
vậy nghiệm của pt là x=200
