Giải phương trình \(\frac{x+1}{99}+\frac{x+3}{97}+\frac{x+5}{95}=\frac{x+7}{93}+\frac{x+9}{91}+\frac{x+11}{89}\)
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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
13 tháng 2 2018
Ta có :
\(\frac{x+1}{65}+\frac{x+3}{63}< \frac{x+5}{61}+\frac{x+7}{59}\)
\(\Leftrightarrow\)\(\left(\frac{x+1}{65}+1\right)+\left(\frac{x+3}{63}+1\right)< \left(\frac{x+5}{61}+1\right)+\left(\frac{x+7}{59}+1\right)\)
\(\Leftrightarrow\)\(\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+5}{61}-\frac{x+7}{59}< 0\)
\(\Leftrightarrow\)\(\left(x+66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)< 0\)
Vì \(\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)< 0\)
\(\Rightarrow\)\(x+66>0\)
\(\Rightarrow\)\(x>-66\)
Vậy \(x>-66\)
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
TN
24 tháng 11 2016
\(\frac{x+1}{94}+\frac{x+2}{93}+\frac{x+3}{92}=\frac{x+4}{91}+\frac{x+5}{90}+\frac{x+6}{89}\)
\(\Leftrightarrow\frac{x+1}{94}+1+\frac{x+2}{93}+1+\frac{x+3}{92}+1=\frac{x+4}{91}+1+\frac{x+5}{90}+1+\frac{x+6}{89}+1\)
\(\Leftrightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}=\frac{x+95}{91}+\frac{x+95}{90}+\frac{x+95}{89}\)
\(\Leftrightarrow\frac{x+95}{94}+\frac{x+95}{93}+\frac{x+95}{92}-\frac{x+95}{91}-\frac{x+95}{90}-\frac{x+95}{89}=0\)
\(\Leftrightarrow\left(x+95\right)\left(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\right)=0\)
\(\Leftrightarrow x+95=0\).Do \(\frac{1}{94}+\frac{1}{93}+\frac{1}{92}-\frac{1}{91}-\frac{1}{90}-\frac{1}{89}\ne0\)
\(\Leftrightarrow x=-95\)
TT
10 tháng 1 2017
(x+1)/94 + ( x+2)/93 + ( x+3)/92.......
= ................ + ( x+6)/89
<=> (x+1)/94 + 1 + ( x+2)/93 +1 .........
=.............. cộng 1 nhá
<=> (x+95)/94 + ( x+96) / 93 + ( x+95)/92
= ( x+95)/91 + ( x+95)/90 + ( x+95)/89
<=> ( x+95) ( 1/94 +1/93 +1/92 )
= ( x+95) ( 1/91 +1/90 +1/89)
<=> ( x+95) ( 1/94 +1/93 +1/92 - 1/91 - 1/90 - 1/89 )
<=> x+95 =0
<=>x = -95
Vậy :x = -95
LN
26 tháng 1 2018
c, Trừ hai vế cho 6
Vế trái thì lấy từng số hạng trừ 1 là được
23 tháng 4 2020
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: \(\frac{x-1}{99}+\frac{x-3}{97}+\frac{x-5}{95}+\frac{x-7}{93}+\frac{x-95}{5}+x=105\)
\(\Leftrightarrow\frac{x-1}{99}+\frac{x-3}{97}+\frac{x-5}{95}+\frac{x-7}{93}+\frac{x-95}{5}+x-105=0\)
\(\Leftrightarrow\frac{x-1}{99}-1+\frac{x-3}{97}-1+\frac{x-5}{95}-1+\frac{x-7}{93}-1+\frac{x-95}{5}-1+x-100=0\)
\(\Leftrightarrow\frac{x-100}{99}+\frac{x-100}{97}+\frac{x-100}{95}+\frac{x-100}{93}+\frac{x-100}{5}+\frac{x-100}{1}=0\)
\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}+\frac{1}{93}+\frac{1}{5}+1\right)=0\)
mà \(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}+\frac{1}{93}+\frac{1}{5}+1\ne0\)
nên x-100=0
hay x=100
Vậy: x=100
25 tháng 2 2019
\(\frac{x+5}{95}+\frac{x+3}{97}+\frac{x+1}{99}=\frac{x+15}{85}+\frac{x+20}{80}+\frac{x+25}{75}.\)
\(\frac{x+5}{95}+1+\frac{x+3}{97}+1+\frac{x+1}{99}+1-\frac{x+15}{85}-1-\frac{x+20}{80}-1-\frac{x+25}{75}-1=0\)
\(\frac{x+100}{95}+\frac{x+100}{97}+\frac{x+100}{99}-\frac{x+100}{85}-\frac{x+100}{80}-\frac{x+100}{75}=0\)
\(\left(x+100\right).\left(\frac{1}{95}+\frac{1}{97}+\frac{1}{99}-\frac{1}{85}-\frac{1}{80}-\frac{1}{75}\right)=0\)
\(\Rightarrow x+100=0\Rightarrow x=-100\)
\(\frac{1}{95}+\frac{1}{97}+\frac{1}{99}-\frac{1}{85}-\frac{1}{80}-\frac{1}{75}\ne0\)
2 tháng 2 2017
(x+1)/99+(x+2)/98+(x+3)/97=(x+4)/96+(x+5)/95+(x+6)/94
[(x+1)/99 +1]+[(x+2)/98 +1]+[(x+3)/97 +1]-3=[(x+4)/96 +1]+[(x+5)/95 +1]+[(x+6)/94 +1]-3
[(x+1+99)/99+(x+2+98)/98+(x+3+97)/97]-3=[(x+4+96)/96+(x+5+95)/95+(x+6+94)/94]-3
(x+100)/99+(x+100)/98+(x+100)/97=(x+100)/96+(x+100)/95+(x+100)/94
(x+100)(1/99+1/98+1/97)=(x+100)(1/96+1/95+1/94)
(x+100)(1/99+1/98+1/97)-(x+100)(1/96+1/95+1/94)=0
(x+100)(1/99+1/98+1/97-1/96-1/95-1/94)=0
Ma : 1/99+1/98+1/97-1/96-1/95-1/94 \(\ne\)0
=>x+100=0
=>x=-100
k mk nha khong hieu noi mk nha.
2 tháng 2 2017
1/3x-1/2=(3/5-4x)15/7
1/3x-1/2=9/7-60/7x
1/3x+60/7x=1/2+9/7
187/21x=25/14
x=75/374
k mk nha ban.
TX
1
31 tháng 8 2016
\(\frac{x-11}{95}+\frac{x-13}{93}=\frac{x-15}{91}+\frac{x-17}{89}\) => \(\frac{x-11}{95}-1+\frac{x-13}{93}-1=\frac{x-15}{91}-1+\frac{x-17}{89}-1\)
=>\(\frac{x-106}{95}+\frac{x-106}{93}=\frac{x-106}{91}+\frac{x-106}{89}\)
=>\(\left(\frac{1}{95}+\frac{1}{93}\right)\left(x-106\right)-\left(\frac{1}{91}+\frac{1}{89}\right)\left(x-106\right)=0\)
<=>\(\left[\left(\frac{1}{95}+\frac{1}{93}\right)-\left(\frac{1}{91}+\frac{1}{89}\right)\right]\left(x-106\right)=0\).Vì\(\frac{1}{95}< \frac{1}{91};\frac{1}{93}< \frac{1}{89}\) nên\(\frac{1}{95}+\frac{1}{93}< \frac{1}{91}+\frac{1}{89}\)
=>\(\left(\frac{1}{95}+\frac{1}{93}\right)-\left(\frac{1}{91}+\frac{1}{89}\right)< 0\) hay khác 0.Vậy x - 106 = 0, tìm được x = 106
\(\frac{x+1}{99}+\frac{x+3}{97}+\frac{x+5}{95}=\frac{x+7}{93}+\frac{x+9}{91}+\frac{x+11}{89}\)
\(\Leftrightarrow\frac{x+1}{99}+1+\frac{x+3}{97}+1+\frac{x+5}{95}+1=\frac{x+7}{93}+1+\frac{x+9}{91}+1+\frac{x+11}{89}+1\)
\(\Leftrightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)=\left(\frac{x+7}{93}+1\right)+\left(\frac{x+9}{91}+1\right)+\left(\frac{x+11}{89}+1\right)\)
\(\Leftrightarrow\frac{x+1+99}{99}+\frac{x+3+97}{97}+\frac{x+5+95}{95}=\frac{x+7+93}{93}+\frac{x+9+91}{91}+\frac{x+11+89}{89}\)
\(\Leftrightarrow\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}=\frac{x+100}{93}+\frac{x+100}{91}+\frac{x+100}{89}\)
\(\Leftrightarrow\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}-\frac{x+100}{93}-\frac{x+100}{91}-\frac{x+100}{89}=0\)
\(\Leftrightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{93}-\frac{1}{91}-\frac{1}{98}\right)=0\)
\(\Leftrightarrow x+100=0\)
\(\Leftrightarrow x=-100\)
\(\frac{x+1}{99}+\frac{x+3}{97}+\frac{x+5}{95}=\frac{x+7}{93}+\frac{x+9}{91}+\frac{x+11}{89}\)
\(\Leftrightarrow\frac{x+1}{99}+1+\frac{x+3}{97}+1+\frac{x+5}{95}+1=\frac{x+7}{93}+1+\frac{x+9}{91}+1+\frac{x+11}{89}+1\)
\(\Leftrightarrow\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}=\frac{x+100}{93}+\frac{x+100}{91}+\frac{x+100}{89}\)
\(\Leftrightarrow x+100=0\)
\(\Leftrightarrow x=-100\)