Giải phương trình :
a) \(\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}=\frac{x+4}{96}+\frac{x+5}{95}+\frac{x+6}{94}\)
b) \(\frac{1}{3}.x-\frac{1}{2}=\left(\frac{3}{5}-4.x\right).\frac{15}{7}\)
K
Khách
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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 \(x=-100\)
T
7 tháng 1 2016
a) \(\left(\frac{x+2}{98}+1\right)+\left(\frac{x+3}{97}+1\right)=\left(\frac{x+4}{96}+1\right)+\left(\frac{x+5}{95}+1\right)\)
=> \(\frac{x+2+98}{98}+\frac{x+3+97}{97}=\frac{x+4+96}{96}+\frac{x+5+95}{95}\)
=> \(\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}\)
=> \(\frac{x+100}{98}+\frac{x+100}{97}-\frac{x+100}{96}-\frac{x+100}{95}=0\)
=> \(\left(x+100\right)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)
Ta có : \(\frac{1}{98}+\frac{1}{97}\ne\frac{1}{96}+\frac{1}{95}\) => \(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\)
=> \(x+100=0\)
=> \(x=-100\)
MT
13 tháng 1 2016
Nhiều thế bạn tách từng câu ra mik giải cho (olm ko dc trừ điểm câu này của e)
13 tháng 1 2016
Phần b bạn tự làm nhé, chỉ cần quy đồng lên lấy MC = 105 là được mà
Phần a mình giải ntn:
PT \(\Leftrightarrow\) \(\frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)
\(\Leftrightarrow\)\(x+100=0\)
\(\Leftrightarrow x=-100\)
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
29 tháng 3 2020
Câu 6 :
a, Ta có : \(x+\frac{2x+\frac{x-1}{5}}{3}=1-\frac{3x-\frac{1-2x}{3}}{5}\)
=> \(\frac{15x}{15}+\frac{5\left(2x+\frac{x-1}{5}\right)}{15}=\frac{15}{15}-\frac{3\left(3x-\frac{1-2x}{3}\right)}{15}\)
=> \(15x+5\left(2x+\frac{x-1}{5}\right)=15-3\left(3x-\frac{1-2x}{3}\right)\)
=> \(15x+10x+\frac{5\left(x-1\right)}{5}=15-9x+\frac{3\left(1-2x\right)}{3}\)
=> \(15x+10x+x-1=15-9x+1-2x\)
=> \(15x+10x+x-1-15+9x-1+2x=0\)
=> \(37x-17=0\)
=> \(x=\frac{17}{37}\)
Vậy phương trình trên có nghiệm là \(S=\left\{\frac{17}{37}\right\}\)
Bài 7 :
a, Ta có : \(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
=> \(\frac{x-23}{24}+\frac{x-23}{25}-\frac{x-23}{26}-\frac{x-23}{27}=0\)
=> \(\left(x-23\right)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)
=> \(x-23=0\)
=> \(x=23\)
Vậy phương trình trên có nghiệm là \(S=\left\{23\right\}\)
c, Ta có : \(\frac{x+1}{2004}+\frac{x+2}{2003}=\frac{x+3}{2002}+\frac{x+4}{2001}\)
=> \(\frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)
=> \(\frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)
=> \(\frac{x+2005}{2004}+\frac{x+2005}{2003}-\frac{x+2005}{2002}-\frac{x+2005}{2001}=0\)
=> \(\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
=> \(x+2005=0\)
=> \(x=-2005\)
Vậy phương trình trên có nghiệm là \(S=\left\{-2005\right\}\)
e, Ta có : \(\frac{x-45}{55}+\frac{x-47}{53}=\frac{x-55}{45}+\frac{x-53}{47}\)
=> \(\frac{x-45}{55}-1+\frac{x-47}{53}-1=\frac{x-55}{45}-1+\frac{x-53}{47}-1\)
=> \(\frac{x-100}{55}+\frac{x-100}{53}=\frac{x-100}{45}+\frac{x-100}{47}\)
=> \(\frac{x-100}{55}+\frac{x-100}{53}-\frac{x-100}{45}-\frac{x-100}{47}=0\)
=> \(\left(x-100\right)\left(\frac{1}{55}+\frac{1}{53}-\frac{1}{45}-\frac{1}{47}\right)=0\)
=> \(x-100=0\)
Vậy phương trình trên có nghiệm là \(S=\left\{100\right\}\)
AH
Akai Haruma
Giáo viên
7 tháng 2 2020
a)
\(\frac{x-23}{24}+\frac{x-23}{25}=\frac{x-23}{26}+\frac{x-23}{27}\)
\(\Leftrightarrow (x-23)\left(\frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\right)=0\)
Dễ thấy: \(\frac{1}{24}>\frac{1}{26}; \frac{1}{25}>\frac{1}{27}\Rightarrow \frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}>0\)
$\Rightarrow \frac{1}{24}+\frac{1}{25}-\frac{1}{26}-\frac{1}{27}\neq 0$
Do đó $x-23=0\Rightarrow x=23$
b)
PT \(\Leftrightarrow \frac{x+100}{98}+\frac{x+100}{97}=\frac{x+100}{96}+\frac{x+100}{95}\)
\(\Leftrightarrow (x+100)\left(\frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)
Dễ thấy: $\frac{1}{98}< \frac{1}{96}; \frac{1}{97}< \frac{1}{95}$
$\Rightarrow \frac{1}{98}+\frac{1}{97}-\frac{1}{96}-\frac{1}{95}< 0$ hay khác $0$
$\Rightarrow x+100=0\Rightarrow x=-100$
AH
Akai Haruma
Giáo viên
7 tháng 2 2020
c)
PT \(\Leftrightarrow \frac{x+1}{2004}+1+\frac{x+2}{2003}+1=\frac{x+3}{2002}+1+\frac{x+4}{2001}+1\)
\(\Leftrightarrow \frac{x+2005}{2004}+\frac{x+2005}{2003}=\frac{x+2005}{2002}+\frac{x+2005}{2001}\)
\(\Leftrightarrow (x+2005)\left(\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}\right)=0\)
Dễ thấy $\frac{1}{2004}+\frac{1}{2003}-\frac{1}{2002}-\frac{1}{2001}<0$ hay khác $0$
Do đó $x+2005=0\Rightarrow x=-2005$
d)
PT \(\Leftrightarrow \frac{201-x}{99}+1+\frac{203-x}{97}+1+\frac{205-x}{96}+1=0\)
\(\Leftrightarrow \frac{300-x}{99}+\frac{300-x}{97}+\frac{300-x}{96}=0\)
\(\Leftrightarrow (300-x)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{96}\right)=0\)
Dễ thấy \(\frac{1}{99}+\frac{1}{97}+\frac{1}{96}>0\) hay khác $0$
Do đó $300-x=0\Rightarrow x=300$
21 tháng 2 2016
\(\frac{x+1}{99}+\frac{x+3}{97}+\frac{x+5}{95}=\frac{x+2}{98}+\frac{x+4}{96}+\frac{x+6}{94}\)
\(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)=\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)+\left(\frac{x+6}{94}+1\right)\)
\(\left(\frac{x+1}{99}+\frac{99}{99}\right)+\left(\frac{x+3}{97}+\frac{97}{97}\right)+\left(\frac{x+5}{95}+\frac{95}{95}\right)=\left(\frac{x+2}{98}+\frac{98}{98}\right)+\left(\frac{x+4}{96}+\frac{96}{96}\right)+\left(\frac{\left(x+6\right)}{94}+\frac{94}{94}\right)\)
\(\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}=\frac{x+100}{92}+\frac{x+100}{94}+\frac{x+100}{96}\)
\(\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}-\frac{x+100}{92}-\frac{x+100}{94}-\frac{x+100}{96}=0\)
\(\left(x+100\right).\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{92}-\frac{1}{94}-\frac{1}{96}\right)=0\)
\(Mà\) \(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{92}-\frac{1}{94}-\frac{1}{96}\ne0\)
Nên x+ 100 = 0
x = 0 - 100 = -100
Vậy x= -100
HP
21 tháng 2 2016
cộng 1 vào mỗi tỉ số,ta được:
\(\left(\frac{x+1}{99}+1\right)+\left(\frac{x+3}{97}+1\right)+\left(\frac{x+5}{95}+1\right)=\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)+\left(\frac{x+6}{94}+1\right)\)\(\Rightarrow\frac{x+1+99}{99}+\frac{x+3+97}{97}+\frac{x+5+95}{95}=\frac{x+2+98}{98}+\frac{x+4+96}{96}+\frac{x+6+94}{94}\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}=\frac{x+100}{98}+\frac{x+100}{96}+\frac{x+100}{94}\)
\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{97}+\frac{x+100}{95}-\frac{x+100}{98}-\frac{x+100}{96}-\frac{x+100}{94}=0\)
\(\Rightarrow\left(x+100\right)\left(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{98}-\frac{1}{96}-\frac{1}{94}\right)\)
Vì \(\frac{1}{99}+\frac{1}{97}+\frac{1}{95}-\frac{1}{98}-\frac{1}{96}-\frac{1}{94}\ne0\)
=>x+100=0
=>x=-100
Vậy x=-100
PT
28 tháng 9 2017
a) \(\dfrac{x+5}{5}+\dfrac{x+5}{7}+\dfrac{x+5}{9}=\dfrac{x+5}{11}+\dfrac{x+5}{13}\)
\(\Rightarrow\left(x+5\right)\left(\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{9}\right)=\left(x+5\right)\left(\dfrac{1}{11}+\dfrac{1}{13}\right)\)
\(\Rightarrow\dfrac{143}{315}\left(x+5\right)=\dfrac{24}{143}\left(x+5\right)\)
\(\Rightarrow\dfrac{143}{315}\left(x+5\right)-\dfrac{24}{143}\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(\dfrac{143}{315}-\dfrac{24}{143}\right)=0\)
\(\Rightarrow x+5=0\Rightarrow x=-5\)
b) \(\dfrac{x+2}{100}+\dfrac{x+3}{99}+\dfrac{x+4}{98}=\dfrac{x+5}{97}+\dfrac{x+6}{96}+\dfrac{x+7}{95}\)
\(\Rightarrow\)\(3+\dfrac{x+2}{100}+\dfrac{x+3}{99}+\dfrac{x+4}{98}=3+\dfrac{x+5}{97}+\dfrac{x+6}{96}+\dfrac{x+7}{95}\)
\(\Rightarrow\)\(1+\dfrac{x+2}{100}+1+\dfrac{x+3}{99}+1+\dfrac{x+4}{98}=1+\dfrac{x+5}{97}+1+\dfrac{x+6}{96}+1+\dfrac{x+7}{95}\)
\(\Rightarrow\)\(\dfrac{100}{100}+\dfrac{x+2}{100}+\dfrac{99}{99}+\dfrac{x+3}{99}+\dfrac{98}{98}+\dfrac{x+4}{98}=\dfrac{97}{97}+\dfrac{x+5}{97}+\dfrac{96}{96}+\dfrac{x+6}{96}+\dfrac{95}{95}+\dfrac{x+7}{95}\)\(\Rightarrow\)\(\dfrac{x+102}{100}+\dfrac{x+102}{99}+\dfrac{x+102}{98}=\dfrac{x+102}{97}+\dfrac{x+102}{96}+\dfrac{x+102}{95}\)
\(\Rightarrow\)\(\left(x+102\right)\left(\dfrac{1}{100}+\dfrac{1}{99}+\dfrac{1}{98}\right)=\left(x+102\right)\left(\dfrac{1}{97}+\dfrac{1}{96}+\dfrac{1}{95}\right)\)
\(\Rightarrow\)\(x+102=0\)
\(\Rightarrow x=-102\)
PT
28 tháng 9 2017
c) \(\left(x+2\right)-\left(x+3\right)>0\)
\(\Rightarrow x+2-x-3>0\Rightarrow-1>0\)
\(\Rightarrow x\in\varnothing\)
d) \(\left(x-5\right)\left(x+\dfrac{7}{3}\right)\ge0\)
TH1: \(\left\{{}\begin{matrix}x-5\ge0\\x+\dfrac{7}{3}\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge5\\x\ge\dfrac{-7}{3}\end{matrix}\right.\)
\(\Rightarrow x\ge\dfrac{-7}{3}\)
TH2: \(\left\{{}\begin{matrix}x-5\le0\\x+\dfrac{7}{3}\le0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\le5\\x\le\dfrac{-7}{3}\end{matrix}\right.\)
\(\Rightarrow x\le5\)
TH3: \(\left[{}\begin{matrix}x-5=0\\x+\dfrac{7}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-7}{3}\end{matrix}\right.\)
(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.
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.