Tìm số hữu tỉ , biết rằng: \(\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\)
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21 tháng 8 2017
\(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
21 tháng 8 2017
a, \(\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}=\dfrac{x+1}{13}+\dfrac{x+1}{14}\)
\(\Rightarrow\dfrac{x+1}{10}+\dfrac{x+1}{11}+\dfrac{x+1}{12}-\dfrac{x+1}{13}-\dfrac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\right)=0\)
Do \(\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}-\dfrac{1}{13}-\dfrac{1}{14}\ne0\)
\(\Rightarrow x+1=0\Rightarrow x=-1\)
Vậy x = -1
b, \(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
Vì \(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
Vậy...
KT
1
25 tháng 5 2021
`(2-x)/2002-1=(1-x)/2003-x/2004`
`<=>(2-x)/2002-1+(x-1)/2003+x/2004=0`(chuyển vế)
`<=>(2-x)/2002+1+(x-1)/2003-1+x/2004-1=0`
`<=>(2004-x)/2002+(x-2004)/2003+(x-2004)/2004=0`
`<=>(x-2004)(1/2003+1/2004-1/2002)=0`
`<=>x=2004` do `1/2003+1/2004-1/2002 ne 0`
Vậy `x=2004`
KT
1
MD
1
28 tháng 3 2017
\(\dfrac{x+1}{2004}+\dfrac{x+2}{2003}+\dfrac{x+3}{2002}+35=2^5\)
\(pt\Leftrightarrow\dfrac{x+1}{2004}+\dfrac{x+2}{2003}+\dfrac{x+3}{2002}+3=0\)
\(\Leftrightarrow\dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1+\dfrac{x+3}{2002}+1=0\)
\(\Leftrightarrow\dfrac{x+1}{2004}+\dfrac{2004}{2004}+\dfrac{x+2}{2003}+\dfrac{2003}{2003}+\dfrac{x+3}{2002}+\dfrac{2002}{2002}=0\)
\(\Leftrightarrow\dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}+\dfrac{x+2005}{2002}=0\)
\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}+\dfrac{1}{2002}\right)=0\)
\(\Rightarrow x+2005=0\). Do \(\dfrac{1}{2004}+\dfrac{1}{2003}+\dfrac{1}{2002}\ne0\)
\(\Rightarrow x=-2005\)
RS
0
5 tháng 3 2022
a, \(\dfrac{x-45}{55}-1+\dfrac{x-47}{53}-1=\dfrac{x-55}{45}-1+\dfrac{x-53}{47}-1\)
\(\Leftrightarrow\dfrac{x-100}{55}+\dfrac{x-100}{53}=\dfrac{x-100}{45}+\dfrac{x-100}{47}\)
\(\Leftrightarrow\left(x-100\right)\left(\dfrac{1}{55}+\dfrac{1}{53}-\dfrac{1}{45}-\dfrac{1}{47}\ne0\right)=0\Leftrightarrow x=100\)
b, \(\dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1=\dfrac{x+3}{2002}+1+\dfrac{x+4}{2001}+1\)
\(\Leftrightarrow\dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}=\dfrac{x+2005}{2002}+\dfrac{x+2005}{2001}\)
\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}-\dfrac{1}{2002}-\dfrac{1}{2001}\ne0\right)=0\Leftrightarrow x=-2005\)
DT
5 tháng 3 2022
a. lấy mỗi phân số e cộng vs 2 là bt làm ra liền
b, - 1 hoặc + 1 vs mỗi phân số nha
UC
2
15 tháng 4 2017
Câu hỏi của Lan Anh - Toán lớp 8 | Học trực tuyến, bạn sửa số 9 thành số 0 ở VP ở dòng 2 nhé
15 tháng 4 2017
ta có \(\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\)
<=>\(\dfrac{2-x}{2002}+1-2=\dfrac{1-x}{2003}+1+1-\dfrac{x}{2004}-2\)
<=>\(\dfrac{2004-x}{2002}=\dfrac{2004-x}{2003}+\dfrac{2004-x}{2004}\)
<=>\(\dfrac{2004-x}{2002}-\dfrac{2004-x}{2003}-\dfrac{2004-x}{2004}=0\)
<=>\(\left(2004-x\right)\left(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\right)=0\)
Vì\(\dfrac{1}{2002}-\dfrac{1}{2003}-\dfrac{1}{2004}\ne0\Rightarrow2004-x=0\Rightarrow x=2004\)
Vậy nghiệm của phương trình là x=2004
BN
2
3 tháng 7 2017
\(\dfrac{x}{2000}+\dfrac{x+1}{2001}+\dfrac{x+2}{2002}+\dfrac{x+3}{2003}+\dfrac{x+4}{2004}=5\)
\(\Leftrightarrow\dfrac{x}{2000}-1+\dfrac{x+1}{2001}-1+\dfrac{x+2}{2002}-1+\dfrac{x+3}{2003}-1+\dfrac{x+4}{2004}-1=0\)
\(\Leftrightarrow\dfrac{x-2000}{2000}+\dfrac{x-2000}{2001}+\dfrac{x-2000}{2002}+\dfrac{x-2000}{2003}+\dfrac{x-2000}{2004}=0\)
\(\Leftrightarrow\left(x-2000\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}\right)=0\)
Mà \(\dfrac{1}{2000}+\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}+\dfrac{1}{2004}>0\)
\(\Leftrightarrow x-2000=0\Leftrightarrow x=2000\)
Vậy x = 2000
\(\Leftrightarrow\dfrac{2-x}{2002}=\dfrac{1-x}{2003}+1-\dfrac{x}{2004}\)
\(\Leftrightarrow\left(\dfrac{2-x}{2002}+1\right)=\left(\dfrac{1-x}{2003}+1\right)+\left(1-\dfrac{x}{2004}\right)\)
=>2004-x=0
=>x=2004