\(\dfrac{x+1006}{1007}+\dfrac{x+1005}{1008}=\dfrac{x+1004}{1009}+\dfr...">
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3 tháng 3 2017

\(\frac{x+1006}{1007}+\frac{x+1005}{1008}=\frac{x+1004}{1009}+\frac{x+1003}{1010}\)

\(\Rightarrow\left(\frac{x+1006}{1007}+1\right)+\left(\frac{x+1005}{1008}+1\right)=\left(\frac{x+1004}{1009}+1\right)+\left(\frac{x+1003}{1010}+1\right)\)

\(\Rightarrow\frac{x+2013}{1007}+\frac{x+2013}{1008}=\frac{x+2013}{1009}+\frac{x+2013}{1010}\)

\(\Rightarrow\frac{x+2013}{1007}+\frac{x+2013}{1008}-\frac{x+2013}{1009}-\frac{x+2013}{1010}=0\)

\(\Rightarrow\left(x+2013\right)\left(\frac{1}{1007}+\frac{1}{1008}-\frac{1}{1009}-\frac{1}{1010}\right)=0\)

\(\frac{1}{1007}+\frac{1}{1008}-\frac{1}{1009}-\frac{1}{1010}\ne0\)

\(\Rightarrow x+2013=0\)

\(\Rightarrow x=-2013\)

Vậy x = -2013

3 tháng 3 2017

thks

25 tháng 3 2018

\(\dfrac{x-1001}{1006}+\dfrac{x-1003}{1004}+\dfrac{x-1005}{1002}+\dfrac{x-1007}{1000}=4\)

\(\Leftrightarrow\dfrac{x-1001}{1006}-1+\dfrac{x-1003}{1004}-1+\dfrac{x-1005}{1002}-1+\dfrac{x-1007}{1000}-1=0\)

\(\Leftrightarrow\dfrac{x-2007}{1006}+\dfrac{x-2007}{1004}+\dfrac{x-2007}{1002}+\dfrac{x-2007}{1000}=0\)

\(\Leftrightarrow\left(x-2007\right)\left(\dfrac{1}{1006}+\dfrac{1}{1004}+\dfrac{1}{1002}+\dfrac{1}{1000}=0\right)\)

\(\Leftrightarrow x-2007=0\)

\(\Leftrightarrow x=2007\)

Ta có: \(\dfrac{x+1006}{1007}+\dfrac{x+1005}{1008}=\dfrac{x+1004}{1009}+\dfrac{x+1003}{1010}\)

\(\Leftrightarrow\dfrac{x+1006}{1007}+1+\dfrac{x+1005}{1008}+1=\dfrac{x+1004}{1009}+1+\dfrac{x+1003}{1010}+1\)

\(\Leftrightarrow\dfrac{x+2013}{1007}+\dfrac{x+2013}{1008}=\dfrac{x+2013}{1009}+\dfrac{x+2013}{1010}\)

\(\Leftrightarrow\dfrac{x+2013}{1007}+\dfrac{x+2013}{1008}-\dfrac{x+2013}{1009}-\dfrac{x+2013}{1010}=0\)

\(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{1007}+\dfrac{1}{1008}-\dfrac{1}{1009}-\dfrac{1}{1010}\right)=0\)

mà \(\dfrac{1}{1007}+\dfrac{1}{1008}-\dfrac{1}{1009}-\dfrac{1}{1010}\ne0\)

nên x+2013=0

hay x=-2013

Vậy: S={-2013}

19 tháng 5 2018

\(\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}=7\)

\(\Leftrightarrow\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}-7=0\)

\(\Leftrightarrow\left(\dfrac{x-1009}{1001}-1\right)+\left(\dfrac{x-4}{1003}-2\right)+\left(\dfrac{x+2010}{1005}-4\right)=0\)

\(\Leftrightarrow\dfrac{x-2010}{1001}+\dfrac{x-2010}{1003}+\dfrac{x-2010}{1005}=0\)

\(\Leftrightarrow\left(x-2010\right)\left(\dfrac{1}{1001}+\dfrac{1}{1003}+\dfrac{1}{1005}\right)=0\)

\(\Leftrightarrow x-2010=0\)

\(\Rightarrow x=2010\)

Vậy....

21 tháng 7 2018

\(\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}=7\)

\(\dfrac{x-1009}{1001}+\dfrac{x-4}{1003}+\dfrac{x+2010}{1005}-7=0\)

\(\left(\dfrac{x-1009}{1001}-1\right)+\left(\dfrac{x-4}{1003}-2\right)+\left(\dfrac{x+2010}{1005}-4\right)=0\)

\(\dfrac{x-2010}{1001}+\dfrac{x-2010}{1003}+\dfrac{x-2010}{1005}=0\)

\(\left(x-2010\right)\left(\dfrac{1}{1001}+\dfrac{1}{1003}+\dfrac{1}{1005}\right)=0\)

\(x-2010=0\)

\(x=2010\)

Vậy x = 2010

NM
18 tháng 3 2022

ta có : 

\(\frac{x-1009}{1001}-1+\frac{x-4}{1003}-2+\frac{x+2010}{1005}-4=0\)

hay \(\frac{x-2010}{1001}+\frac{x-2010}{1003}+\frac{x-2010}{1005}=0\Leftrightarrow x-2010=0\)

hay x =2010

Vậy phương trình có nghiệm x = 2010

7 tháng 11 2019

Câu a : B-A= 10012 +10022 +10042 +10072 -10002-10032-10052-10062

=(10012-10002)+(10022-10032) (10042-10052)+(10072-10062)

HĐT số 3= (1001-1000)*(1001+1000)+(1002-1003)*(1002+1003)+(1004-1005)*(1004+1005)+(1007-1006)*(1007+1006)

=2001 -2005-2009+2013=0

vậy A=B

22 tháng 2 2019

\(a.\Leftrightarrow\frac{3\left(x-2\right)-\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\frac{-9}{\left(x+1\right)\left(x-2\right)}.DKXD:x\ne-1;x\ne2\)

\(\Rightarrow3x-6-x-1=-9\)

\(\Leftrightarrow2x=-2\)

\(\Leftrightarrow x=-1\)

\(b.\frac{\left(x-4\right)\left(x+1\right)+\left(x+4\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}.DKXDx\ne1;-1\)

\(\Rightarrow x^2+x-4x-4+x^2-x+4x-4=2x^2+2x-2x-2\)

\(\Leftrightarrow-6=0\left(voly\right)\)

vay \(S=\varnothing\)

22 tháng 4 2017

a) 1x13x2x31=2xx2+x+11x−1−3x2x3−1=2xx2+x+1

Ta có: x31=(x1)(x2+x+1)x3−1=(x−1)(x2+x+1)

=(x1)[(x+12)2+34]=(x−1)[(x+12)2+34] cho nên x3 – 1 ≠ 0 khi x – 1 ≠ 0⇔ x ≠ 1

Vậy ĐKXĐ: x ≠ 1

Khử mẫu ta được:

x2+x+13x2=2x(x1)2x2+x+1=2x22xx2+x+1−3x2=2x(x−1)⇔−2x2+x+1=2x2−2x

4x23x1=0⇔4x2−3x−1=0

4x(x1

22 tháng 4 2017

Giải bài 52 trang 33 SGK Toán 8 Tập 2 | Giải toán lớp 8

Giải bài 52 trang 33 SGK Toán 8 Tập 2 | Giải toán lớp 8