\(\dfrac{x-3}{2011}+\dfrac{x-2}{2012}=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}\)

NT

Ngọc Thảo

22 tháng 2 2019

\(\dfrac{x-3}{2011}+\dfrac{x-2}{2012}=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}\)

#Toán lớp 8

1

AT

An Trần

22 tháng 2 2019

\(\dfrac{x-3}{2011}+\dfrac{x-2}{2012}=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}\)

\(\Leftrightarrow\dfrac{x-3}{2011}-1+\dfrac{x-2}{2012}-1=\dfrac{x-2012}{2}-1+\dfrac{x-2011}{3}-1\)

\(\Leftrightarrow\dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}=\dfrac{x-2014}{2}+\dfrac{x-2014}{3}\)

\(\Leftrightarrow\dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}-\dfrac{x-2014}{2}-\dfrac{x-2014}{3}=0\)

\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)

\(\Leftrightarrow x-2014=0\)

\(\Leftrightarrow x=2014\)

Vậy x = 2014.

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BC

Big City Boy

2 tháng 3 2021

Cho 2 số sau: \(x=\dfrac{2011^3-1}{2011^2+2012}\) và \(y=\dfrac{2012^3+1}{2012^2-2011}\). Tính x+y

#Toán lớp 8

1

H

hnamyuh

2 tháng 3 2021

\(x= \dfrac{2011^3-1}{2011^2+2012} = \dfrac{(2011-1)(2011^2+2011+1)}{2011^2 + 2011 + 1} = 2010\)

\(y = \dfrac{2012^3+1}{2012^2-2011} = \dfrac{(2012+1)(2012^2-2012+1)}{2012^2-2012 + 1} = 2013\)

Suy ra:

x + y = 2010 + 2013 = 4023

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SB

Silver Bullet

3 tháng 3 2017

\(\dfrac{x-3}{2001}+\dfrac{x-2}{2012}=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}\)

#Toán lớp 8

2

TT

Trần Thiên Kim

3 tháng 3 2017

Đề sai rồi bạn!

Mình sửa lại đề nha!

\(\dfrac{x-3}{2011}+\dfrac{x-2}{2012}=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}\)

\(\Leftrightarrow\left(\dfrac{x-3}{2011}-1\right)+\left(\dfrac{x-2}{2012}-1\right)=\left(\dfrac{x-2012}{2}-1\right)+\left(\dfrac{x-2011}{3}-1\right)\)

\(\Leftrightarrow\dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}=\dfrac{x-2014}{2}+\dfrac{x-2014}{3}\)

\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)

\(\Leftrightarrow x-2014=0\)

\(\Leftrightarrow x=2014\)

Vậy S={2014}

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NH

Nguyễn Huy Tú

3 tháng 3 2017

\(\frac{x-3}{2011}+\frac{x-2}{2012}=\frac{x-2012}{2}+\frac{x-2011}{3}\)

\(\Rightarrow\left(\frac{x-3}{2011}-1\right)+\left(\frac{x-2}{2012}-1\right)=\left(\frac{x-2012}{2}-1\right)+\left(\frac{x-2011}{3}-1\right)\)

\(\Rightarrow\frac{x-2014}{2011}+\frac{x-2014}{2012}=\frac{x-2014}{2}+\frac{x-2014}{3}\)

\(\Rightarrow\frac{x-2014}{2011}+\frac{x-2014}{2012}-\frac{x-2014}{2}-\frac{x-2014}{3}=0\)

\(\Rightarrow\left(x-2014\right)\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2}-\frac{1}{3}\right)=0\)

Mà \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2}-\frac{1}{3}\ne0\)

\(\Rightarrow x-2014=0\)

\(\Rightarrow x=2014\)

Vậy x = 2014

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TT

Thanh Thúy

8 tháng 3 2018

\(\dfrac{x-3}{2011}\)+\(\dfrac{x-2}{2012}\)=\(\dfrac{x-2012}{2}\)+\(\dfrac{x-2011}{3}\)

#Toán lớp 8

3

CT

Cao Thị Hương Giang

8 tháng 3 2018

\(\dfrac{x-3}{2011}+\dfrac{x-2}{2012}=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}\)

\(\Leftrightarrow\dfrac{x-3}{2011}+\dfrac{x-2}{2012}-2=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}-2\)

\(\Leftrightarrow\left(\dfrac{x-3}{2011}-1\right)+\left(\dfrac{x-2}{2012}-1\right)=\left(\dfrac{x-2012}{2}-1\right)+\left(\dfrac{x-2011}{3}-1\right)\)

\(\Leftrightarrow\dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}-\dfrac{x-2014}{2}-\dfrac{x-2014}{3}=0\)

\(\Leftrightarrow\left(x-2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)

\(\Leftrightarrow x-2014=0\)

\(\Leftrightarrow x=0\)

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HA

Hoàng Anh Thư

8 tháng 3 2018

\(\dfrac{x-3}{2011}+\dfrac{x-2}{2012}=\dfrac{x-2012}{2}+\dfrac{x-2011}{3}\)

<=>\(\dfrac{x-3}{2011}-1+\dfrac{x-2}{2012}-1=\dfrac{x-2012}{2}-1+\dfrac{x-2011}{3}-1\)

<=>\(\dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}=\dfrac{x-2014}{2}+\dfrac{x-2014}{3}\)

<=>\(\dfrac{x-2014}{2011}+\dfrac{x-2014}{2012}-\dfrac{x-2014}{2}-\dfrac{x-2014}{3}=0\)

<=>\(\left(x-2014\right)\left(\dfrac{1}{2011}+\dfrac{1}{2012}-\dfrac{1}{2}-\dfrac{1}{3}\right)=0\)

vì 1/2011+1/2012-1/2-1/3 khác 0

=>x-2014=0<=>x=2014

vậy....................

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AP

Anh Pha

20 tháng 1 2018

Giải pt sau : \(\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}=2012\)

#Toán lớp 8

1

AH

Akai Haruma

Giáo viên

20 tháng 1 2018

Lời giải:

Ta có:

\(\frac{x-1}{2012}+\frac{x-2}{2011}+\frac{x-3}{2010}+...+\frac{x-2012}{1}=2012\)

\(\Leftrightarrow \left(\frac{x-1}{2012}-1\right)+\left(\frac{x-2}{2011}-1\right)+\left(\frac{x-3}{2010}-1\right)+...+\left(\frac{x-2012}{1}-1\right)=0\)

\(\Leftrightarrow \frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)

\(\Leftrightarrow (x-2013)\left(\frac{1}{2012}+\frac{1}{2011}+...+1\right)=0\)

Dễ thấy \(\frac{1}{2012}+\frac{1}{2011}+...+1\neq 0\Rightarrow x-2013=0\)

\(\Leftrightarrow x=2013\)

Vậy PT có nghiệm \(x=2013\)

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D

DPKhanh

24 tháng 2 2022

\(\dfrac{x+1}{2012}+\dfrac{x+2}{2011}+\dfrac{x+3}{2010}=\dfrac{x-1}{2014}+\dfrac{x-2}{2015}+\dfrac{x-3}{2016}\)

#Toán lớp 8

2

NL

Nguyễn Lê Phước Thịnh

24 tháng 2 2022

\(\Leftrightarrow\dfrac{x+1}{2012}+1+\dfrac{x+2}{2011}+1+\dfrac{x+3}{2010}+1=\dfrac{x-1}{2014}+1+\dfrac{x-2}{2015}+1+\dfrac{x-3}{2016}+1\)

=>x+2013=0

hay x=-2013

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NH

Nguyễn Huy Tú

24 tháng 2 2022

\(\dfrac{x+1}{2012}+1+\dfrac{x+2}{2011}+1+\dfrac{x+3}{2010}+1=\dfrac{x-1}{2014}+1+\dfrac{x-2}{2015}+1+\dfrac{x-3}{2016}+1\)

\(\Leftrightarrow\left(x+2013\right)\left(\dfrac{1}{2022}+\dfrac{1}{2011}+\dfrac{2}{2010}-\dfrac{1}{2014}-\dfrac{1}{2015}-\dfrac{1}{2016}\ne0\right)=0\Leftrightarrow x=-2013\)

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BD

Bướm Đêm Sát Thủ

10 tháng 4 2018

giải phương trình:

\(\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}=2012\)

#Toán lớp 8

4

HA

Hoàng Anh Thư

10 tháng 4 2018

\(\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}=2012\)

<=>\(\dfrac{x-1}{2012}-1+\dfrac{x-2}{2011}-1+\dfrac{x-3}{2010}-1+...+\dfrac{x-2012}{1}-1=0\)

<=>\(\dfrac{x-2013}{2012}+\dfrac{x-2013}{2011}+\dfrac{x-2013}{2010}+...+\dfrac{x-2013}{1}=0\)

<=>\(\left(x-2013\right)\left(\dfrac{1}{2012}+\dfrac{1}{2011}+...+1\right)=0\)

do 1/2012+1/2011....+1 khác 0 =>x-2013=0<=>x=2013

vậy..........................

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TM

Trần Minh Hoàng

10 tháng 4 2018

\(\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}=2012\)

\(\left(\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}\right)-2012=0\)

\(\Rightarrow\dfrac{x-2013}{2012}+\dfrac{x-2013}{2011}+\dfrac{x-2013}{2010}+...+\dfrac{x-2013}{1}=0\)

\(\Rightarrow x-2013\left(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}+...+\dfrac{1}{1}\right)=0\)

Vì \(x-2013\left(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}+...+\dfrac{1}{1}\right)=0\)nên x - 2013 hoặc \(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}+...+\dfrac{1}{1}\) = 0. Nhưng \(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}+...+\dfrac{1}{1}\ne0\) nên x - 2013 = 0. Vì vậy x = 2013.

Vậy...

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TD

Trần Đức Mạnh

9 tháng 11 2017

Giải phương trình:

\(A=\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}=2012\)

#Toán lớp 8

1

LF

Lightning Farron

9 tháng 11 2017

\(\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}=2012\)

\(\Leftrightarrow\dfrac{x-1}{2012}-1+\dfrac{x-2}{2011}-1+...+\dfrac{x-2012}{1}-1=0\)

\(\Leftrightarrow\dfrac{x-2013}{2012}+\dfrac{x-2013}{2011}+...+\dfrac{x-2013}{1}=0\)

\(\Leftrightarrow\left(x-2013\right)\left(\dfrac{1}{2012}+\dfrac{1}{2011}+...+\dfrac{1}{1}\right)=0\)

Dễ thấy: \(\dfrac{1}{2012}+\dfrac{1}{2011}+...+\dfrac{1}{1}>0\)

\(\Rightarrow x-2013=0\Rightarrow x=2013\)

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NT

Nguyễn tấn tài

12 tháng 2 2019

Sao lại trừ 1 vậy bạn ??? mình không hiểu cho lắm mong bạn giúp đỡ

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TM

Trà My Nguyễn Thị

25 tháng 7 2018

Tìm x biết :

\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}+..._{ }\)\(+\dfrac{x-2012}{2}=2012\)

#Toán lớp 8

1

Y

Yukru

23 tháng 8 2018

\(\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}+...+\dfrac{x-2012}{2}=2012\)

\(\Rightarrow\dfrac{x-1}{2013}+\dfrac{x-2}{2012}+\dfrac{x-3}{2011}+...+\dfrac{x-2012}{2}-2012=0\)

\(\Rightarrow\dfrac{x-1}{2013}-1+\dfrac{x-2}{2012}-1+\dfrac{x-3}{2011}-1+...+\dfrac{x-2012}{2}-1=0\)

\(\Rightarrow\dfrac{x-2014}{2013}+\dfrac{x-2014}{2012}+\dfrac{x-2014}{2011}+...+\dfrac{x-2014}{2}=0\)

\(\Rightarrow\left(x-2014\right)\left(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}+...+\dfrac{1}{2}\right)=0\)

Mà \(\dfrac{1}{2013}+\dfrac{1}{2012}+\dfrac{1}{2011}+...+\dfrac{1}{2}\ne0\)

\(\Rightarrow x-2014=0\)

\(\Rightarrow x=2014\)

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