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Những câu hỏi liên quan
hoshimiya ichigo
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Edogawa Conan
9 tháng 1 2020 lúc 21:09

\(\frac{x+2}{2015}+\frac{x+1}{2016}=\frac{x+3}{2014}+\frac{x+4}{2013}\)

=> \(\left(\frac{x+2}{2015}+1\right)+\left(\frac{x+1}{2016}+1\right)=\left(\frac{x+3}{2014}+1\right)+\left(\frac{x+4}{2013}+1\right)\)

=> \(\frac{x+2017}{2015}+\frac{x+2017}{2016}=\frac{x+2017}{2014}+\frac{x+2017}{2013}\)

=> (x + 2017)(1/2015 + 1/2016 - 1/2014 - 1/2013) = 0

=> x + 2017 = 0 

=> x = -2017

Khách vãng lai đã xóa
zZz Cool Kid_new zZz
9 tháng 1 2020 lúc 21:09

\(\frac{x+2}{2015}+\frac{x+1}{2016}=\frac{x+3}{2014}+\frac{x+4}{2013}\)

\(\Leftrightarrow\frac{x+2}{2015}+1+\frac{x+1}{2016}+1=\frac{x+3}{2014}+1+\frac{x+4}{2013}+1\)

\(\Leftrightarrow\frac{x+2017}{2015}+\frac{x+2017}{2016}=\frac{x+2017}{2014}+\frac{x+2017}{2013}\)

\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2014}-\frac{1}{2013}\right)=0\)

Dễ thấy cái ngoặc to < 0

=> x=-2017

Khách vãng lai đã xóa
holicuoi
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Ác Mộng
11 tháng 6 2015 lúc 15:35

\(\frac{x+2}{2013}+\frac{x+1}{2014}=\frac{x}{2015}+\frac{x-1}{2016}\)

\(\Leftrightarrow\)\(\frac{x+2}{2013}+1+\frac{x+1}{2014}+1=\frac{x}{2015}+1+\frac{x-1}{2016}+1\)

\(\Leftrightarrow\frac{x+2015}{2013}+\frac{x+2015}{2014}=\frac{x+2015}{2015}+\frac{x+2015}{2016}\)

\(\Leftrightarrow\left(x+2015\right)\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\right)=0\)

Do\(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}>0\)

=>x+2015=0

<=>x=-2015

Trần Thị Loan
11 tháng 6 2015 lúc 15:38

=> \(\frac{x+2015-2013}{2013}+\frac{x+2015-2014}{2014}=\frac{x+2015-2015}{2015}+\frac{x+2015-2016}{2016}\)

<=> \(\frac{x+2015}{2013}-1+\frac{x+2015}{2014}-1=\frac{x+2015}{2015}-1+\frac{x+2015}{2016}-1\)

<=> \(\frac{x+2015}{2013}+\frac{x+2015}{2014}-\frac{x+2015}{2015}-\frac{x+2015}{2016}=0\)

<=> \(\left(x+2015\right).\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\right)=0\)

<=> x + 2015 = 0 Vì \(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\ne0\)

<=> x = -2015

Yasuo
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Nguyễn Trần An Thanh
24 tháng 1 2017 lúc 15:18

Đề bạn hình như hơi sai thì phải, nhưng nếu tìm x thì mình giải như sau

Ta có: \(\frac{x-1}{2016}+\frac{x-2}{2015}-\frac{x-3}{2014}=\frac{x-4}{2013}\)

\(\Rightarrow\frac{x-1}{2016}+\frac{x-2}{2015}=\frac{x-4}{2013}+\frac{x-3}{2014}\)

\(\Rightarrow\frac{x-1}{2016}-1+\frac{x-2}{2015}-1=\frac{x-4}{2013}-1+\frac{x-3}{2014}-1\)

\(\Rightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}=\frac{x-2017}{2013}+\frac{x-2017}{2014}\)

\(\Rightarrow\frac{x-2017}{2016}+\frac{x-2017}{2015}-\frac{x-2017}{2014}-\frac{x-2017}{2013}=0\)

\(\Rightarrow\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\right)=0\)

Vì \(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}< 0\)

\(\Rightarrow x-2017=0\)

\(\Rightarrow x=2017\)

Limited Edition
Xem chi tiết
Lê Anh Duy
11 tháng 4 2019 lúc 12:29

\(\Leftrightarrow\left(\frac{x+4}{2013}+1\right)+\left(\frac{x+3}{2014}+1\right)=\left(\frac{x+2}{2015}+1\right)+\left(\frac{x+1}{2016}+1\right)\)

\(\Leftrightarrow\frac{x+2017}{2013}+\frac{x+2017}{2014}-\frac{x+2017}{2015}-\frac{x+2017}{2016}=0\)

\(\Leftrightarrow\left(x+2017\right)\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\right)=0\)

\(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\ne0\)

\(\Rightarrow x+2017=0\Rightarrow x=-2017\)

Phạm Tuấn Kiệt
Xem chi tiết
Devil
12 tháng 4 2016 lúc 19:26

\(\frac{x-1}{2016}+\frac{x-2}{2015}-\frac{x-3}{2014}=\frac{x-4}{2013}\)

\(\left(\frac{x-1}{2016}-1\right)+\left(\frac{x-2}{2015}-1\right)-\left(\frac{x-3}{2014}-1\right)-\left(\frac{x-4}{2013}-1\right)=0\)

\(\frac{x-2017}{2016}+\frac{x-2017}{2015}-\frac{x-2017}{2014}-\frac{x-2017}{2013}=0\)

\(\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\right)=0\)

\(x-2017=0\left(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\ne0\right)\)

\(x=2017\)

Thánh Lạnh Lùng
12 tháng 4 2016 lúc 19:30

x-1/2016+x-2/2015-x-3/2014=x-4/2013

<=> x-1/2016+x-2/2015-x-3/2014-x-4/2013=0

trừ mỗi phân số thêm 1 ta được

x-2017/2016+x-2017/2015-x-2017/2014-x-2017/2013=0

<=>(x-2017).(1/2016+1/2015+1/2014+1/2013)=0

(1/2016+1/2015+1/2014+1/2013) khác 0

=>x-2017=0

<=>x=2017

vậy x = 2017

Trần jenny
25 tháng 2 2018 lúc 12:58

X=2017

nanami
Xem chi tiết
Hoàng Thủy Tiên
2 tháng 8 2016 lúc 10:37

\(\frac{x+4}{2013}+\frac{x+3}{2014}=\frac{x+2}{2015}+\frac{x+1}{2016}\)

\(\Rightarrow\frac{x+4}{2013}+1+\frac{x+3}{2014}+1=\frac{x+2}{2015}+1+\frac{x+1}{2016}+1\)

\(\Rightarrow\frac{x+2017}{2013}+\frac{x+2017}{2014}=\frac{x+2017}{2015}+\frac{x+2017}{2016}\)

\(\Rightarrow\frac{x+2017}{2013}+\frac{x+2017}{2014}-\frac{x+2017}{2015}-\frac{x+2017}{2016}=0\)

\(\Rightarrow\left(x+2017\right)\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\right)=0\)

\(Do\)\(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}-\frac{1}{2016}\ne0\)

\(\Rightarrow x+2017=0\)

\(\Rightarrow x=-2017\)

Vậy \(x=-2017\)

anh ta là ai chấm com ch...
2 tháng 8 2016 lúc 10:15

bạn bấm vào "đúng 0" là sẽ có đáp án hiện ra

Hoàng Nguyễn Bảo Trâm
5 tháng 5 2017 lúc 14:32

- 2017 nha bạn

Nguyễn Linh Nhi
Xem chi tiết
Isolde Moria
10 tháng 9 2016 lúc 20:19

\(\frac{x+4}{2012}+\frac{x+3}{2013}=\frac{x+2}{2014}+\frac{x+1}{2015}\)

\(\Rightarrow\frac{x+4}{2012}+1+\frac{x+3}{2013}+1=\frac{x+2}{2014}+1+\frac{x+1}{2015}+1\)

\(\Rightarrow\frac{x+2016}{2012}+\frac{x+2016}{2013}=\frac{x+2016}{2014}+\frac{x+2016}{2015}\)

\(\Rightarrow\frac{x+2016}{2012}+\frac{x+2016}{2013}-\left(\frac{x+2016}{2014}+\frac{x+2016}{2015}\right)=0\)

\(\Rightarrow\left(x+2016\right)\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)=0\)

Vì \(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\ne0\)

\(\Rightarrow x+2016=0\)

\(\Rightarrow x=-2016\)

Isolde Moria
10 tháng 9 2016 lúc 20:14

Bạn ơi ghi sai đề rồi kìa

Ngọc Mai
Xem chi tiết
Nguyen
17 tháng 3 2019 lúc 21:38

Phải là \(\frac{x}{2012}\)

\(\frac{x}{2012}-1+\frac{x+1}{2013}-1+\frac{x+2}{2014}-1+\frac{x+3}{2015}-1+\frac{x+4}{2016}-1=0\)

\(\Leftrightarrow\left(x-2012\right)\left(\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}+\frac{1}{2015}+\frac{1}{2016}\right)=0\)

\(\Leftrightarrow x=2012\)

Vậy ...

Lý Mạnh Trường
Xem chi tiết
Tibber Bear
12 tháng 4 2017 lúc 14:25

CÓ: \(\frac{x-1}{2015}+\frac{x-2}{2014}-\frac{x-3}{2013}-\frac{x-4}{2012}=0\)\(0\)

<=>\(\left(\frac{x-1}{2015}-1\right)+\left(\frac{x-2}{2014}-1\right)-\left(\frac{x-3}{2013}-1\right)-\left(\frac{x-4}{2012}-1\right)=0\)

<=>\(\frac{x-2016}{2015}+\frac{x-2016}{2014}-\frac{x-2016}{2013}-\frac{x-2016}{2012}=0\)

<=>\(\left(x-2016\right)\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)=0\)

Do:\(\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{2013}-\frac{1}{2012}\right)\ne0\)

=>\(x-2016=0\)

<=>\(x=2016\)