K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
13 tháng 8 2019
a) \(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\frac{x+2015}{5}+\frac{5}{5}+\frac{x+2016}{4}+\frac{4}{4}=\frac{x+2017}{3}+\frac{3}{3}+\frac{x+2018}{2}+\frac{2}{2}\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2002}{2}\)
\(\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right).\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)
\(\Leftrightarrow x=-2020\)
Vậy : \(x=-2020\)
Chúc bạn học tốt !!
13 tháng 8 2019
a) \(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\\ \left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\\ \frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2020}{2}\\ \frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\\ \left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\\ \Rightarrow x+2020=0\\ \Rightarrow x=-2020\)
Vậy x = -2020
b) \(\frac{x+2015}{5}+\frac{x+2016}{6}=\frac{x+2017}{7}+\frac{x+2018}{8}\\ \left(\frac{x+2015}{5}-1\right)+\left(\frac{x+2016}{6}-1\right)=\left(\frac{x+2017}{7}-1\right)+\left(\frac{x+2018}{8}-1\right)\\ \frac{x+2010}{5}+\frac{x+2010}{6}=\frac{x+2010}{7}+\frac{x+2010}{8}\\ \frac{x+2010}{5}+\frac{x+2010}{6}-\frac{x+2010}{7}-\frac{x+2010}{8}=0\\ \left(x+2010\right)\left(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\right)=0\\ \Rightarrow x+2010=0\\ \Rightarrow x=-2010\)
Vậy x = -2010
10 tháng 7 2017
a) \(\frac{x+2015}{5}+\frac{x+2015}{6}=\frac{x+2015}{7}+\frac{x+2015}{8}\)
\(\frac{x+2015}{5}+\frac{x+2015}{6}-\frac{x+2015}{7}-\frac{x+2015}{8}=0\)
\(\left(x+2015\right).\left(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\right)=0\)
vì \(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\ne0\)
\(\Rightarrow\)x + 2015 = 0
\(\Rightarrow\)x = -2015
b) Tương tự
H
2
DT
13 tháng 7 2016
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\frac{12\left(x+2015\right)}{60}+\frac{15\left(x+2016\right)}{60}=\frac{20\left(x+2017\right)}{60}+\frac{30\left(x+2018\right)}{60}\)
\(\Rightarrow12x+24180+15x+30240=20x+40340+30x+60540\)
\(\Leftrightarrow-23x=22460\Leftrightarrow x=-\frac{22460}{23}\)
H
1
13 tháng 7 2016
\(\frac{x+2015}{7}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Rightarrow\frac{x+2015}{7}+\frac{7}{7}+\frac{x+2016}{4}+\frac{4}{4}=\frac{x+2017}{3}+\frac{3}{3}+\frac{x+2018}{2}+\frac{2}{2}\)
\(\Rightarrow\frac{x+2020}{7}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2020}{2}\)
\(\Rightarrow\frac{x+2020}{7}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Rightarrow\left(x+2020\right)\left(\frac{1}{7}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
Mà \(\frac{1}{7}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\ne0\)
\(\Rightarrow x+2020=0\)
\(\Rightarrow x=-2020\)
TT
1
S
18 tháng 10 2017
\(\dfrac{x-2}{2018}=\dfrac{x-3}{2017}=\dfrac{x-4}{2016}=\dfrac{x-5}{2015}\)
\(\dfrac{x-2}{2018}+\dfrac{x-3}{2017}=\dfrac{x-4}{2016}+\dfrac{x-5}{2015}\)
\(\left(\dfrac{x-2}{2018}-1\right)+\left(\dfrac{x-3}{2017}-1\right)=\left(\dfrac{x-4}{2016}-1\right)+\left(\dfrac{x-5}{2015}-1\right)\)
\(\dfrac{x-2020}{2018}+\dfrac{x-2020}{2017}=\dfrac{x-2020}{2016}+\dfrac{x-2020}{2015}\)
\(\dfrac{x-2020}{2018}+\dfrac{x-2020}{2017}-\dfrac{x-2020}{2016}-\dfrac{x-2020}{2015}=0\)
\(\left(x-2020\right)\left(\dfrac{1}{2018}+\dfrac{1}{2017}-\dfrac{1}{2016}-\dfrac{1}{2015}\right)=0\)
\(\dfrac{1}{2018};\dfrac{1}{2017};\dfrac{1}{2016};\dfrac{1}{2015}>0\)
Nên \(x-2020=0\)
\(x=0+2020\)
\(x=2020\)
Vậy x bằng 2020
8 tháng 7 2017
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
TT
1
HH
24 tháng 6 2018
Giải:
\(\dfrac{x+2015}{5}+\dfrac{x+2016}{4}=\dfrac{x+2017}{3}+\dfrac{x+2018}{2}\)
\(\Leftrightarrow2+\dfrac{x+2015}{5}+\dfrac{x+2016}{4}=2+\dfrac{x+2017}{3}+\dfrac{x+2018}{2}\)
\(\Leftrightarrow\dfrac{x+2015}{5}+1+\dfrac{x+2016}{4}+1=\dfrac{x+2017}{3}+1+\dfrac{x+2018}{2}+1\)
\(\Leftrightarrow\dfrac{x+2015+5}{5}+\dfrac{x+2016+4}{4}=\dfrac{x+2017+3}{3}+\dfrac{x+2018+2}{2}\)
\(\Leftrightarrow\dfrac{x+2020}{5}+\dfrac{x+2020}{4}=\dfrac{x+2020}{3}+\dfrac{x+2020}{2}\)
\(\Leftrightarrow\dfrac{x+2020}{5}+\dfrac{x+2020}{4}-\dfrac{x+2020}{3}-\dfrac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\dfrac{1}{5}+\dfrac{1}{4}-\dfrac{1}{3}-\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)
\(\Leftrightarrow x=-2020\)
Vậy ...
DT
2
MT
11 tháng 7 2015
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\frac{x+2015}{5}+1+\frac{x+2016}{4}+1=\frac{x+2017}{3}+1+\frac{x+2018}{2}+1\)
\(\frac{x+2015}{5}+\frac{5}{5}+\frac{x+2016}{4}+\frac{4}{4}=\frac{x+2017}{3}+\frac{3}{3}+\frac{x+2018}{2}+\frac{2}{2}\)
\(\frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2020}{2}\)
\(\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\left(x+2020\right).\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
mà \(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\ne0\)nên
\(x+2020=0\)
\(x=-2020\)
11 tháng 7 2015
Cộng 1 vào 2 vế ta có
\(\frac{x+2015}{5}+1+\frac{x+2016}{4}+1=\frac{x+2017}{3}+1+\frac{x+2018}{2}+1\)
\(\left(\frac{x+2015}{5}+\frac{5}{5}\right)+\left(\frac{x+2016}{4}+\frac{4}{4}\right)=\left(\frac{x+2017}{3}+\frac{3}{3}\right)+\left(\frac{x+2018}{2}+\frac{2}{2}\right)\)
\(\frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2020}{2}\)
\(\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
Vì \(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\ne0\)
nên \(x+2020=0\Rightarrow x=-2020\)
x+2015/5 + x+2016/4=x+2017/3 + x+2018/2
\(\Rightarrow\frac{x+2015}{5}+1+\frac{x+2016}{4}+1=\frac{x+2017}{3}+1+\frac{x+2018}{2}+1\)
\(\Rightarrow\frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2020}{2}\)
\(\Rightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Rightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Rightarrow x+2020=0\).Do \(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\ne0\)
\(\Rightarrow x=-2020\)