A = 2017/2018 x 7/8 + 2017/2018 x 3/8 - 2017/2018 x 1/4
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2 tháng 4 2018
\(\left|3x-2018\right|+\left|x-2017\right|=\left|2x-1\right|\)
\(\Rightarrow\orbr{\begin{cases}3x-2018+x-2017=2x-1\\-\left(3x-2018\right)+\left[-\left(x-2017\right)\right]=2x-1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}4x-4035=2x-1\\\left(-3x-x\right)+\left(2018+2017\right)=2x-1\end{cases}}\)
Làm tiếp
TH2:
\(\left|3x-2018\right|+\left|x-2017\right|=\left|2x-1\right|\)
\(\Rightarrow\orbr{\begin{cases}3x-2018+x-2017=-2x+1\\-\left(3x-2018\right)+\left[-\left(x-2017\right)\right]=-2x+1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}4x-4035=-2x+1\\\left(-3x-x\right)+\left(2018+2017\right)=-2x+1\end{cases}}\)
Tự tiếp tiếp nha bạn
Bài sau cũng tg tự vậy mà làm
24 tháng 10 2021
\(\frac{2017}{2018}\)x\(\frac{7}{8}\)+\(\frac{2017}{2018}\)x\(\frac{3}{8}\)-\(\frac{2017}{2018}\)x\(\frac{1}{4}\)
= \(\frac{2017}{2018}\)x (\(\frac{7}{8}\)+\(\frac{3}{8}\)-\(\frac{1}{4}\))
= \(\frac{2017}{2018}\)x ( \(\frac{10}{8}\)- \(\frac{1}{4}\))
= \(\frac{2017}{2018}\)x ( \(\frac{10}{8}\)- \(\frac{2}{8}\))
= \(\frac{2017}{2018}\)x 1
= \(\frac{2017}{2018}\)
Chúc em học tốt nhé :>
24 tháng 10 2021
=2017/2018*(7/8+3/8)-2017*1/4
=2017/2018*5/4+2017*-1/4
=2017/2018*(5/4-1/4)
=2017/2018*1
=2017/2018
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
Ta có : A =\(\frac{2017}{2018}\)x \(\frac{7}{8}\)+ \(\frac{2017}{2018}\)x \(\frac{3}{8}\)- \(\frac{2017}{2018}\)x \(\frac{1}{4}\)
= \(\frac{2017}{2018}\) x ( \(\frac{7}{8}+\frac{3}{8}-\frac{1}{4}\))
= \(\frac{2017}{2018}\)x 1
=\(\frac{2017}{2018}\)
Vậy A= : \(\frac{2017}{2018}\)
Bài giải
\(A=\frac{2017}{2018}\text{ x }\frac{7}{8}+\frac{2017}{2018}\text{ x }\frac{3}{8}-\frac{2017}{2018}\text{ x }\frac{1}{4}\)
\(A=\frac{2017}{2018}\text{ x }\frac{1}{4}\left(\frac{7}{2}+\frac{3}{2}-1\right)=\frac{2017}{2018}\text{ x }\frac{1}{4}\text{ x }4==\frac{2017}{2018}\text{ x }1=\frac{2017}{2018}\)