tim min
B=/X+3/+/X+4/+/X+5/
A=/X-2017/+/X-2018/
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\(\frac{x+4}{2018}+\frac{x+5}{2017}+\frac{x+6}{2016}+\frac{x+7}{2015}=-4\)
\(\Rightarrow\left(\frac{x+4}{2018}+1+\frac{x+5}{2017}+1+\frac{x+6}{2016}+1+\frac{x+7}{2015}+1\right)=-4+4=0\)
\(\Rightarrow\frac{x+2022}{2018}+\frac{x+2022}{2017}+\frac{x+2022}{2016}+\frac{x+2022}{2015}=0\)
\(\Rightarrow\left(x+2022\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}\right)=0\)
\(\Rightarrow x+2022=0\Leftrightarrow x=-2022\)
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
Min A, Min B chứ em.
Định làm nhưng lười với quên béng mất cách trình bày ==''
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 !!
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
\(\left(\frac{5}{4}-\frac{2}{5}\right)\times\frac{2017}{2018}+\left(\frac{3}{4}-\frac{3}{5}\right)\times\frac{2017}{2018}\)
\(=\left[\left(\frac{5}{4}-\frac{2}{5}\right)+\left(\frac{3}{4}-\frac{3}{5}\right)\right]\times\frac{2017}{2018}\)
\(=\left[\left(\frac{5}{4}+\frac{3}{4}\right)-\left(\frac{2}{5}+\frac{3}{5}\right)\right]\times\frac{2017}{2018}\)
\(=\left[2-1\right]\times\frac{2017}{2018}\)
\(=1\times\frac{2017}{2018}\)
\(=\frac{2017}{2018}\)
\(\left(\frac{5}{4}-\frac{2}{5}\right)\cdot\frac{2017}{2018}-\left(\frac{3}{4}-\frac{3}{5}\right)\cdot\frac{2017}{2018}\)
\(=\frac{2017}{2018}\cdot\left(\frac{5}{4}-\frac{2}{5}+\frac{3}{4}-\frac{3}{5}\right)\)
\(=\frac{2017}{2018}.\left(2+-1\right)\)
\(=\frac{2017}{2018}.1=\frac{2017}{2018}\)