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\(\Rightarrow\frac{x+5}{2015}+1+\frac{x+4}{2016}+1+\frac{x+3}{2017}+1=\frac{x+2015}{5}+1+\frac{x+2016}{4}+1+\frac{x+2017}{3}+1\)
\(\Rightarrow\frac{x+2020}{2015}+\frac{x+2020}{2016}+\frac{x+2020}{2017}=\frac{x+2020}{5}+\frac{x+2020}{4}+\frac{x+2020}{3}\)
\(\Rightarrow\left(x+2020\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}-\frac{1}{5}-\frac{1}{4}-\frac{1}{3}\right)=0\)
\(\Rightarrow x=-2020\)
Ta có : x/3=y/2 = x/12 = y /8
y/4=z/5 = y/8 = z/10 ( mình biến đổi sao cho y có mẫu chung là 8 ý bạn )
=> x/12=y/8=z/10 = -x-y+z/ -12-8+10
= -10/-10 =1
=> x = 1.12=12
y=1.8=8
z=1.10=10
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
Ta có : \(\frac{x}{2}+\frac{x}{4}+\frac{x}{2016}=\frac{x}{3}+\frac{x}{5}+\frac{x}{2017}\)
\(\Rightarrow\frac{x}{2}+\frac{x}{4}+\frac{x}{2016}-\frac{x}{3}-\frac{x}{5}-\frac{x}{2017}=0\)
\(\Leftrightarrow x\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{2016}-\frac{1}{3}-\frac{1}{5}-\frac{1}{2017}\right)\)
Vì : \(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{2016}-\frac{1}{3}-\frac{1}{5}-\frac{1}{2017}\right)\ne0\)
Nên x = 0
\(\frac{x}{2}+\frac{x}{4}+\frac{x}{2016}=\frac{x}{3}+\frac{x}{5}+\frac{x}{2017}\)
\(\Rightarrow x.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{2016}\right)=x.\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{2017}\right)\)
\(\Rightarrow x.\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{2016}\right)-x.\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{2017}\right)\)
\(\Rightarrow x.\left[\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{2016}\right)-\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{2017}\right)\right]=0\)
\(\Rightarrow x=0\)\(\left(vi\left[\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{2016}\right)-\left(\frac{1}{3}+\frac{1}{5}+\frac{1}{2017}\right)\right]\right)\ne0\)
Tham khảo ở đây nhé bn: olm.vn/hoi-dap/question/686545.html, mk lm r`
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
\(\frac{x+2}{2017}+\frac{x+3}{2016}+\frac{x+4}{2015}+\frac{x+5}{1007}+\frac{x+2074}{11}=0\)
\(\Leftrightarrow\frac{x+2}{2017}+1+\frac{x+3}{2016}+1+\frac{x+4}{2015}+1+\frac{x+5}{1007}+2+\frac{x+2074}{11}-5=0\)
\(\Leftrightarrow\frac{x+2019}{2017}+\frac{x+2019}{2016}+\frac{x+2019}{2015}+\frac{x+2019}{1007}+\frac{x+2019}{11}=0\)
\(\Leftrightarrow\left(x+2019\right)\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{1007}+\frac{1}{11}\right)=0\)
\(\Leftrightarrow\left(x+2019\right)=0vì\left(\frac{1}{2017}+\frac{1}{2016}+\frac{1}{2015}+\frac{1}{1007}+\frac{1}{11}\right)\ne0\)
\(\Leftrightarrow x=-2019\)
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\frac{12\left(x+2015\right)}{60}+\frac{15\left(x+2016\right)}{60}=\frac{20\left(x+2017\right)}{60}+\frac{30\left(x+2018\right)}{60}\)
\(\Rightarrow12x+24180+15x+30240=20x+40340+30x+60540\)
\(\Leftrightarrow-23x=22460\Leftrightarrow x=-\frac{22460}{23}\)
\(\frac{x+2015}{7}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Rightarrow\frac{x+2015}{7}+\frac{7}{7}+\frac{x+2016}{4}+\frac{4}{4}=\frac{x+2017}{3}+\frac{3}{3}+\frac{x+2018}{2}+\frac{2}{2}\)
\(\Rightarrow\frac{x+2020}{7}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2020}{2}\)
\(\Rightarrow\frac{x+2020}{7}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Rightarrow\left(x+2020\right)\left(\frac{1}{7}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
Mà \(\frac{1}{7}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\ne0\)
\(\Rightarrow x+2020=0\)
\(\Rightarrow x=-2020\)
Đề bài y/c gì bạn
Nếu tìm x thì là
\(\frac{x}{2}\)+\(\frac{x}{4}\)+\(\frac{x}{2016}\)-\(\frac{x}{3}\)-\(\frac{x}{5}\)-\(\frac{x}{2017}\)=0 (quy tắc chuyển vế)
=> x(\(\frac{1}{2}\)+\(\frac{1}{4}\)+\(\frac{1}{2016}\)-\(\frac{1}{3}\)-\(\frac{1}{5}\)-\(\frac{1}{2017}\))=0 => x=0 hoăc (........)=0
mà (\(\frac{1}{2}\)+\(\frac{1}{4}\)+\(\frac{1}{2016}\)-\(\frac{1}{3}\)-\(\frac{1}{5}\)-\(\frac{1}{2017}\)) khác 0 ( vì \(\frac{1}{2}\)>\(\frac{1}{3}\);\(\frac{1}{4}\)>\(\frac{1}{5}\);;\(\frac{1}{2016}\)>\(\frac{1}{2017}\))
=>x=0