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Ta có :
\(x=\frac{2016^{2017}+1}{2016^{2016}+1}\)
\(\frac{1}{2016}x=\frac{2016^{2017}+1}{2016^{2017}+2016}=\frac{2016^{2017}+2016-2015}{2016^{2017}+2016}\)
\(\Rightarrow\frac{1}{2006}x=1-\frac{2015}{2016^{2017}+2016}\)
Ta lại có :
\(y=\frac{2016^{2016}+1}{2016^{2015}+1}\)
\(\Rightarrow\frac{1}{2016}y=\frac{2016^{2016}+1}{2016^{2016}+2016}=\frac{2016^{2016}+2016-2015}{2016^{2016}+2016}\)
\(\Rightarrow\frac{1}{2016}y=1-\frac{2015}{2016^{2016}+2016}\)
Mà \(\frac{2015}{2016^{2017}+2016}< \frac{2015}{2016^{2016}+2016}\)(so sánh mẫu)
\(\Rightarrow1-\frac{2015}{2016^{2017}+2016}>1-\frac{2015}{2016^{2016}+2016}\)
\(\Rightarrow\frac{1}{2016}x>\frac{1}{2016}y\)
\(\Rightarrow x>y\)
DÀI QUÁ KHÔNG TÍNH ĐƯỢC. CÁI NÀY CÓ MÀ ĐI HỎI THẦN ĐỒNG VỀ MÔN TOÁN ĐI
\(\frac{x+2015}{2016}+\frac{x+2016}{2015}+\frac{x+2017}{2014}=-3\)
\(\Leftrightarrow\frac{x+2015}{2016}+1+\frac{x+2016}{2015}+1+\frac{x+2017}{2014}+1=0\)
\(\Leftrightarrow\frac{x+4031}{2016}+\frac{x+4031}{2015}+\frac{x+4031}{2014}=0\)
\(\Leftrightarrow\left(x+4031\right)\left(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}\right)=0\)
Có: \(\frac{1}{2016}+\frac{1}{2015}+\frac{1}{2014}\ne0\)
\(\Rightarrow x+4031=0\)
\(\Rightarrow x=-4031\)
\(2\left(x-y\right)^2=\left(z-x\right)\left(z-y\right)\Leftrightarrow\frac{2\left(x-y\right)^2}{\left(z-x\right)\left(z-y\right)}=1\)
\(\frac{2\left(z-y\right)^2}{\left(z-x\right)\left(z-y\right)}=\frac{\left(x-y\right)^2}{z\left(x-y\right)}=\frac{x-y}{z}\Rightarrow x-y=z\)
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
\(\frac{x+2015}{x-2015}=\frac{y+2017}{y-2017}\)
\(\frac{x+2015}{y+2017}=\frac{x-2015}{y-2017}\)
Áp dụng tính chất của dãy tỉ số bằng nhau,ta có :
\(\frac{x+2015}{y+2017}=\frac{x-2015}{y-2017}=\frac{\left(x+2015\right)-\left(x-2015\right)}{\left(y+2017\right)-\left(y-2017\right)}=\frac{2015}{2017}\)( 1 )
\(\frac{x+2015}{y+2017}=\frac{x-2015}{y-2017}=\frac{\left(x+2015\right)+\left(x-2015\right)}{\left(y+2017\right)+\left(y-2017\right)}=\frac{x}{y}\)( 2 )
Từ ( 1 ) và ( 2 ) \(\Rightarrow\frac{x}{y}=\frac{2015}{2017}\)