Giá trị của x thỏa mãn: (Do công thức bị lỗi, các bn thông cảm!!!)
\(|^x_{2015}+\frac{x}{2016}=|^x_{2016}+\frac{x}{2017}\)
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Có : đenta = (-m)2 -4(m-1) = m2 -4m + 4 = (m-2)2 >= 0
Áp dụng hệ thức Vi-ét, ta có : x1 + x2 = m
x1.x2 = m-1
Có:\(\frac{1}{x_{ }_{ }1}+\frac{1}{x2}=\frac{x1.x2}{2011}\)
<=> \(\frac{x1+x2}{x1.x2}=\frac{x1.x2}{2011}\)
<=> \(\frac{m}{m-1}=\frac{m-1}{2011}\)
<=> 2011m = (m-1)2
<=> 2011m = m2-2m + 1
<=> m2-2013m + 1 =0
Giải pt ra
\(\frac{x}{2015}+\frac{x}{2016}=\frac{x}{2016}+\frac{x}{2017}\)
\(\Rightarrow\frac{x}{2015}+\frac{x}{2016}-\frac{x}{2016}-\frac{x}{2017}=0\)
\(\Rightarrow\frac{x}{2015}-\frac{x}{2017}=0\)
\(\Rightarrow x.\left(\frac{1}{2015}-\frac{1}{2017}\right)=0\)
Mà ta thấy \(\frac{1}{2015}-\frac{1}{2017}\ne0\Rightarrow x=0\)
Vậy \(x=0\)
\(\frac{x}{2015}+\frac{x}{2016}=\frac{x}{2016}+\frac{x}{2017}\)
\(\Leftrightarrow\frac{x}{2015}+\frac{x}{2016}-\frac{x}{2016}-\frac{x}{2017}=0\)
\(\Leftrightarrow x\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2016}-\frac{1}{2017}\right)=0\)
\(\Leftrightarrow x=0\).Do \(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2016}-\frac{1}{2017}\ne0\)
Vậy giá trị của x là x=0
\(\left|\frac{x}{2015}+\frac{x}{2016}\right|=\left|\frac{x}{2016}+\frac{x}{2017}\right|\)
<=>\(\left|x\right|.\left|\frac{1}{2015}+\frac{1}{2016}\right|=\left|x\right|.\left|\frac{1}{2016}+\frac{1}{2017}\right|\)
<=>\(\left|x\right|.\left(\frac{1}{2015}+\frac{1}{2016}\right)=\left|x\right|.\left(\frac{1}{2016}+\frac{1}{2017}\right)\)
<=>\(\left|x\right|.\left(\frac{1}{2015}+\frac{1}{2016}\right)-\left|x\right|.\left(\frac{1}{2016}+\frac{1}{2017}\right)=0\)
<=>\(\left|x\right|.\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2016}-\frac{1}{2017}\right)=0\)
<=>\(\left|x\right|.\left(\frac{1}{2015}-\frac{1}{2017}\right)=0\)
Vì \(\frac{1}{2015}-\frac{1}{2017}\ne0\Rightarrow\left|x\right|=0\Rightarrow x=0\)
Vậy x=0
\(\left|\frac{x}{2015}+\frac{x}{2016}\right|=\left|\frac{x}{2016}+\frac{x}{2017}\right|\)
\(\Rightarrow\left|x.\left(\frac{1}{2015}+\frac{1}{2016}\right)\right|=\left|x.\left(\frac{1}{2016}+\frac{1}{2017}\right)\right|\)
\(\Rightarrow\left|x\right|.\left|\frac{1}{2015}+\frac{1}{2016}\right|=\left|x\right|.\left|\frac{1}{2016}+\frac{1}{2017}\right|\)
\(\Rightarrow\left|x\right|.\left(\frac{1}{2015}+\frac{1}{2016}\right)=\left|x\right|.\left(\frac{1}{2016}+\frac{1}{2017}\right)\)
Mà \(\frac{1}{2015}+\frac{1}{2016}>\frac{1}{2016}+\frac{1}{2017}\)
=> |x| = 0
=> x = 0
Vậy x = 0
\(\frac{x}{2015}+\frac{x}{2016}=\frac{x}{2016}+\frac{x}{2017}\)
=>\(\frac{x}{2015}=\frac{x}{2017}\)
Vì 2015 khác 2017. Nên x=0