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Áp dụng tc dtsbn:
\(\dfrac{x}{2013}=\dfrac{y}{2014}=\dfrac{z}{2015}=\dfrac{x-z}{-2}=\dfrac{y-z}{-1}=\dfrac{x-y}{-1}\\ \Leftrightarrow\dfrac{x-z}{2}=\dfrac{y-z}{1}=\dfrac{x-y}{1}\\ \Leftrightarrow x-z=2\left(y-z\right)=2\left(x-y\right)\\ \Leftrightarrow\left(x-z\right)^3=8\left(x-y\right)^3=8\left(x-y\right)^2\left(x-y\right)=8\left(x-y\right)^2\left(y-z\right)\)
\(\frac{x}{2014}=\frac{y}{2015}=\frac{z}{2016}=\frac{x-z}{-2}=\frac{x-y}{-1}=\frac{y-z}{-1}\Leftrightarrow x-z=2\left(x-y\right)=2\left(y-z\right)\)
\(\Leftrightarrow\left(x-z\right)^3=\left(x-z\right)^2\left(x-z\right)=\left[2\left(x-y\right)\right]^2.\left[2\left(y-z\right)\right]=8\left(x-y\right)^2\left(y-z\right)\)
Giải :
Đặt \(\frac{x}{2013}=\frac{y}{2014}=\frac{z}{2015}=k\Rightarrow\hept{\begin{cases}x=2013k\\y=2014k\\z=2015k\end{cases}}\)
Khi đó, ta có : 4(2013k - 2014k)(2014k - 2015k) = 4. (-k).(-k) = 4.k2 (1)
(2015k - 2013k)2 = (2k)2 = 22.k2 = 4k2 (2)
Từ (1) và (2) suy ta 4(x - y)(y - z) = (z - x)2
\(2\left(x-y\right)^2=\left(z-x\right)\left(z-y\right)\Leftrightarrow\frac{2\left(x-y\right)^2}{\left(z-x\right)\left(z-y\right)}=1\)
\(\frac{2\left(z-y\right)^2}{\left(z-x\right)\left(z-y\right)}=\frac{\left(x-y\right)^2}{z\left(x-y\right)}=\frac{x-y}{z}\Rightarrow x-y=z\)
a)
Ta có: \(\frac{x+y}{2014}\ne\frac{x-y}{2016}\)
\(\Leftrightarrow2016x+2016y=2014x-2014y\)
\(\Leftrightarrow2x=-4030y\)
\(\Leftrightarrow x=-2015y\)
Thay \(x=-2015y\)vào \(\frac{x+y}{2014}=\frac{xy}{2015}\)ta được:
\(\Leftrightarrow\frac{-2015+y}{2014}=\frac{-2015y}{2015}\)
\(\Leftrightarrow\frac{-2014y}{2014}=\frac{-2015y^2}{2015}\)
\(\Leftrightarrow-y=-y^2\)
\(\Leftrightarrow y-y^2=0\)
\(\Leftrightarrow y\left(1-y\right)=0\)
\(\Rightarrow\orbr{\begin{cases}y=0\\1-y=0\end{cases}}\Rightarrow\orbr{\begin{cases}y=0\\y=1\end{cases}}\)
Trường hợp \(y=0\):
\(y=0\Rightarrow x.y=-2015.0=0\)
Trường hợp \(y=1\):
\(y=1\Rightarrow x.y=-2015.1=-2015\)
Đặt \(\frac{x}{2015}=\frac{y}{2016}=\frac{z}{2017}=k\)
\(\Rightarrow x=2015k;y=2016k;z=2017k\)
Ta có:
\(\left(x-z\right)^3=\left(2015k-2017k\right)^3=-8k^3\left(1\right)\)
Mặt khác:
\(-8\left(x-y\right)^2\left(z-y\right)=-8\left(2015k-2016k\right)^2\left(2017k-2016k\right)\)
\(=-8k^2\cdot k=-8k^3\left(2\right)\)
Từ ( 1 );( 2 ) suy ra đpcm
Đặt t=x−z, dễ thấy 0≤t≤x−y⇒t=k(x−y),k∈[0;1]. Ta có:
f(x)+f(y)−f(z)−f(x+y−z)=f(x)+f(y)−f(x−t)−f(y+t)=f(x)+f(y)−f(x−k(x−y))−f(y+k(x−y))=f(x)+f(y)−f((1−k)x+ky)−f(kx+(1−k)y)≥f(x)+f(y)−(1−k)f(x)−kf(y)−kf(x)−(1−k)f(y)=0(Q.E.D