Điều kiện: \(x\ge2012;y\ge2013;z\ge2014\)

Áp dụng bất đẳng thức Cauchy, ta có:

\(\left\{{}\begin{matrix}\dfrac{\sqrt{x-2012}-1}{x-2012}=\dfrac{\sqrt{4\left(x-2012\right)}-2}{2\left(x-2012\right)}\le\dfrac{\dfrac{4+x-2012}{2}-2}{2\left(x-2012\right)}=\dfrac{1}{4}\\\dfrac{\sqrt{y-2013}-1}{y-2013}=\dfrac{\sqrt{4\left(y-2013\right)}-2}{2\left(y-2013\right)}\le\dfrac{\dfrac{4+y-2013}{2}-2}{2\left(y-2013\right)}=\dfrac{1}{4}\\\dfrac{\sqrt{z-2014}-1}{z-2014}=\dfrac{\sqrt{4\left(z-2014\right)}-2}{2\left(z-2014\right)}\le\dfrac{\dfrac{4+z-2014}{2}-2}{2\left(z-2014\right)}=\dfrac{1}{4}\end{matrix}\right.\)

Cộng vế theo vế, ta được:

\(\dfrac{\sqrt{x-2012}-1}{x-2012}+\dfrac{\sqrt{y-2013}-1}{y-2013}+\dfrac{\sqrt{z-2014}-1}{z-2014}\le\dfrac{3}{4}\)

Đẳng thức xảy ra khi \(x=2016;y=2017;z=2018\)

Vậy....

Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:

\(\dfrac{x+y}{2012}=\dfrac{xy}{2013}=\dfrac{x-y}{2014}=\dfrac{x+y+x-y}{2012+2014}=\dfrac{2x}{4026}=\dfrac{x}{2013}\)

\(hay:\dfrac{xy}{2013}=\dfrac{x}{2013}\Rightarrow xy=x\Rightarrow y=1\)

Ta có:

\(\dfrac{x+y}{2012}=\dfrac{x}{2013}\) (c/m trên)

\(\Rightarrow\left(x+y\right)2013=2012x\\ hay:\left(x+1\right)2013=2012x\\ \Rightarrow2013x+2013=2012x\\ \Rightarrow2013x-2012x=-2013\\ \Rightarrow x=-2013\)

Vậy: x=-2013

Có \(VT=\dfrac{x^2}{x^3-xyz+2013x}+\dfrac{y^2}{y^3-xyz+2013y}+\dfrac{z^2}{z^3-xyz+2013z}\)

\(\ge\dfrac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2013\left(x+y+z\right)}\)

\(=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left[x^2+y^2+z^2-\left(xy+yz+zx\right)\right]+2013\left(x+y+z\right)}\)

\(=\dfrac{x+y+z}{x^2+y^2+z^2-\left(xy+yz+zx\right)+3\left(xy+yz+zx\right)}\) 

(vì \(2013=3.671=3\left(xy+yz+zx\right)\))

\(=\dfrac{x+y+z}{x^2+y^2+z^2+2\left(xy+yz+zx\right)}\)

\(=\dfrac{x+y+z}{\left(x+y+z\right)^2}\)

\(=\dfrac{1}{x+y+z}\)

ĐTXR \(\Leftrightarrow\dfrac{1}{x^2-yz+2013}=\dfrac{1}{y^2-zx+2013}=\dfrac{1}{z^2-xy+2013}\)

\(\Leftrightarrow x^2-yz=y^2-zx=z^2-xy\)

\(\Leftrightarrow x=y=z\) (với \(x,y,z>0\))

Vậy ta có đpcm.

Ta có:

\(VT=\dfrac{x^2}{x^3-xyz-2013x}+\dfrac{y^2}{y^3-xyz-2013y}+\dfrac{z^2}{z^3-xyz-2013z}\ge\dfrac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz-2013.\left(z+y+z\right)}\)

\(VT=\dfrac{\left(x+y+x\right)^2}{x^3+y^3+z^3+3\left[\left(x+y+z\right).\left(xy+yz+xz\right)-xyz\right]}\)

\(VT=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^3}\)

\(VT=\dfrac{1}{x+y+z}=VP\)

\(\Rightarrow\) Đpcm.

ĐKXĐ : \(\left\{{}\begin{matrix}x\ge2011\\y\ge2012\\z\ge2013\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}a=\sqrt{x-2011}\ge0\\b=\sqrt{y-2012}\ge0\\c=\sqrt{z-2013}\ge0\end{matrix}\right.\) ta có :

\(\frac{a-1}{a^2}+\frac{b-1}{b^2}+\frac{c-1}{c^2}=\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{a^2}-\frac{1}{a}+\frac{1}{4}+\frac{1}{b^2}-\frac{1}{b}+\frac{1}{4}+\frac{1}{c^2}-\frac{1}{c}+\frac{1}{4}=0\)

\(\Leftrightarrow\left(\frac{1}{a}-\frac{1}{2}\right)^2+\left(\frac{1}{b}-\frac{1}{2}\right)^2+\left(\frac{1}{c}-\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow a=b=c=2\Leftrightarrow\left\{{}\begin{matrix}x=2015\\y=2016\\z=2017\end{matrix}\right.\)

Áp dụng tc dtsbn:

\(\dfrac{x}{2013}=\dfrac{y}{2014}=\dfrac{z}{2015}=\dfrac{x-z}{-2}=\dfrac{y-z}{-1}=\dfrac{x-y}{-1}\\ \Leftrightarrow\dfrac{x-z}{2}=\dfrac{y-z}{1}=\dfrac{x-y}{1}\\ \Leftrightarrow x-z=2\left(y-z\right)=2\left(x-y\right)\\ \Leftrightarrow\left(x-z\right)^3=8\left(x-y\right)^3=8\left(x-y\right)^2\left(x-y\right)=8\left(x-y\right)^2\left(y-z\right)\)

2) \(\sum\dfrac{x}{x^2-yz+2013}=\sum\dfrac{x^2}{x^3-xyz+2013x}\ge\dfrac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2013\left(x+y+z\right)}=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^3}=\dfrac{1}{x+y+z}\left(đpcm\right)\)

Còn câu 1 nữa ạ, ai giải giúp em vớii

phần a

vì x/2= y/3

y/5= z/4

=>x/2 nhân 1.5 = y/3 nhân 1/5

=> y/5 nhân 1/3 = z/4 nhân 1/3

=>x/10 = y/15 (1)

=>y/15 = z/12 (2)

Từ (1) , (2) ta có :

x/10 = y/15 = z/12

áp dụng t/c......

=>x/10 = y/15 = z/12

=>x+y+z/10+15+12

=> -49/37

b lm tiếp bc tiếp theo nhé✔

Vì mk cmt đầu tiên lên b tích dùm m☢