ai đó làm giúp mình , mình tích cho

nhân 2 vế cho 2

=>2x²+2y²+2z²=2xy+2yz+2zx

=>2x²+2y²+2z²-2xy-2yz-2zx=0

=>(2x²-2xy)+(2y²-2yz)+(2z²-2zx)=0

=>(x-y)²+(y-z)²+(z-x)²=0

mà (x-y)² >= 0 với mọi x,y

(y-z)² >= 0 với mọi y,z

(z-x)² >=0 với mọi z,x

=>(x-y)²+(y-z)²+(z-x)² >= 0

mà theo đề:(x-y)²+(y-z)²+(z-x)²=0

=>(x-y)²=(y-z)²=(z-x)²=0

=>x=y

   y=z

   z=x

hay x=y=z

do đó x²⁰¹⁵+y²⁰¹⁵+z²⁰¹⁵=3²⁰¹⁶

<=>x²⁰¹⁵+x²⁰¹⁵+x²⁰¹⁵=3²⁰¹⁶

<=>3x²⁰¹⁵=3²⁰¹⁶<=>x²⁰¹⁵=3²⁰¹⁶:3=3²⁰¹⁵<=>x=2015

Vậy x=y=z=2015

cau a ban de o hang dang thuc (x-y-z)^2 di

\(x^2+y^2+z^2=xy+yz+xz\)

\(2x^2+2y^2+2z^2=2xy+2yz+2xz\)

\(2x^2+2y^2+2z^2-2xy-2yz-2xz=0\)

\(\left(x^2-2xy+y^2\right)+\left(y^2-2yz+z^2\right)+\left(z^2-2xz+x^2\right)=0\)

\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2=0\)

Vì mũ chẵn luôn lớn hơn hoặc bằng 0

\(\Rightarrow\hept{\begin{cases}x-y=0\\y-z=0\\z-x=0\end{cases}\Rightarrow\hept{\begin{cases}x=y\\y=z\\z=x\end{cases}\Rightarrow}}x=y=z\)

\(\Rightarrow x^{2015}+y^{2015}+z^{2015}=x^{2015}+x^{2015}+x^{2015}=3x^{2015}\)

\(\Rightarrow3x^{2015}=3^{2016}\)

\(\Rightarrow x^{2015}=3^{2015}\)

\(\Rightarrow x=3\)

Vậy \(x=y=z=3\)

ko bít làm à

k bik nên mới hỏi

Ta có: \(\left(xy+2016z\right)\left(yz+2016z\right)\left(zx+2016y\right)\\ =\left(xy+\left(x+y+z\right)z\right)\left(yz+\left(x+y+z\right)x\right)\left(zx+\left(x+y+z\right)y\right)\\ =\left(xy+zx+zy+z^2\right)\left(yz+x^2+xy+xz\right)\left(zx+xỹ+y^2+yz\right)\\ =\left(y+z\right)\left(x+z\right)\left(x+z\right)\left(y+x\right)\left(z+y\right)\left(x+y\right)\\ =\left(y+z\right)^2\left(x+y\right)^2\left(z+x\right)^2\\ \Rightarrow\frac{\left(xy+2016z\right)\left(yz+2016z\right)\left(zx+2016y\right)}{\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}\\ =\frac{\left(y+z\right)^2\left(x+y\right)^2\left(z+x\right)^2}{\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}\\ =1\)

Thay x = 0; y = -z = 1, thỏa mãn đề bài nhưng:

0²⁰¹⁶ + 1²⁰¹⁶ + (-1)²⁰¹⁶ không bằng ( 0 + 1 - 1)²⁰¹⁶

=> xem lại đề.

Ta có: x²+y²+z²=xy+yz+zx (gt)

\(\Leftrightarrow\)2x²+2y²+2z²=2xy+2yz+2zx

\(\Leftrightarrow\)x²-2xy+y²+y²-2yz+z²+z²-2zx+x²=0

\(\Leftrightarrow\)(x-y)²+(y-z)²+(z-x)²=0

\(\Leftrightarrow\)x=y,y=z,z=x

\(\Leftrightarrow\)x=y=z

Khi đó:x²⁰¹⁶+y²⁰¹⁶+z²⁰¹⁶=3²⁰¹⁷

\(\Leftrightarrow\)3.x²⁰¹⁶=3²⁰¹⁷

\(\Leftrightarrow\)x²⁰¹⁶=3²⁰¹⁶

\(\Leftrightarrow\)x=\(\pm\)3

Vậy:x=y=z=3 hoặc x=y=z=-3

Ta có : \(x^2+y^2+z^2=xy+yz+xz\Leftrightarrow x^2+y^2+z^2-xy-yz-xz=0\)

\(\Leftrightarrow2\left(x^2+y^2+z^2-xy-yz-xz\right)=0\)

\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)

\(\Leftrightarrow x=y=z\)

Mà \(x^{2016}+y^{2016}+z^{2016}=3^{2017}\)

\(x^{2016}=y^{2016}=z^{2016}=\frac{3^{2017}}{3}=3^{2016}\)

\(\Rightarrow x=y=z=\sqrt[2016]{3^{2016}}=3\)

\(VT=\sqrt{\dfrac{yz}{x^2+xy+yz+xz}}+\sqrt{\dfrac{xy}{y^2+xy+yz+xz}}+\sqrt{\dfrac{xz}{z^2+xy+yz+xz}}\)

\(VT=\sqrt{\dfrac{yz}{\left(x+y\right)\left(x+z\right)}}+\sqrt{\dfrac{xy}{\left(y+z\right)\left(x+y\right)}}+\sqrt{\dfrac{xz}{\left(x+z\right)\left(y+z\right)}}\)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{yz}{\left(x+y\right)\left(x+z\right)}}\le\dfrac{\dfrac{y}{x+y}+\dfrac{z}{x+z}}{2}\\\sqrt{\dfrac{xy}{\left(y+z\right)\left(x+y\right)}}\le\dfrac{\dfrac{x}{x+y}+\dfrac{y}{y+z}}{2}\\\sqrt{\dfrac{xz}{\left(x+z\right)\left(y+z\right)}}\le\dfrac{\dfrac{x}{x+z}+\dfrac{z}{y+z}}{2}\end{matrix}\right.\)

\(\Rightarrow VT\le\dfrac{\left(\dfrac{x}{x+y}+\dfrac{y}{x+y}\right)+\left(\dfrac{y}{y+z}+\dfrac{z}{y+z}\right)+\left(\dfrac{z}{x+z}+\dfrac{x}{x+z}\right)}{2}\)

\(\Rightarrow VT\le\dfrac{\dfrac{x+y}{x+y}+\dfrac{y+z}{y+z}+\dfrac{x+z}{x+z}}{2}=\dfrac{3}{2}\)

\(\Leftrightarrow\sqrt{\dfrac{yz}{x^2+2016}}+\sqrt{\dfrac{xy}{y^2+2016}}+\sqrt{\dfrac{xz}{z^2+2016}}\le\dfrac{3}{2}\) ( đpcm )

Dấu " = " xảy ra khi \(x=y=z=4\sqrt{42}\)

Sửa đề:\(\sqrt{\dfrac{yz}{x^2+2016}}+\sqrt{\dfrac{xy}{z^2+2016}}+\sqrt{\dfrac{xz}{y^2+2016}}\le\dfrac{3}{2}\)

Giải

Ta có:

\(\sqrt{\dfrac{xy}{z^2+2016}}=\sqrt{\dfrac{xy}{z^2+xy+xz+yz}}=\sqrt{\dfrac{xy}{\left(x+z\right)\left(y+z\right)}}\)

Áp dụng BĐT AM-GM ta có:

\(\sqrt{\dfrac{xy}{z^2+2016}}=\sqrt{\dfrac{xy}{\left(x+z\right)\left(y+z\right)}}\le\dfrac{1}{2}\left(\dfrac{x}{x+z}+\dfrac{y}{y+z}\right)\)

Tương tự cho 2 BĐT còn lại ta có:

\(\sqrt{\dfrac{yz}{x^2+2016}}\le\dfrac{1}{2}\left(\dfrac{y}{x+y}+\dfrac{z}{x+z}\right);\sqrt{\dfrac{xz}{y^2+2016}}\le\dfrac{1}{2}\left(\dfrac{x}{x+y}+\dfrac{z}{y+z}\right)\)

Cộng theo vế 3 BĐT trên ta có:

\(\Sigma\sqrt{\dfrac{xy}{z^2+2016}}\le\dfrac{1}{2}\Sigma\left(\dfrac{x}{x+z}+\dfrac{y}{y+z}\right)=\dfrac{1}{2}\Sigma\left(\dfrac{x}{x+z}+\dfrac{z}{x+z}\right)=\dfrac{3}{2}\)

Đẳng thức xảy ra khi \(x=y=z=4\sqrt{42}\)

xí bài này nhé, lát nữa hoặc mai giải