\(\dfrac{x}{10}=\dfrac{y}{14}=\dfrac{z}{15}=t\)

\(10.14.t^2+14.15.t^2+10.15.t^2=-2000\) < 0 loai

Vay ko co gt nao .....

Lời giải:

Dễ thấy:

$|7x-5y|\geq 0$ với mọi $x,y$

$|2z-3x|\geq 0$ với mọi $x,z$

$|xy+yz+xz-2000|\geq 0$ với mọi $x,y,z$

Do đó để tổng của chúng bằng $0$ thì:

$|7x-5y|=|2z-3x|=|xy+yz+xz-2000|=0$

\(\Rightarrow \left\{\begin{matrix} 7x=5y\\ 2z=3x\\ xy+yz+xz=2000\end{matrix}\right.\Rightarrow \left\{\begin{matrix} \frac{x}{10}=\frac{y}{14}=\frac{z}{15}\\ xy+yz+xz=2000\end{matrix}\right.\)

Đặt \(\frac{x}{10}=\frac{y}{14}=\frac{z}{15}=t\Rightarrow x=10t; y=14t; z=15t\)

\(\Rightarrow 2000=xy+yz+xz=10t.14t+10t.15t+14t.15t\)

\(\Leftrightarrow 2000=500t^2\Rightarrow t^2=4\Rightarrow t=\pm 2\)

\(\Rightarrow (x,y,z)=(20; 28; 30); (-20; -28; -30)\)

Vậy.......

Lời giải:
Dễ thấy:

$|7x-5y|\geq 0$ với mọi $x,y$

$|2z-3x|\geq 0$ với mọi $x,z$

$|xy+yz+xz-2000|\geq 0$ với mọi $x,y,z$

Do đó để tổng của 3 số trên bằng $0$ thì:

$|7x-5y|=|2z-3x|=|xy+yz+xz-2000|=0$

\(\Leftrightarrow \left\{\begin{matrix} 7x=5y\\ 2z=3x\\ xy+yz+xz=2000\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} \frac{x}{10}=\frac{y}{14}=\frac{z}{15}\\ xy+yz+xz=2000(*)\end{matrix}\right.\)

Đặt $\frac{x}{10}=\frac{y}{14}=\frac{z}{15}=t\Rightarrow x=10t; y=14t; z=15t$

Thay vào $(*)\Leftrightarrow 500t^2=2000\Rightarrow t=\pm 2$

$\Rightarrow (x,y,z)=(\pm 20,\pm 28, \pm 30)$

Lời giải:

Ta thấy:

$(7x-5y)^{2018}\geq 0, \forall x,y$

$(3x-2z)^{2020}\geq 0, \forall x,z$

$(xy+yz+xz-4500)^{2022}\geq 0, \forall x,y,z$

Do đó để tổng $(7x-5y)^{2018}+(3x-2z)^{2020}+(xy+yz+xz-4500)^{2022}=0$ thì:

$(7x-5y)^{2018}=(3x-2z)^{2020}=(xy+yz+xz-4500)^{2022}=0$

$\Leftrightarrow$ \(\left\{\begin{matrix} 7x=5y(1)\\ 3x=2z(2)\\ xy+yz+xz=4500(3)\end{matrix}\right.\)

Từ $(1);(2)\Rightarrow y=\frac{7}{5}x; z=\frac{3}{2}x$

Thay vào $(3)$:

$x.\frac{7}{5}x+\frac{7}{5}x.\frac{3}{2}x+x.\frac{3}{2}x=4500$

$\Leftrightarrow x^2=900\Rightarrow x=\pm 30$

Nếu $x=30\Rightarrow y=42; z=45$

Nếu $x=-30\Rightarrow y=-42; z=-45$

!

Cách khác:

\(\left(7x-5y\right)^{2018}+\left(3x-2z\right)^{2020}+\left(xy+yz+zx-4500\right)^{2022}=0\)

Ta có:

\(\left\{{}\begin{matrix}\left(7x-5y\right)^{2018}\ge0\\\left(3x-2z\right)^{2020}\ge0\\\left(xy+yz+zx-4500\right)^{2022}\ge0\end{matrix}\right.\forall x,y,z.\)

\(\Rightarrow\left(7x-5y\right)^{2018}+\left(3x-2z\right)^{2020}+\left(xy+yz+zx-4500\right)^{2022}\ge0\) \(\forall x,y,z.\)

\(\Rightarrow\left(7x-5y\right)^{2018}+\left(3x-2z\right)^{2020}+\left(xy+yz+zx-4500\right)^{2022}=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(7x-5y\right)^{2018}=0\\\left(3x-2z\right)^{2020}=0\\\left(xy+yz+zx-4500\right)^{2022}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}7x-5y=0\\3x-2z=0\\xy+yz+zx-4500=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}7x=5y\\3x=2z\\xy+yz+zx=4500\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\frac{x}{5}=\frac{y}{7}\\\frac{x}{2}=\frac{z}{3}\\xy+yz+zx=4500\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{10}=\frac{y}{14}\\\frac{x}{10}=\frac{z}{15}\\xy+yz+zx=4500\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\frac{x}{10}=\frac{y}{14}=\frac{z}{15}\\xy+yz+zx=4500\end{matrix}\right.\)

Đặt \(\frac{x}{10}=\frac{y}{14}=\frac{z}{15}=k\Rightarrow\left\{{}\begin{matrix}x=10k\\y=14k\\z=15k\end{matrix}\right.\)

Có: \(xy+yz+zx=4500\)

\(\Rightarrow10k.14k+14k.15k+15k.10k=4500\)

\(\Rightarrow140.k^2+210.k^2+150.k^2=4500\)

\(\Rightarrow k^2.\left(140+210+150\right)=4500\)

\(\Rightarrow k^2.500=4500\)

\(\Rightarrow k^2=4500:500\)

\(\Rightarrow k^2=9\)

\(\Rightarrow k=\pm3.\)

+ TH1: \(k=3.\)

\(\Rightarrow\left\{{}\begin{matrix}x=10.3=30\\y=14.3=42\\z=15.3=45\end{matrix}\right.\)

+ TH2: \(k=-3.\)

\(\Rightarrow\left\{{}\begin{matrix}x=10.\left(-3\right)=-30\\y=14.\left(-3\right)=-42\\z=15.\left(-3\right)=-45\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(30;42;45\right),\left(-30;-42;-45\right).\)

Chúc bạn học tốt!

Giá trị nhỏ nhất của A là 0

Ta có : (7x - 5y)²⁰¹⁸ + (3x - 2z)²⁰²⁰ + (xy + yz + xz - 4500)²⁰¹⁸ = 0

Ta có : \(\hept{\begin{cases}\left(7x-5y\right)^{2018}\ge0\\\left(3x-2z\right)^{2020}\ge0\\\left(xy+yz+xz-4500\right)^{2018}\ge0\end{cases}}\)

 \(\Rightarrow\left(7x-5y\right)^{2018}+\left(3x-2z\right)^{2020}+\left(xy+yz+xz-4500\right)^{2018}\ge0\)

Dấu bằng xảy ra <=> 

\(\begin{cases}7x=5y\\3x=2z\\xy+yz+xz=4500\end{cases}\Rightarrow\hept{\begin{cases}\frac{x}{5}=\frac{y}{7}\\\frac{x}{2}=\frac{z}{3}\\xy+yz+xz=4500\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{10}=\frac{y}{14}\\\frac{x}{10}=\frac{z}{15}\\xy+yz+xz=4500\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\frac{x}{10}=\frac{y}{14}=\frac{z}{15}\\x+y+z=4500\end{cases}}\)

Đặt \(\frac{x}{10}=\frac{y}{14}=\frac{z}{15}=k\Rightarrow\hept{\begin{cases}x=10k\\y=14k\\z=15k\end{cases}}\)

=> xy + yz + xz = 4500

<=> 10k.14k + 14k.15k + 10k.15k = 4500

=> 140.k² + 210.k² + 150.k² = 4500

=> k².(140 + 210 + 150) = 4500

=> k² . 500 = 4500

=> k² = 9

=> k = \(\pm3\)

Nếu k = 3

=> \(\hept{\begin{cases}x=30\\y=42\\z=45\end{cases}}\)

Nếu k = - 3

=> \(\hept{\begin{cases}x=-30\\y=-42\\z=-45\end{cases}}\)

\(\left(xy+yz+zx\right)^2\ge3xyz\left(x+y+z\right)=9\Rightarrow xy+yz+zx\ge3\)

\(2\left(x^2+y^2\right)-xy\ge\left(x+y\right)^2-\dfrac{1}{4}\left(x+y\right)^2=\dfrac{3}{4}\left(x+y\right)^2\)

Tương tự và nhân vế với vế:

\(VT\ge\dfrac{27}{64}\left[\left(x+y\right)\left(y+z\right)\left(z+x\right)\right]^2\)

Mặt khác ta có:

\(\left(x+y\right)\left(y+z\right)\left(z+x\right)=\left(x+y+z\right)\left(xy+yz+zx\right)-xyz\)

\(\ge\left(x+y+z\right)\left(xy+yz+zx\right)-\sqrt[3]{xyz}.\sqrt[3]{xy.yz.zx}\)

\(\ge\left(x+y+z\right)\left(xy+yz+xz\right)-\dfrac{1}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\)

\(=\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\ge\dfrac{8}{9}\sqrt{3\left(xy+yz+zx\right)}.\left(xy+yz+zx\right)\)

\(\Rightarrow VT\ge\dfrac{27}{64}.\dfrac{64}{81}.3\left(xy+yz+zx\right)^3\ge3^3=27\) (đpcm)

\(\left(x+y\right)\left(y+z\right)\left(z+x\right)=\left(x+y+z\right)\left(xy+yz+zx\right)-xyz\)

\(=\left(x+y+z\right)\left(xy+yz+zx\right)-\sqrt[3]{xyz}.\sqrt[3]{xy.yz.zx}\)

\(\ge\left(x+y+z\right)\left(xy+yz+zx\right)-\dfrac{1}{3}.\left(x+y+z\right).\dfrac{1}{3}\left(xy+yz+zx\right)\)

\(=\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\)

\(\ge\dfrac{8}{9}\sqrt{3\left(xy+yz+zx\right)}.\left(xy+yz+zx\right)\)

\(=\dfrac{8}{9}\sqrt{3\left(xy+yz+zx\right)^3}\)

\(\Rightarrow3\left(xy+yz+zx\right)^3\le\left(\dfrac{9}{8}\right)^2\)

\(\Rightarrow\left(xy+yz+zx\right)^3\le\dfrac{27}{64}\)

\(\Rightarrow xy+yz+zx\le\dfrac{3}{4}\)

Từ giả thiết \(xy+yz+zx=5\)

ta có \(x^2+5=x^2+xy+yz+zx=\left(x+y\right)\left(z+x\right)\)

Áp dụng BĐT AM-GM , ta có

\(\sqrt{6\left(x^2+5\right)}=\sqrt{6\left(x+y\right)\left(z+x\right)}\le\frac{3\left(x+y\right)+2\left(z+x\right)}{2}=\frac{5x+3y+2z}{2}\)

CM tương tự ta được \(\sqrt{6\left(y^2+5\right)}\le\frac{3x+5y+2z}{2};\sqrt{z^2+5}\le\frac{x+y+2z}{2}\)

Cộng zế zới zế BĐt trên ta đc

\(\sqrt{6\left(x^2+5\right)}+\sqrt{6\left(y^2+5\right)}+\sqrt{z^2+5}\le\frac{9x+9y+6z}{2}\)

\(=>P=\frac{3x+3y+2z}{\sqrt{6\left(x^2+5\right)}+\sqrt{6\left(y^2+5\right)}+\sqrt{x^2+5}}\ge\frac{2\left(3x+3y+2z\right)}{9x+9y+6z}=\frac{2}{3}\)

=> \(GTNN\left(P\right)=\frac{2}{3}khi\left(x=y=1;z=2\right)\)

Ta có \(\sqrt{6\left(x^2+5\right)}+\sqrt{6\left(y^2+5\right)}+\sqrt{z^2+5}=\sqrt{6\left(x+y\right)\left(x+z\right)}+\sqrt{6\left(y+z\right)\left(y+x\right)}\)\(+\sqrt{6\left(z+x\right)\left(z+y\right)}\)

\(\le\frac{3\left(x+y\right)+2\left(x+z\right)}{2}+\frac{3\left(x+y\right)+2\left(y+z\right)}{2}+\frac{\left(z+x\right)+\left(z+y\right)}{2}\le\frac{9x+9y+6z}{2}=\frac{3}{2}\)\(\left(3x+3y+2z\right)\)

\(\Rightarrow P=\frac{3x+3y+2z}{\sqrt{6\left(x^2+5\right)}+\sqrt{6\left(y^2+5\right)}+\sqrt{z^2+5}}\ge\frac{2}{3}\)

dấu "=" xảy ra \(\Leftrightarrow x=y=1;z=2\)

Vậy \(P_{min}=\frac{2}{3}\Leftrightarrow x=y=1;z=2\)