ĐKXĐ: \(x;y;z\ge0\)

Đặt \(\left(\dfrac{\sqrt{x}}{5};\dfrac{\sqrt{y}}{4};\dfrac{\sqrt{z}}{3}\right)=\left(a;b;c\right)>0\)

\(\Rightarrow\left\{{}\begin{matrix}5a+4b+3c=12\\10a+20b+30c=60abc\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}5a+4b+3c=12\\a+2b+3c=6abc\end{matrix}\right.\)

Ta có:

\(12=\left(a+a+a+a+a\right)+\left(b+b+b+b\right)+\left(c+c+c\right)\ge12\sqrt[12]{a^5b^4c^3}\)

\(\Rightarrow a^5b^4c^3\le1\) (1)

\(6abc=a+b+b+c+c+c\ge6\sqrt[6]{ab^2c^3}\)

\(\Rightarrow a^6b^6c^6\ge ab^2c^3\Rightarrow a^5b^4c^3\ge1\) (2)

(1);(2) \(\Rightarrow a^5b^4c^3=1\)

Đẳng thức xảy ra khi và chỉ khi \(a=b=c=1\)

\(\Rightarrow\left(x;y;z\right)=\left(25;16;9\right)\)

ĐKXĐ:...

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y}+\sqrt{z}=12\\\frac{\sqrt{x}}{5}+\frac{\sqrt{y}}{2}+\sqrt{z}=\frac{\sqrt{x}}{5}.\frac{\sqrt{y}}{2}.\sqrt{z}\end{matrix}\right.\)

Đặt \(\left(\frac{\sqrt{x}}{5};\frac{\sqrt{y}}{4};\frac{\sqrt{z}}{3}\right)=\left(a;b;c\right)\)

\(\Rightarrow\left\{{}\begin{matrix}5a+4b+3c=12\\a+2b+3c=6abc\end{matrix}\right.\)

Từ pt đầu ta có:

\(12=5a+4b+3c\ge12\sqrt[12]{a^5.b^4.c^3}\Leftrightarrow a^5b^4c^3\le1\) (1)

Từ pt sau:

\(6abc=a+2b+3c\ge6\sqrt[6]{ab^2c^3}\Leftrightarrow abc\ge\sqrt[6]{ab^2c^3}\)

\(\Leftrightarrow a^6b^6c^6\ge ab^2c^3\Leftrightarrow a^5b^4c^3\ge1\) (2)

Từ (1) và (2) \(\Rightarrow a^5b^4c^3=1\)

Dấu "=" xảy ra khi và chỉ khi \(a=b=c=1\)

\(\Rightarrow\left(\sqrt{x};\sqrt{y};\sqrt{z}\right)=\left(5;4;3\right)\Rightarrow\left(x;y;z\right)=\left(25;16;9\right)\)

Lời giải:
ĐK: $x,y,z\geq 0$

Áp dụng BĐT Cô-si:

\(\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}\geq 3\sqrt[3]{\frac{xyz}{(x+1)(y+1)(z+1)}}\)

\(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\geq 3\sqrt[3]{\frac{1}{(x+1)(y+1)(z+1)}}\)

Cộng theo vế và thu gọn:

\(3\geq 3.\frac{\sqrt[3]{xyz}+1}{\sqrt[3]{(x+1)(y+1)(z+1)}}\Leftrightarrow (x+1)(y+1)(z+1)\geq (1+\sqrt[3]{xyz})^3\)

Dấu "=" xảy ra khi $x=y=z$

Thay vào pt $(1)$ thì suy ra $x=y=z=1$

1) Ta có: \(\left\{{}\begin{matrix}3\sqrt{x}-\sqrt{y}=5\\2\sqrt{x}+3\sqrt{y}=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}9\sqrt{x}-3\sqrt{y}=15\\2\sqrt{x}+3\sqrt{y}=18\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}11\sqrt{x}=33\\3\sqrt{x}-\sqrt{y}=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=3\\\sqrt{y}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9\\y=16\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=9\\y=16\end{matrix}\right.\)

2) Ta có: \(\left\{{}\begin{matrix}\sqrt{x+3}-2\sqrt{y+1}=2\\2\sqrt{x+3}+\sqrt{y+1}=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-2\sqrt{x+3}+4\sqrt{y+1}=-4\\2\sqrt{x+3}+\sqrt{y+1}=4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}5\sqrt{y+1}=0\\\sqrt{x+3}-2\sqrt{y+1}=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{y+1}=0\\\sqrt{x+3}=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y+1=0\\x+3=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=1\end{matrix}\right.\)

Vậy: Hệ phương trình có nghiệm duy nhất là \(\left\{{}\begin{matrix}x=1\\y=-1\end{matrix}\right.\)

4. Đk: \(x,y\ge0\)

\(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y+1}=1\\\sqrt{y}+\sqrt{x+1}=1\end{matrix}\right.\left(1\right)\)

Ta có: \(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y+1}\ge0+1=1\\\sqrt{y}+\sqrt{x+1}\ge0+1=1\end{matrix}\right.\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow\left\{{}\begin{matrix}\sqrt{x}=0,\sqrt{x+1}=1\\\sqrt{y}=0,\sqrt{y+1}=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\)<tmđk>

Vậy hệ pt có nghiệm \(\left(x,y\right)=\left(0;0\right)\)

a, Áp dụng bất đẳng thức Holder cho 2 bộ số \(\left(x,y,z\right)\left(3;3;3\right)\) ta có:

\(\left(x+3\right)\left(y+3\right)\left(z+3\right)\ge\left(\sqrt[3]{xyz}+\sqrt[3]{3.3.3}\right)^3=\left(\sqrt[3]{xyz}+3\right)\)

\(\sqrt[3]{\left(x+3\right)\left(y+3\right)\left(z+3\right)}\ge3+\sqrt[3]{xyz}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z\)

\(\Rightarrow\sqrt{x}+\sqrt{y}+\sqrt{z}=3\sqrt{x}=\sqrt{2017}\)

\(\Rightarrow x=\frac{\sqrt{2017}}{3}\)

\(\Rightarrow\left(x,y,z\right)=\left(\frac{\sqrt{2017}}{3},\frac{\sqrt{2017}}{3},\frac{\sqrt{2017}}{3}\right)\)

P/s: Không chắc cho lắm ạ.

Vũ Minh Tuấn, Hoàng Tử Hà, đề bài khó wá, Lê Gia Bảo, Aki Tsuki, Nguyễn Việt Lâm, Lê Thị Thục Hiền,

Học 24h, @tth_new, @Akai Haruma, Nguyễn Trúc Giang, Băng Băng 2k6

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