Áp dụng bất đẳng thức Bunyakovsky:

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

\(\le\left(1^2+1^2+1^2\right)\left(2x+yz+2y+xz+2z+xy\right)\)

\(=3\left(4+xy+yz+xz\right)=12+3\left(xy+yz+xz\right)\)

Mặt khác,theo AM-GM:

\(3\left(xy+yz+xz\right)\le\left(x+y+z\right)^2=4\)

\(\Rightarrow12+3\left(xy+yz+xz\right)\le12+4=16\)

\(\Rightarrow P^2\le16\Leftrightarrow P\le4\)

Dấu "=" xảy ra khi: \(x=y=z=\dfrac{2}{3}\)

<=>\(\left(x-19\right)-2\sqrt{x-19}+1+\left(y-7\right)+4\sqrt{y-7}+4\)+\(+\left(z-1997\right)-6\sqrt{z-1997}+9=0\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-19}=1\\\sqrt{y-7}=2\\\sqrt{z-1997}=3\end{cases}\Leftrightarrow\hept{\begin{cases}x=20\\y=11\\z=2006\end{cases}}}\)

vay...

\(\Leftrightarrow\left(x-19\right)2\sqrt{x-19}+1+\left(y-7\right)+4+\left(z-1997\right)+9=0\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x-19}=1\\\sqrt{y-7}=2\\\sqrt{z-1997}=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=20\\y=11\\z=2006\end{cases}}\)

Chúc bạn học tốt!

\(a.\) 

\(\text{*)}\) Áp dụng bđt  \(AM-GM\)  cho hai số thực dương  \(x,y,\)  ta có:

\(x+y\ge2\sqrt{xy}=2\)  (do  \(xy=1\)  )

\(\Rightarrow\)  \(3\left(x+y\right)\ge6\)

nên  \(D=x^2+y^2+\frac{9}{x^2+y^2+1}+3\left(x+y\right)\ge x^2+y^2+\frac{9}{x^2+y^2+1}+6\)

\(\Rightarrow\)  \(D\ge\left[\left(x^2+y^2+1\right)+\frac{9}{x^2+y^2+1}\right]+5\)

\(\text{*)}\)  Tiếp tục áp dụng bđt  \(AM-GM\)  cho bộ số loại hai số không âm gồm \(\left(x^2+y^2+1;\frac{9}{x^2+y^2+1}\right),\)  ta có:

\(\left[\left(x^2+y^2+1\right)+\frac{9}{x^2+y^2+1}\right]\ge2\sqrt{\left(x^2+y^2+1\right).\frac{9}{\left(x^2+y^2+1\right)}}=6\)

Do đó,  \(D\ge6+5=11\)

Dấu  \("="\)  xảy ra khi  \(x=y=1\)

Vậy,  \(D_{min}=11\)  \(\Leftrightarrow\)  \(x=y=1\)

\(b.\) Bạn tìm điểm rơi rồi báo lại đây

b

\(8\sqrt{x-1}=4.2.\sqrt{x-1}.1\le4.\left(x-1+1\right)=4x\)

\(x.\sqrt{16-3x^2}\le\frac{x^2+16-3x^2}{2}=8-x^2\)

\(\Rightarrow y\le4x-x^2+8=-\left(x-2\right)^2+12\le12\)

Dấu bằng xảy ra khi \(x=2\)

+ ĐKXĐ : \(\left\{{}\begin{matrix}x\ge-3\\y\ge-4\end{matrix}\right.\)

\(gt\Rightarrow x+y=6\left(\sqrt{x+3}+\sqrt{4+y}\right)\le6\sqrt{2\left(x+y+7\right)}\)

\(\Rightarrow\left(x+y\right)^2\le72\left(x+y+7\right)\)

\(\Rightarrow\left(x+y\right)^2-72\left(x+y\right)-504\le0\)

\(\Rightarrow\left(x+y-36\right)^2\le1800\Rightarrow P\le36+30\sqrt{2}\)

max \(P=36+30\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x+3=y+4\\x+y=36+30\sqrt{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\frac{37}{2}+15\sqrt{2}\\y=\frac{35}{2}+15\sqrt{2}\end{matrix}\right.\)

+ \(x+y=6\left(\sqrt{x+3}+\sqrt{y+4}\right)\)

\(\Rightarrow\left(x+y\right)^2=36\left(x+y+7+2\sqrt{\left(x+3\right)\left(y+4\right)}\right)\)

\(\Rightarrow\left(x+y\right)^2-36\left(x+y\right)-252=72\sqrt{\left(x+3\right)\left(y+4\right)}\ge0\)

\(\Rightarrow\left(x+y-42\right)\left(x+y+6\right)\ge0\Rightarrow x+y\ge42\)

Min \(P=42\Leftrightarrow\left\{{}\begin{matrix}\sqrt{\left(x+3\right)\left(y+4\right)}=0\\x+y=42\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-3\\y=45\end{matrix}\right.\\\left\{{}\begin{matrix}x=46\\y=-4\end{matrix}\right.\end{matrix}\right.\)

xin lỗi em mới lớp 8 ko trả lời dc

Ta có \(2x+y-xy=5\Leftrightarrow xy-2x-y+5=0\Leftrightarrow x\left(y-2\right)-\left(y-2\right)+3=0\Leftrightarrow\left(x-1\right)\left(y-2\right)=-3\).

Ta có bảng:

 \(y=\sqrt[3]{18+\sqrt{x+100}}+\sqrt[3]{18-\sqrt{x+100}}\) (Điều kiện xác định : \(x\ge-100\))

Ta có : \(36=\left(18+\sqrt{x+100}\right)+\left(18-\sqrt{x+100}\right)=\left(\sqrt[3]{18+\sqrt{x+100}}\right)^3+\left(\sqrt[3]{18-\sqrt{x+100}}\right)^3\)

Đặt \(a=\sqrt[3]{18+\sqrt{x+100}}\) ; \(b=\sqrt[3]{18-\sqrt{x+100}}\)

\(\Rightarrow a^3+b^3=36\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)=36\). Vì \(a+b\in Z^+\) nên a+b \(\in\) Ư(36)

\(\Rightarrow a+b\in\left\{1;2;3;4;6;9;12;18;36\right\}\)

Giải từng trường hợp , được x = 225 , y = 3 thoả mãn đề bài.