đặt \(P=\frac{1}{\sqrt{x^5-x^2+3xy+6}}+\frac{1}{\sqrt{y^5-y^2+3yz+6}}+\frac{1}{\sqrt{z^5-z^2+3zx+6}}\)

ta có:\(\left(x^3+2x^2+3x+3\right)\left(x-1\right)^2\ge0\)

\(\Leftrightarrow x^5-x^2\ge3x-3\)

cmtt=>\(y^5-y^2\ge3y-3;z^5-z^2\ge3z-3\)

\(\Rightarrow P\le\frac{1}{\sqrt{3x-3+3xy+6}}+\frac{1}{\sqrt{3y-3+3yz+6}}+\frac{1}{\sqrt{3z-3+3zx+6}}\)

\(=\frac{1}{\sqrt{3\left(x+xy+1\right)}}+\frac{1}{\sqrt{3\left(y+yz+1\right)}}+\frac{1}{\sqrt{3\left(z+zx+1\right)}}\)

áp dụng bunhia ta có:

\(3\left(x+xy+1\right)\ge\left(\sqrt{x}+\sqrt{xy}+1\right)^2\)

cmtt\(\Rightarrow P\le\frac{1}{\sqrt{x}+\sqrt{xy}+1}+\frac{1}{\sqrt{y}+\sqrt{yz}+1}+\frac{1}{\sqrt{z}+\sqrt{zx}+1}\)

đặt \(\sqrt{x}=a;\sqrt{y}=b;\sqrt{z}=c\)

\(\Rightarrow\frac{1}{\sqrt{x}+\sqrt{xy}+1}+\frac{1}{\sqrt{y}+\sqrt{yz}+1}+\frac{1}{\sqrt{z}+\sqrt{zx}+1}=\frac{1}{a+ab+1}+\frac{1}{b+bc+1}+\frac{1}{c+ca+1}\)

\(=\frac{abc}{a+ab+abc}+\frac{1}{b+bc+1}+\frac{b}{bc+abc+b}=\frac{bc}{bc+b+1}+\frac{b}{bc+b+1}+\frac{1}{bc+b+1}=1\)

\(\Rightarrow P\le1\)

Theo bdt bunhiacopxki

\(T^2=\left(x\sqrt{4-y^2}+y\sqrt{4-z^2}+z\sqrt{4-x^2}\right)^2\le\left(x^2+y^2+z^2\right)\left(12-x^2-y^2-z^2\right)\)

Đặt \(x^2+y^2+z^2=t\), bài toán ban đầu trở thành 

Cho t thuộc khoảng [0;12] tìm max của \(T^2=t\left(12-t\right)\)

Ta có\(\left(t-6\right)^2\ge0\Leftrightarrow t^2-12t+36\ge0\Rightarrow t^2-12t\ge-36\Rightarrow-t^2+12t\le36\)

\(\Rightarrow t\left(12-t\right)\le36\Leftrightarrow T^2\le36\Rightarrow T\le6\) Vậy max=6, dấu bằng xảy ra khi \(x=y=z=\sqrt{2}\)

ta có P=\(\frac{x^2}{x\sqrt{y+3}}+\frac{y^2}{y\sqrt{z+3}}+\frac{z^2}{z\sqrt{x+3}}\ge\frac{\left(x+y+z\right)^2}{x\sqrt{y+3}+y\sqrt{z+3}+z\sqrt{x+3}}\)

mà \(\left(x\sqrt{y+3}+...\right)^2\le\left(x+y+z\right)\left(xy+yz+zx+3x+3y+3z\right)\le3\left(9+3\right)=36\) ( vì xy+yz+zx<=3)

=>\(x\sqrt{y+3}+...\le6\Rightarrow P\ge\frac{9}{6}=\frac{3}{2}\)

dấu = xảy ra <=> x=y=z=1

Tìm các giới hạn sau:

\(\lim\limits_{x\rightarrow-\infty}\) \(\dfrac{\sqrt{x^6+2}}{3\text{x}^3-1}\)

\(\lim\limits_{x\rightarrow+\infty}\) \(\dfrac{\sqrt{x^6+2}}{3\text{x}^3-1}\)

\(\lim\limits_{x\rightarrow-\infty}\) \(\left(\sqrt{2\text{x}^2+1}+x\right)\)

\(\lim\limits_{x\rightarrow1}\) \(\dfrac{2\text{x}^3-5\text{x}-4}{\left(x+1\right)^2}\)

Tim x, y, z

1/ \(\sqrt{x-2}+\sqrt{y-2008}+\sqrt{z-2009}=\dfrac{1}{2}\left(x+y+z\right)\)

2/ \(x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{x-5}\)

3/ Tinh T = \(x^2+y^2+z^2-7\) biet x-y-z = \(2\sqrt{x-34}+4\sqrt{y-21}+6\sqrt{z-4}+45\)

4/ \(2x^2+9y^2-6xy-12y-6x+29=0\)

5/\(4x^2+3y-4x+4xy-10y+9=0\)

ta có bđt cần chứng minh 

\(\frac{\sqrt{xy+z}+\sqrt{2x^2+2y^2}}{1+\sqrt{xy}}\ge1\Leftrightarrow\sqrt{xy+z}+\sqrt{2\left(x^2+y^2\right)}\ge1+\sqrt{xy}\)

Áp dụng bđt bu nhi ta có 

\(\sqrt{2\left(x^2+y^2\right)}\ge x+y\) (1)

mà x+y+z=1\(\Rightarrow xy+z=xy+z\left(x+y+z\right)=\left(z+x\right)\left(z+y\right)\)

áp dụng bu nhi a ta có \(\sqrt{\left(z+x\right)\left(z+y\right)}\ge z+\sqrt{xy}\) (2)

từ (1) và (2) => \(\sqrt{xy+z}+\sqrt{2x^2+2y^2}\ge x+y+z+\sqrt{xy}=1+\sqrt{xy}\)