- Bổ sung đề: CMR \(\dfrac{1+3x}{1+y^2}+\dfrac{1+3y}{1+z^2}+\dfrac{1+3z}{1+x^2}\ge6\).

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

- Tương tự, ta cũng có:

\(\dfrac{1+3y}{1+z^2}=1+3y-\dfrac{z}{2}-\dfrac{3yz}{2}\left(2\right)\), \(\dfrac{1+3z}{1+x^2}=1+3z-\dfrac{x}{2}-\dfrac{3zx}{2}\left(3\right)\)

- Lấy \(\left(1\right)+\left(2\right)+\left(3\right)\), ta được:

\(\dfrac{1+3x}{1+y^2}+\dfrac{1+3y}{1+z^2}+\dfrac{1+3z}{1+x^2}\ge3+\dfrac{5}{2}\left(x+y+z\right)-\dfrac{3}{2}\left(xy+yz+zx\right)=3+\dfrac{5}{2}\left(x+y+z\right)-\dfrac{3}{2}.3=\dfrac{5}{2}\left(x+y+z\right)-\dfrac{3}{2}\left(4\right)\)

- Mặt khác: \(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)=3.3=9\Rightarrow x+y+z\ge3\left(5\right)\)

- Từ (4), (5) ta có:

\(\dfrac{1+3x}{1+y^2}+\dfrac{1+3y}{1+z^2}+\dfrac{1+3z}{1+x^2}\ge\dfrac{5}{2}.3-\dfrac{3}{2}=6\left(đpcm\right)\)

- Dấu "=" xảy ra khi \(x=y=z=1\)

đk a > 0

\(=\dfrac{4+4\sqrt{a}+a-a+6\sqrt{a}-9}{\sqrt{a}\left(2\sqrt{a}-1\right)}\\ =\dfrac{10\sqrt{a}-5}{\sqrt{a}\left(2\sqrt{a}-1\right)}\\ =\dfrac{5\left(2\sqrt{a}-1\right)}{\sqrt{a}\left(2\sqrt{a}-1\right)}\\ =\dfrac{5}{\sqrt{a}}\)

\(\sqrt{x-1}\le2\\ < =>0\le x-1\le4\\ < =>1\le x\le5\)

\(\sqrt{3x-2}=2-x\left(x\ge\dfrac{2}{3}\right)\\ < =>\left\{{}\begin{matrix}2-x\ge0\\3x-2=\left(2-x\right)^2=x^2-4x+4\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}-x\ge-2\\x^2-7x+6=0\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}x\le2\\\left(x-6\right)\left(x-1\right)=0\end{matrix}\right.\\ < =>\left\{{}\begin{matrix}x\le2\\\left[{}\begin{matrix}x=6\\x=1\end{matrix}\right.\end{matrix}\right.< =>x=1\left(TM\right)\)

    \(\sqrt{3+2\sqrt{2}}\) + \(\sqrt{6-4\sqrt{2}}\)

=   \(\sqrt{2+2\sqrt{2}+1}\)  + \(\sqrt{4-4\sqrt{2}+2}\)

= \(\sqrt{\left(\sqrt{2}+1\right)^2}\)   + \(\sqrt{\left(2-\sqrt{2}\right)^2}\)

= \(\sqrt{2}\) + 1 + 2 - \(\sqrt{2}\)

= 3

\(P=\dfrac{-3}{\sqrt{x}+3}=\dfrac{-\left(\sqrt{x}+3\right)+\sqrt{x}}{\sqrt{x}+3}=-1+\dfrac{\sqrt{x}}{\sqrt{x}+3}\ge-1\\ Dấu"="xảyrakhi:\\ \sqrt{x}=0\Rightarrow x=0\)

Đề có gì đó sai sai 

Thử với  n = 1 được 

P = 17n² + 7n + 1 = 25 (tm)

Nhưng 8n + 3 = 11 không là hợp số