a) đk: x\(\ge0\);

P = \(\left[\dfrac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}+1}\right].\dfrac{4\sqrt{x}}{3}\)

= \(\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\dfrac{4\sqrt{x}}{3}\)

= \(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\dfrac{4\sqrt{x}}{3}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)

b) Để P = \(\dfrac{8}{9}\)

<=> \(\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\dfrac{8}{9}\)

<=> \(\dfrac{\sqrt{x}}{x-\sqrt{x}+1}=\dfrac{2}{3}\)

<=> \(\dfrac{3\sqrt{x}-2x+2\sqrt{x}-2}{3\left(x-\sqrt{x}+1\right)}=0\)

<=> \(-2x+5\sqrt{x}-2=0\)

<=> \(\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)

<=> \(\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{1}{4}\left(tm\right)\end{matrix}\right.\)

c)

Đặt \(\sqrt{x}=a\) (\(a\ge0\))

P = \(\dfrac{4a}{3\left(a^2-a+1\right)}\)

Xét P + \(\dfrac{4}{9}\) = \(\dfrac{4a}{3a^2-3a+3}+\dfrac{4}{9}=\dfrac{12a+4a^2-4a+4}{9\left(a^2-a+1\right)}=\dfrac{4a^2+8a+4}{9\left(a^2-a+1\right)}=\dfrac{4\left(a+1\right)^2}{9\left(a^2-a+1\right)}\ge0\)

Dấu "=" <=> a = -1 (loại)

=> Không tìm được Min của P

Xét P - \(\dfrac{4}{3}\) = \(\dfrac{4a}{3\left(a^2-a+1\right)}-\dfrac{4}{3}=\dfrac{4a-4a^2+4a-4}{3\left(a^2-a+1\right)}=\dfrac{-4a^2+8a-4}{3\left(a^2-a+1\right)}=\dfrac{-4\left(a-1\right)^2}{3\left(a^2-a+1\right)}\le0\)

<=> \(P\le\dfrac{4}{3}\)

Dấu "=" <=> a = 1 <=> x = 1 (tm)

a) ĐKXĐ: \(x\ge0\)

b) Ta có: \(P=\left(\dfrac{x+2}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right)\cdot\dfrac{4\sqrt{x}}{3}\)

\(=\left(\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\right)\cdot\dfrac{4\sqrt{x}}{3}\)

\(=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\dfrac{4\sqrt{x}}{3}\)

\(=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)

Ta có: \(P=\dfrac{8}{9}\)

nên \(36\sqrt{x}=27\left(x-\sqrt{x}+1\right)\)

\(\Leftrightarrow27x-27\sqrt{x}+27-36\sqrt{x}=0\)

\(\Leftrightarrow27x-63\sqrt{x}+27=0\)

\(P=\dfrac{\sqrt{x}-3+2}{\sqrt{x}-3}=1+\dfrac{2}{\sqrt{x}-3}\)

P lớn nhất khi \(\dfrac{2}{\sqrt{x}-3}\) lớn nhất

\(\Rightarrow\sqrt{x}-3\) là số dương nhỏ nhất

\(\Rightarrow x\) là số nguyên dương nhỏ nhất thỏa mãn \(\sqrt{x}-3\) dương

\(\sqrt{x}-3>0\Rightarrow x>9\)

\(\Rightarrow x_{min}=10\)

Khi đó \(P_{max}=\dfrac{\sqrt{10}-1}{\sqrt{10}-3}\)

$A=2x-\sqrt{x}=2(x-\frac{1}{2}\sqrt{x}+\frac{1}{4^2})-\frac{1}{8}$

$=2(\sqrt{x}-\frac{1}{4})^2-\frac{1}{8}$

$\geq \frac{-1}{8}$

Vậy $A_{\min}=-\frac{1}{8}$. Giá trị này đạt tại $x=\frac{1}{16}$

$B=x+\sqrt{x}$

Vì $x\geq 0$ nên $B\geq 0+\sqrt{0}=0$

Vậy $B_{\min}=0$. Giá trị này đạt tại $x=0$

Vì $2-x\geq 0$ (theo ĐKXĐ) nên $C=1+\sqrt{2-x}\geq 1$

Vậy $C_{\min}=1$. Giá trị này đạt tại $2-x=0\Leftrightarrow x=2$

ĐKXĐ: x>=0

\(M=\dfrac{\sqrt{x}+2+1}{\sqrt{x}+2}=1+\dfrac{1}{\sqrt{x}+2}\)

\(\sqrt{x}+2>=2\)

=>\(\dfrac{1}{\sqrt{x}+2}< =\dfrac{1}{2}\)

=>\(M=1+\dfrac{1}{\sqrt{x}+2}< =\dfrac{1}{2}+1=\dfrac{3}{2}\)

Dấu = xảy ra khi x=0

Dấu = xảy ra khi x=0

1. 1/x + 2/1-x = (1/x - 1) + (2/1-x - 2) + 3

= 1-x/x + (2-2(1-x))/1-x  + 3

= 1-x/x + 2x/1-x + 3    >= 2√2 + 3

Dấu "=" xảy ra khi x =√2 - 1

2. a = √z-1, b = √x-2, c = √y-3 (a,b,c >=0)

=> P = √z-1 / z + √x-2 / x + √y-3 / y 

= a/a^2+1 + b/b^2+2 + c/c^2+3

a^2+1 >= 2a              => a/a^2+1 <= 1/2

b^2+2 >= 2√2 b          => b/b^2+2 <= 1/2√2

c^2+3 >= 2√3 c            => c/c^2+3 <= 1/2√3

=> P <= 1/2 + 1/2√2 + 1/2√3

Dấu = xảy ra khi a^2 = 1, b^2 = 2, c^2 =3

<=> z-1 = 1, x-2 = 2, y-3 = 3

<=> x=4, y=6, z=2

Lời giải:

Câu a:

Áp dụng BĐT Cô-si ngược dấu ta có:

\(\sqrt{3(x-3)}\leq \frac{3+(x-3)}{2}=\frac{x}{2}\)

\(\Rightarrow \sqrt{x-3}\leq \frac{x}{2\sqrt{3}}\Rightarrow \frac{\sqrt{x-3}}{x}\leq \frac{1}{2\sqrt{3}}\)

Hoàn toàn tương tự: \(\frac{\sqrt{y-3}}{y}\leq \frac{1}{2\sqrt{3}}\)

\(\Rightarrow p=\frac{\sqrt{x-3}}{x}+\frac{\sqrt{y-3}}{y}\leq \frac{1}{2\sqrt{3}}+\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{3}\)

Dấu "=" xảy ra khi \(3=x-3; 3=y-3\Rightarrow x=y=6\)

Vậy \(p_{\max}=\frac{\sqrt{3}}{3}\Leftrightarrow x=y=6\)

Câu b: Các phân thức của $q$ là nghịch đảo của $p$ nên $q$ có min thôi em nhé. Nếu tìm min thì tương tự như câu a.

\(P=\dfrac{x}{\sqrt{2.\dfrac{1}{2}x+2yz}}+\dfrac{y}{\sqrt{2.\dfrac{1}{2}y+zx}}+\dfrac{z}{\sqrt{2.\dfrac{1}{2}z+xy}}\)

\(=\dfrac{x}{\sqrt{2x\left(x+y+z\right)+yz}}+\dfrac{y}{\sqrt{2y\left(x+y+z\right)+2zx}}+\dfrac{z}{\sqrt{2z\left(x+y+z\right)+2xy}}\)

\(=\dfrac{x}{\sqrt{2\left(x+y\right)\left(x+z\right)}}+\dfrac{y}{\sqrt{2\left(x+y\right)\left(y+z\right)}}+\dfrac{z}{\sqrt{2\left(x+z\right)\left(y+z\right)}}\)

\(=\dfrac{1}{2\sqrt{2}}.2\sqrt{\dfrac{x}{x+y}}.\sqrt{\dfrac{x}{x+z}}+\dfrac{1}{2\sqrt{2}}.2\sqrt{\dfrac{y}{x+y}}.\sqrt{\dfrac{y}{y+z}}+\dfrac{1}{2\sqrt{2}}.2\sqrt{\dfrac{z}{x+z}}.\sqrt{\dfrac{z}{y+z}}\)

\(\le\dfrac{1}{2\sqrt{2}}\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}+\dfrac{y}{x+y}+\dfrac{y}{y+z}+\dfrac{z}{x+z}+\dfrac{z}{y+z}\right)\)

\(=\dfrac{3}{2\sqrt{2}}\)

Dấu "=" xảy ra tại  \(x=y=z=\dfrac{1}{6}\)

\(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\left(x\ge0,x\ne1\right)\)

\(=\left(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right).\dfrac{2}{\sqrt{x}-1}\)

\(=\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)

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

\(=\dfrac{2}{x+\sqrt{x}+1}\)

b) Ta có: \(x\ge0\Rightarrow x+\sqrt{x}+1\ge1\Rightarrow\dfrac{2}{x+\sqrt{x}+1}\le2\)

\(\Rightarrow max=2\) khi \(x=0\)

Ta có: \(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)

\(=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)

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

\(=\dfrac{2}{x+\sqrt{x}+1}\)