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

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

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

\(P\ge\dfrac{\sqrt{3\sqrt[3]{x^3y^3}}}{xy}+\dfrac{\sqrt{3\sqrt[3]{y^3z^3}}}{yz}+\dfrac{\sqrt{3\sqrt[3]{z^3x^3}}}{zx}\)

\(P\ge\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\ge\sqrt{3}.3\sqrt[3]{\dfrac{1}{\sqrt{xy.yz.zx}}}=3\sqrt{3}\)

Dấu "=" xảy ra khi \(x=y=z=1\)

Ta có bất đẳng thức sau \(x^3+y^3\ge xy\left(x+y\right)\Leftrightarrow\left(x+y\right)\left(x-y\right)^2\ge0.\)

Do đó:

\(P=\sum\dfrac{\sqrt{1+x^3+y^3}}{xy}\ge\sum\dfrac{\sqrt{xyz+xy\left(x+y\right)}}{xy}\)

\(=\sqrt{x+y+z}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\ge\sqrt{3\sqrt[3]{xyz}}\cdot3\sqrt[3]{\dfrac{1}{\sqrt{xy}}\cdot\dfrac{1}{\sqrt{yz}}\cdot\dfrac{1}{\sqrt{zx}}}=3\sqrt{3}\)

Đẳng thức xảy ra khi $x=y=z=1.$

Áp dụng BĐT AM-GM:

\(P=\dfrac{\sqrt{1+x^3+y^3}}{xy}+\dfrac{\sqrt{1+y^3+z^3}}{yz}+\dfrac{\sqrt{1+z^3+x^3}}{zx}\)

\(\ge\dfrac{\sqrt{3xy}}{xy}+\dfrac{\sqrt{3yz}}{yz}+\dfrac{\sqrt{3zx}}{zx}\)

\(=\sqrt{3}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{zx}}\right)\)

\(\ge\sqrt{3}.3\sqrt[3]{\dfrac{1}{xyz}}=3\sqrt{3}\)

\(minP=3\sqrt{3}\Leftrightarrow x=y=z\)

Lời giải:
a. \(B=\frac{3(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}-\frac{\sqrt{x}+5}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{3(\sqrt{x}+1)-(\sqrt{x}+5)}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{2(\sqrt{x}-1)}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{2}{\sqrt{x}+1}\)

b.

\(P=2AB+\sqrt{x}=2.\frac{\sqrt{x}+1}{\sqrt{x}+2}.\frac{2}{\sqrt{x}+1}+\sqrt{x}=\frac{4}{\sqrt{x}+2}+\sqrt{x}\)

Áp dụng BĐT Cô-si:

$P=\frac{4}{\sqrt{x}+2}+(\sqrt{x}+2)-2\geq 2\sqrt{4}-2=2$

Vậy $P_{\min}=2$ khi $\sqrt{x}+2=2\Leftrightarrow x=0$

\(P=\dfrac{x-1+4}{\sqrt{x}+1}=\sqrt{x}-1+\dfrac{4}{\sqrt{x}+1}\)

\(=\sqrt{x}+1+\dfrac{4}{\sqrt{x}+1}-2>=2\cdot2-2=2\)

Dấu = xảy ra khi x=1

Lời giải:

Do $x+y=1$ nên:

$P=\frac{x}{\sqrt{x+y-x}}+\frac{y}{\sqrt{x+y-y}}=\frac{x}{\sqrt{y}}+\frac{y}{\sqrt{x}}$
$=\frac{x^2}{x\sqrt{y}}+\frac{y^2}{y\sqrt{x}}$

$\geq \frac{(x+y)^2}{x\sqrt{y}+y\sqrt{x}}=\frac{1}{x\sqrt{y}+y\sqrt{x}}$ (áp dụng BĐT Cauchy-Schwarz)

Áp dụng BĐT Bunhiacopxky:

$(x\sqrt{y}+y\sqrt{x})^2\leq (x+y)(xy+xy)=2xy(x+y)\leq \frac{(x+y)^2}{2}(x+y)=\frac{1}{2}$

$\Rightarrow x\sqrt{y}+y\sqrt{x}\leq \frac{\sqrt{2}}{2}$

$\Rightarrow P\geq \frac{1}{x\sqrt{y}+y\sqrt{x}}\geq \frac{1}{\frac{\sqrt{2}}{2}}=\sqrt{2}$

Vậy $P_{\min}=\sqrt{2}$. Giá trị này đạt tại $x=y=\frac{1}{2}$.

\(a,=\dfrac{2x+6\sqrt{x}+x-3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}-3}{\sqrt{x}+3}\\ =\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}-5}\\ =\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}-3\right)}\)

a: \(=\dfrac{2x+6\sqrt{x}+x-3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}+3}\)

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

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

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

1: \(=3\left(x+\dfrac{2}{3}\sqrt{x}+\dfrac{1}{3}\right)\)

\(=3\left(x+2\cdot\sqrt{x}\cdot\dfrac{1}{3}+\dfrac{1}{9}+\dfrac{2}{9}\right)\)

\(=3\left(\sqrt{x}+\dfrac{1}{3}\right)^2+\dfrac{2}{3}>=3\cdot\dfrac{1}{9}+\dfrac{2}{3}=1\)

Dấu '=' xảy ra khi x=0

2: \(=x+3\sqrt{x}+\dfrac{9}{4}-\dfrac{21}{4}=\left(\sqrt{x}+\dfrac{3}{2}\right)^2-\dfrac{21}{4}>=-3\)

Dấu '=' xảy ra khi x=0

3: \(A=-2x-3\sqrt{x}+2< =2\)

Dấu '=' xảy ra khi x=0

5: \(=x-2\sqrt{x}+1+1=\left(\sqrt{x}-1\right)^2+1>=1\)

Dấu '=' xảy ra khi x=1