Lời giải:

$A=\frac{\sqrt{x}-1}{\sqrt{x}+4}=1-\frac{5}{\sqrt{x}+4}$

Vì $x\geq 4\Rightarrow \sqrt{x}\geq 2\Rightarrow \sqrt{x}+4\geq 6$

$\Rightarrow \frac{5}{\sqrt{x}+4}\leq \frac{5}{6}$

$\Rightarrow A=1-\frac{5}{\sqrt{x}+4}\geq 1-\frac{5}{6}=\frac{1}{6}$

Vậy $A_{\min}=\frac{1}{6}$. Giá trị này đạt tại $x=4$.

2, rút gọn B=x^2/(y-1)+y^2/(x-1) 

AM-GM : x^2/(y-1)+4(y-1) >/ 4x ; y^2/(x-1)+4(x-1) >/ 4y 

=> B >/ 4x-4(y-1)+4y-4(x-1)=4x-4y+4+4y-4x+4=8 

minB=8 

Câu 1:

Áp dụng BĐT AM-GM ta có: \(x+1\ge2\sqrt{x}\)

\(\Rightarrow x+1+x+1\ge x+2\sqrt{x}+1\)

\(\Rightarrow2x+2\ge\left(\sqrt{x}+1\right)^2\left(1\right)\)

Tương tự cũng có: \(2y+2\ge\left(\sqrt{y}+1\right)^2\left(2\right)\)

Nhân theo vế của \(\left(1\right);\left(2\right)\) ta có:

\(\left(2x+2\right)\left(2y+2\right)\ge\left(\sqrt{x}+1\right)^2\left(\sqrt{y}+1\right)^2\ge16\)

\(\Rightarrow4\left(x+1\right)\left(y+1\right)\ge16\Rightarrow\left(x+1\right)\left(y+1\right)\ge4\)

Lại áp dụng BĐT AM-GM ta có:

\(\left(x+1\right)+\left(y+1\right)\ge2\sqrt{\left(x+1\right)\left(y+1\right)}\ge4\)

\(\Rightarrow x+y\ge2\). Giờ thì áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

\(A=\frac{x^2}{y}+\frac{y^2}{x}\ge\frac{\left(x+y\right)^2}{x+y}=x+y\ge2\)

Đẳng thức xảy ra khi \(x=y=1\)

\(\sqrt{xy}+\sqrt{x}+\sqrt{y}\ge3\)

ÁP DỤNG BĐT COSI
\(\sqrt{xy}+\sqrt{x}+\sqrt{y}\le\frac{x+y}{2}+\frac{x+1}{2}+\frac{y+1}{2}=x+y+1\ge3=>x+y\ge2\)

\(P\ge\frac{\left(x+y\right)^2}{x+y}=2\left(cosi\right)\) vậy min P=2 <=> x=y=1

                      Bài làm :

Ta có :

\(\left(\sqrt{x}+1\right)\left(\sqrt{y}+1\right)\ge4\)

\(\Leftrightarrow\sqrt{xy}+\sqrt{y}+\sqrt{x}+1\ge4\)

\(\Leftrightarrow\sqrt{xy}+\sqrt{x}+\sqrt{y}\ge3\)

Áp dụng BĐT cosi cho các số không âm ; ta được :

\(3\le\sqrt{xy}+\sqrt{x}+\sqrt{y}\le\frac{x+y}{2}+\frac{x+1}{2}+\frac{y+1}{2}=x+y+1\)

\(\Rightarrow x+y\ge2\)

Ta có :

\(P=\frac{x^2}{y}+\frac{y^2}{x}\ge\frac{\left(x+y\right)^2}{x+y}=x+y\)

\(\Rightarrow P\ge2\)

Dấu "=" xảy ra khi x=y=1

Vậy MinP = 2 <=> x=y=1

\(4\le\left(\sqrt{x}+1\right)\left(\sqrt{y}+1\right)\le\dfrac{1}{4}\left(\sqrt{x}+\sqrt{y}+2\right)^2\)

\(\Rightarrow\sqrt{x}+\sqrt{y}+2\ge4\)

\(\Rightarrow2\le\sqrt{x}+\sqrt{y}\le\sqrt{2\left(x+y\right)}\Rightarrow x+y\ge2\)

\(\Rightarrow P\ge\dfrac{\left(x+y\right)^2}{x+y}=x+y\ge2\)

Dấu "=" xảy ra khi \(x=y=1\)

Dạ có thể diễn đạt theo cách dễ hiểu cho đứa ngu lâu dốt bền như em được không ạ ? ._.

a: Ta có: \(A=\dfrac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{3-\sqrt{x}}\)

\(=\dfrac{x\sqrt{x}-3-2\left(x-6\sqrt{x}+9\right)-x-4\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x\sqrt{x}-x-4\sqrt{x}-6-2x+12\sqrt{x}-18}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x\sqrt{x}-3x+8\sqrt{x}-24}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)

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

\(=\dfrac{x+8}{\sqrt{x}+1}\)

Lời giải:
ĐKXĐ: $x\geq 0; x\neq 9$

a. \(A=\frac{x\sqrt{x}-3}{(\sqrt{x}+1)(\sqrt{x}-3)}-\frac{2(\sqrt{x}-3)^2}{(\sqrt{x}+1)(\sqrt{x}-3)}-\frac{(\sqrt{x}+3)(\sqrt{x}+1)}{(\sqrt{x}+1)(\sqrt{x}-3)}\)

\(=\frac{x\sqrt{x}-3x+8\sqrt{x}-24}{(\sqrt{x}+1)(\sqrt{x}-3)}=\frac{(\sqrt{x}-3)(x+8)}{(\sqrt{x}+1)(\sqrt{x}-3)}=\frac{x+8}{\sqrt{x}+1}\)

b.

\(14-6\sqrt{5}=(3-\sqrt{5})^2\Rightarrow \sqrt{x}=3-\sqrt{5}\)

\(A=\frac{14-6\sqrt{5}+8}{3-\sqrt{5}+1}=\frac{22-6\sqrt{5}}{4-\sqrt{5}}=\frac{58-2\sqrt{5}}{11}\)

c. 

Áp dụng BĐT Cô-si:
$x+4\geq 4\sqrt{x}\Rightarrow x+8\geq 4(\sqrt{x}+1)$

$\Rightarrow A=\frac{x+8}{\sqrt{x}+1}\geq 4$

Vậy $A_{\min}=4$. Giá trị này đạt tại $x=4$

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)

Ai bảo cậu là không tìm được minP vậy?