a) \(ĐK:x\ge0,x\ne9\)

Với\(x\ge0,x\ne9\)thì \(B=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right]:\left[\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right]\)\(=\left[\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right]:\left[\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right]\)\(=\left[\frac{2x-6\sqrt{x}}{x-9}+\frac{x+3\sqrt{x}}{x-9}-\frac{3\sqrt{x}+9}{x-9}\right]:\left[\frac{\sqrt{x}+1}{\sqrt{x}-3}\right]\)\(=\left[\frac{3x-6\sqrt{x}-9}{x-9}\right].\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{\left(\sqrt{x}+1\right)\left(3\sqrt{x}-9\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}=\frac{3\sqrt{x}-9}{\sqrt{x}+3}\)

b) \(B< -1\Leftrightarrow\frac{3\sqrt{x}-9}{\sqrt{x}+3}< -1\Leftrightarrow\frac{3\sqrt{x}-9}{\sqrt{x}+3}+1< 0\Leftrightarrow\frac{4\sqrt{x}-6}{\sqrt{x}+3}< 0\)

Mà \(\sqrt{x}+3>0\)nên \(4\sqrt{x}-6< 0\Leftrightarrow\sqrt{x}< \frac{3}{2}\Leftrightarrow x< \frac{9}{4}\)

Vậy với \(0\le x< \frac{9}{4}\)thì B < -1

c) \(B=\frac{4\sqrt{x}-6}{\sqrt{x}+3}=\frac{4\left(\sqrt{x}+3\right)-18}{\sqrt{x}+3}=4-\frac{18}{\sqrt{x}+3}\)

Ta có: \(\sqrt{x}\ge0\Leftrightarrow\sqrt{x}+3\ge3\Leftrightarrow\frac{18}{\sqrt{x}+3}\le6\Leftrightarrow-\frac{18}{\sqrt{x}+3}\ge-6\Leftrightarrow4-\frac{18}{\sqrt{x}+3}\ge-2\)

Vậy \(MinB=-2\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)

Nhìn nhầm câu c)

\(B=\frac{3\sqrt{x}-9}{\sqrt{x}+3}\)làm tương tự

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

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

b: Thay \(x=9+2\sqrt{7}\) vào P, ta được:

\(P=\dfrac{\sqrt{9+2\sqrt{7}}+1}{9+2\sqrt{7}+\sqrt{9+2\sqrt{7}+1}}\simeq0,25\)

đặt x^2-7x=y=> \(y\ge-\frac{49}{4}\) (*)

\(A=y\left(y+12\right)=y^2+12y=\left(y+6\right)^2-36\ge-36\)

đẳng thức khi y=-6 thủa mãn đk (*)

Vậy: GTNN của A=-36 khí y=-6 =>\(\left[\begin{matrix}x=1\\x=6\end{matrix}\right.\)

1.  x≥1 <=> \(\frac{1}{x}\le1\Leftrightarrow\frac{1}{x}+1\le2\Leftrightarrow A\le2\Rightarrow MaxA=2\Leftrightarrow x=1\)

2. Áp dụng bđt cosi cho x>0. ta có: \(x+\frac{1}{x}\ge2\sqrt{x.\frac{1}{x}}=2\Leftrightarrow P\ge2\Rightarrow MinP=2\Leftrightarrow x=\frac{1}{x}\Leftrightarrow x=1\)

3: \(A=\frac{x^2+x+4}{x+1}=\frac{\left(x^2+2x+1\right)-\left(x+1\right)+4}{x+1}=x+1-1+\frac{4}{x+1}\)

áp dụng cosi cho 2 số dương ta có: \(x+1+\frac{4}{x+1}\ge2\sqrt{x+1.\frac{4}{x+1}}=2\Leftrightarrow A+1\ge2\Rightarrow A\ge3\Rightarrow MinA=3\Leftrightarrow x+1=\frac{4}{x+1}\Leftrightarrow x=1\)

x=1 nhe nhap minh di ma ket ban voi minh nhe

a) đk: \(x\ge0;x\ne9\)

Ta có:

\(B=\left[\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{3\left(\sqrt{x}+3\right)}{x-9}\right]\div\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

\(B=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+3\right)\sqrt{x}-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)

\(B=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\div\frac{\sqrt{x}+1}{\sqrt{x}-3}\)

\(B=\frac{3x-6\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}-3}{\sqrt{x}+1}\)

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

\(B=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}=\frac{3\sqrt{x}-9}{\sqrt{x}+3}\)

b) \(B< -1\Leftrightarrow\frac{3\sqrt{x}-9}{\sqrt{x}+3}+1< 0\)

\(\Leftrightarrow\frac{4\sqrt{x}-6}{\sqrt{x}+3}< 0\) , mà \(\sqrt{x}+3\ge3>0\left(\forall x\right)\)

=> \(4\sqrt{x}-6< 0\)

\(\Leftrightarrow4\sqrt{x}< 6\)

\(\Rightarrow\sqrt{x}< \frac{3}{2}\)

\(\Rightarrow x< \frac{9}{4}\)

Vậy \(0\le x< \frac{9}{4}\)

c) Ta có: \(B=\frac{3\sqrt{x}-9}{\sqrt{x}+3}=\frac{3\left(\sqrt{x}+3\right)-18}{\sqrt{x}+3}=3-\frac{18}{\sqrt{x}+3}\)

Vì \(\sqrt{x}+3\ge3\Rightarrow\frac{18}{\sqrt{x}+3}\le6\)

\(\Leftrightarrow3-\frac{18}{\sqrt{x}+3}\ge-3\)

\(\Rightarrow A\ge-3\)

Dấu "=" xảy ra khi: \(\sqrt{x}+3=3\Rightarrow x=0\)

Vậy \(Min_A=-3\Leftrightarrow x=0\)

1. Vì \(x^2\ge0\left(\text{ với mọi x}\right)\)(1)

=>\(x^2+2\ge2>0\)

=>\(\left(x^2+2\right)^2>0\)(2)

Từ (1) và (2) =>\(\frac{x^2}{\left(x^2+2\right)^2}\le\frac{0}{\left(x^2+2\right)^2}=0\) hay A\(\le0\)

=> giá trị lớn nhất của A là 0, khi và chỉ khi \(x^2=0\) <=> x=0.