1, Thay x = 16 vào ta được \(A=\dfrac{4}{4+3}=\dfrac{4}{7}\)

2, \(A+B=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)-3x-9}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{-x+6\sqrt{x}-9}{x-9}=\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{\sqrt{x}-3}{\sqrt{x}+3}=\dfrac{3}{\sqrt{x}+3}\)

Ta có đpcm 

A

Thay x=16 vào biểu thức A , ta có :

A= 

a, Ta có : \(x=9\Rightarrow\sqrt{x}=3\)

Thay vào biểu thức A ta được : \(A=\frac{2}{3-2}=2\)

b, Với \(x\ge0;x\ne4\)

\(B=\frac{\sqrt{x}}{\sqrt{x}+2}+\frac{4\sqrt{x}}{x-4}=\frac{\sqrt{x}\left(\sqrt{x}-2\right)+4\sqrt{x}}{x-4}\)

\(=\frac{x+2\sqrt{x}}{x-4}=\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}=\frac{\sqrt{x}}{\sqrt{x}-2}\)( đpcm )

c, Ta có : \(A+B=\frac{3x}{\sqrt{x}-2}\)hay 

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

\(\Rightarrow2+\sqrt{x}=3x\Leftrightarrow3x-2-\sqrt{x}=0\)

\(\Leftrightarrow\sqrt{x}=3x-2\Leftrightarrow x=9x^2-12x+4\)

\(\Leftrightarrow\left(9x-4\right)\left(x-1\right)=0\Leftrightarrow x=\frac{4}{9}\left(ktm\right);x=1\)( đk : \(x\ge\frac{2}{3}\))

sao cô cho cả đáp án ra lun thế ạ @@

à ko em nhầm nhầm em xin lỗi cô 

1) A = 2 khi x =9

3) 2 + \(\sqrt{x}\)- 3x = 0

-3x +\(\sqrt{x}\)+2 = 0

3x- \(\sqrt{x}\)- 2 =0

Đặt \(\sqrt{x}\)= t ; ĐK : t\(\ge\)0; t \(\ne\) 2

\(^{3t^2-t-2=0}\)

t1 =1 (TMĐK)

t2 = \(\dfrac{-2}{3}\)(Không TMĐK)

Với t = 1 => \(\sqrt{x}=1\Leftrightarrow x=1\)

a, Ta có : \(x=4\Rightarrow\sqrt{x}=2\)

Thay vào biểu thức A ta được : \(\frac{1}{2-1}=1\)

b, Với \(x\ge0;x\ne1\)

\(Q=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2}{x-1}-1=\frac{\sqrt{x}\left(\sqrt{x}+1\right)-2-x+1}{x-1}\)

\(=\frac{x+\sqrt{x}-2-x+1}{x-1}=\frac{\sqrt{x}-1}{x-1}=\frac{1}{\sqrt{x}+1}\)

c, Ta có : \(\frac{1}{Q}+P\le4\)hay\(1:\frac{1}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}\le4\)ĐK : \(x\ne1\)

\(\Leftrightarrow\frac{x-1+1}{\sqrt{x}-1}-4\le0\Leftrightarrow\frac{x-4\sqrt{x}+4}{\sqrt{x}-1}\le0\)

\(\Leftrightarrow\frac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-1}\le0\Rightarrow\sqrt{x}-1\le0\Leftrightarrow\sqrt{x}\le1\Leftrightarrow x\le1\)do \(\left(\sqrt{x}-2\right)^2\ge0\)

Kết hợp với đk, vậy \(x< 1\)

1, thay x=4 (TMĐKXĐ) vào P ta được:

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

vậy khi x=4 thì P =1

2,với x≥0,x≠1:

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

vậy Q=\(\dfrac{-1}{\sqrt{x}-1}\)

3,\(\dfrac{1}{Q}+P\le4\)

⇒1/\(\dfrac{-1}{\sqrt{x}-1}\)+\(\dfrac{1}{\sqrt{x}-1}\)≤4⇔\(\dfrac{-\sqrt{x}-1}{1}+\dfrac{1}{\sqrt{x}-1}\le4\)⇔\(\dfrac{-x+1+1}{\sqrt{x}-1}-4\le0\)⇔\(\dfrac{-x+2-4\sqrt{x}+4}{\sqrt{x}-1}\le0\)⇔\(\dfrac{-x-4\sqrt{x}+6}{\sqrt{x}-1}\le0\)⇔\(\dfrac{x+4\sqrt{x}-6}{\sqrt{x}-1}\le0\)⇔\(\dfrac{x+4\sqrt{x}+4-10}{\sqrt{x}-1}\le0\)

\(\dfrac{ \left(\sqrt{x}+2\right)^2-10}{\sqrt{x}-1}\le0\)⇒\(\sqrt{x}-1\le0\) (vì (\(\sqrt{x}+2\))\(^2\)≥0 ∀ x hay (\(\sqrt{x}+2\))\(^2\)-10>0 ∀ x)

                                ⇔x≤1 (KTM)

vậy không có giá trị nào của x TM  để  \(\dfrac{1}{Q}+P\le4\)

☹

                      Bài làm :

1) Khi x=9 ; giá trị của A là :

\(A=\frac{\sqrt{9}}{\sqrt{9}+2}=\frac{3}{3+2}=\frac{3}{5}\)

2) Ta có :

\(B=...\)

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

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

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

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

\(=\frac{\sqrt{x}}{\sqrt{x}-2}\)

3) Ta có :

\(\frac{A}{B}=\frac{\sqrt{x}}{\sqrt{x}+2}\div\frac{\sqrt{x}}{\sqrt{x}-2}=\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\sqrt{x}}=\frac{\sqrt{x}-2}{\sqrt{x}+2}=\frac{\sqrt{x}+2-4}{\sqrt{x}+2}=1-\frac{4}{\sqrt{x}+2}\)

Xét :

\(\frac{A}{B}+1=\frac{4}{\sqrt{x+2}}>0\Rightarrow\frac{A}{B}>-1\)

=> Điều phải chứng minh

1, thay x=9(TMĐKXĐ) vào A ta đk:

A=\(\dfrac{\sqrt{9}}{\sqrt{9}-2}=3\)

vậy khi x=9 thì A =3

2,với x>0,x≠4 ta đk:

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

vậy B=\(\dfrac{\sqrt{x}}{\sqrt{x}-2}\)

3,\(\dfrac{A}{B}>-1\) (x>0,x≠4)

⇒\(\dfrac{\sqrt{x}}{\sqrt{x}+2}:\dfrac{\sqrt{x}}{\sqrt{x}-2}>-1\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}+2}.\dfrac{\sqrt{x}-2}{\sqrt{x}}>-1\Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+2}>-1\)

⇒\(\sqrt{x}-2>-1\) (vì \(\sqrt{x}+2>0\))

⇔\(\sqrt{x}>1\)⇔x=1 (TM)

vậy x=1 thì \(\dfrac{A}{B}>-1\) với x>0 và x≠4

tự làm đi

a, Ta có : \(x=9\Rightarrow\sqrt{x}=3\)

Thay vào biểu thức A ta được : \(A=\frac{6+1}{3}=\frac{7}{3}\)

b, Với \(x>0;x\ne4\)

\(B=\frac{x-3\sqrt{x}+4}{x-2\sqrt{x}}-\frac{1}{\sqrt{x}-2}=\frac{\sqrt{x}-2}{\sqrt{x}}\)

\(a.x=3-2\sqrt{2}\\ \Rightarrow\sqrt{x}=\sqrt{3-2\sqrt{2}}\\ =\sqrt{2-2\sqrt{2}+1}\\ =\sqrt{\left(\sqrt{2}-1\right)^2}\\ =\left|\sqrt{2}-1\right|\\ =\sqrt{2}-1\left(vì\sqrt{2}>1\right)\)

Thay \(\sqrt{x}=\sqrt{2}-1\) vào A ta được

\(A=\dfrac{\sqrt{2}-1}{1+\sqrt{2}-1}=\dfrac{\sqrt{2}-1}{\sqrt{2}}=\dfrac{\sqrt{2}-2}{2}\)

\(b.B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}-\dfrac{10-5\sqrt{x}}{x-5\sqrt{x}+6}\\ B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}-\dfrac{10-5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ B=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{10-5\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\\ B=\dfrac{x-3\sqrt{x}-\sqrt{x}+3-x+4-10+5\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ B=\dfrac{\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\\ B=\dfrac{1}{\sqrt{x}-2}\)

\(c,P=A:B\\ P=\dfrac{\sqrt{x}}{1+\sqrt{x}}:\dfrac{1}{\sqrt{x}-2}\\ P=\dfrac{x-2\sqrt{x}}{1+\sqrt{x}}\)

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

Có: \(\sqrt{x}\ge0\)

\(\Rightarrow\sqrt{x}+1\ge1\left(I\right)\)

Lại có: \(\sqrt{x}\ge0\)

\(\Rightarrow-\sqrt{x}\le0\\ \Rightarrow-\sqrt{x}+2\le2\)

mà \(-\sqrt{x}\le0\)

\(\Rightarrow-\sqrt{x}\left(-\sqrt{x}+2\right)\ge2\)

Kết hợp với \(\left(I\right)\) \(\Rightarrow\) \(P=\dfrac{-\sqrt{x}\left(-\sqrt{x}+2\right)}{\sqrt{x}+1}\ge2\)

Vậy gtnn của P = \(2\) khi \(x=10+4\sqrt{6}\)

a: Khi \(x=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\) thì 

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

b: \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}+\dfrac{\sqrt{x}+2}{3-\sqrt{x}}-\dfrac{10-5\sqrt{x}}{x-5\sqrt{x}+6}\)

\(=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{5\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+5\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x-4\sqrt{x}+3-x+4+5\sqrt{x}-10}{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}-3\right)}\)

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