a) x = 16 (tm) => A = \(\frac{\sqrt{16}-2}{\sqrt{16}+1}=\frac{4-2}{4+1}=\frac{2}{5}\)

b) B = \(\left(\frac{1}{\sqrt{x}+5}-\frac{x+2\sqrt{x}-5}{25-x}\right):\frac{\sqrt{x}+2}{\sqrt{x}-5}\)

B = \(\frac{\sqrt{x}-5+x+2\sqrt{x}-5}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\frac{\sqrt{x}-5}{\sqrt{x}+2}\)

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

B = \(\frac{x+5\sqrt{x}-2\sqrt{x}-10}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}\)

B = \(\frac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}+2\right)}=\frac{\sqrt{x}-2}{\sqrt{x}+2}\)

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

=> \(P\left(\sqrt{x}+2\right)\ge x+6\sqrt{x}-13\)

<=> \(\frac{\sqrt{x}+1}{\sqrt{x}+2}.\left(\sqrt{x}+2\right)-x-6\sqrt{x}+13\ge0\)

<=> \(-x-6\sqrt{x}+13+\sqrt{x}+1\ge0\)

<=> \(-x-5\sqrt{x}+14\ge0\)

<=> \(x+5\sqrt{x}-14\le0\)

<=> \(x+7\sqrt{x}-2\sqrt{x}-14\le0\)

<=> \(\left(\sqrt{x}+7\right)\left(\sqrt{x}-2\right)\le0\)

Do \(\sqrt{x}+7>0\) với mọi x => \(\sqrt{x}-2\le0\)

<=> \(\sqrt{x}\le2\) <=> \(x\le4\)

Kết hợp với Đk: x \(\ge\)0; x \(\ne\)4; x \(\ne\)25

và x thuộc Z => x = {0; 1; 2; 3}

d) M = \(3P\cdot\frac{\sqrt{x}+2}{x+\sqrt{x}+4}\) <=>M = \(3\cdot\frac{\sqrt{x}+1}{\sqrt{x}+2}\cdot\frac{\sqrt{x}+2}{x+\sqrt{x}+4}\)

M = \(\frac{3\sqrt{x}+3}{x+\sqrt{x}+4}=\frac{x+\sqrt{x}+4-x+2\sqrt{x}-1}{\left(x+\sqrt{x}+\frac{1}{4}\right)+\frac{15}{4}}=1-\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{15}{4}}\le1\)(Do \(\left(\sqrt{x}-1\right)^2\ge0\) và \(\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{15}{4}>0\))

Dấu "=" xảy ra <=> \(\sqrt{x}-1=0\) <=> \(x=1\)

Vậy MaxM = 1 khi x = 1

a) \(\frac{\sqrt{11}}{2}\)

b)ko bt

TL:

\(A=\frac{\sqrt{x+2}}{\sqrt{x-5}}\) mà x = 9

\(A=\frac{\sqrt{0+2}}{\sqrt{9-2}}\)

\(A=\frac{\sqrt{11}}{2}\)

b) chưa bt làm

Ủa dùng máy tính thì nó ra \(\frac{\sqrt{11}}{2}\) mà

c,M =  \(\dfrac{A}{B}\) = \(\dfrac{\sqrt{x}-4}{\sqrt{x}+5}\) :  \(\dfrac{\sqrt{x}+3}{\sqrt{x}+5}\) 

   M =  \(\dfrac{A}{B}\) = \(\dfrac{\sqrt{x}-4}{\sqrt{x}+5}\) \(\times\) \(\dfrac{\sqrt{x}+5}{\sqrt{x}+3}\) 

   M =  \(\dfrac{A}{B}\) = \(\dfrac{\sqrt{x}-4}{\sqrt{x}+3}\) = \(\dfrac{\sqrt{x}+3-7}{\sqrt{x}+3}\)

 M = 1  - \(\dfrac{7}{\sqrt{x}+3}\) 

 M \(\in\) Z ⇔ 7 ⋮ \(\sqrt{x}\) + 3 vì \(\sqrt{x}\) ≥ 0 ⇒ \(\sqrt{x}\) + 3 ≥ 3 ⇒ 0< \(\dfrac{7}{\sqrt{x}+3}\) ≤ \(\dfrac{7}{3}\)

⇒ M Đạt giá trị nguyên lớn nhất ⇔ \(\dfrac{7}{\sqrt{x}+3}\) đạt giá trị nguyên nhỏ nhất ⇔ \(\dfrac{7}{\sqrt{x}+3}\) = 1 ⇔ \(\sqrt{x}\) + 3  = 7 ⇔ \(\sqrt{x}\) = 4 ⇔ \(x\) = 16 

Mnguyên(max)  = 1 - 1 = 0 xảy ra khi \(x\) = 16

học lớp 9 chưa mà đòi đăng ? :))

a) Ta có : \(A=\frac{x+5\sqrt{x}}{x-25}=\frac{\sqrt{x}\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{\sqrt{x}}{\sqrt{x}-5}\)

Để A nhận giá trị = 0 thì \(\sqrt{x}=0\)<=> x = 0 ( tmđk )

Vậy với x = 0 thì A = 0

b) \(B=\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{x+9\sqrt{x}}{x-9}\)

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

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

c) P = B : A = \(\frac{\frac{\sqrt{x}}{\sqrt{x}+3}}{\frac{\sqrt{x}}{\sqrt{x}-5}}=\frac{\sqrt{x}}{\sqrt{x}+3}\div\frac{\sqrt{x}}{\sqrt{x}-5}=\frac{\sqrt{x}}{\sqrt{x}+3}\times\frac{\sqrt{x}-5}{\sqrt{x}}=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)

Xét hiệu P - 1 ta có :

\(\frac{\sqrt{x}-5}{\sqrt{x}+3}-1=\frac{\sqrt{x}-5}{\sqrt{x}+3}-\frac{\sqrt{x}+3}{\sqrt{x}+3}=\frac{\sqrt{x}-5-\sqrt{x}-3}{\sqrt{x}+3}=\frac{-8}{\sqrt{x}+3}\)

Vì \(\hept{\begin{cases}-8< 0\\\sqrt{x}+3>0\end{cases}}\Rightarrow\frac{-8}{\sqrt{x}+3}< 0\)hay P - 1 < 0

=> P < 1 

a) \(A=0\Rightarrow\frac{x+5\sqrt{x}}{x-25}=0\Rightarrow x+5\sqrt{x}=0\Leftrightarrow x=0\)(thỏa mãn).

b) \(B=\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{x+9\sqrt{x}}{x-9}\)

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

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

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

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

\(B=\frac{\sqrt{x}}{\sqrt{x}+3}\)

c) \(P=B\div A=\frac{\sqrt{x}}{\sqrt{x}+3}\div\frac{\sqrt{x}\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}=\frac{\sqrt{x}}{\sqrt{x}+3}.\frac{\sqrt{x}-5}{\sqrt{x}}=\frac{\sqrt{x}-5}{\sqrt{x}+3}=1-\frac{8}{\sqrt{x}+3}< 1\)

game dễ nhỉ 

câu c 1 là xét P-1 

2 là xét 1-P thôi câu a vs b bình thường rồi chịu khó lấy nháp ra mà tính tay đi bro

Trả lời:

a, \(A=\frac{\sqrt{x}}{\sqrt{x}-5}-\frac{10\sqrt{x}}{x-25}-\frac{5}{\sqrt{x}+5}\left(ĐK:x\ge0;x\ne25\right)\)

\(=\frac{\sqrt{x}}{\sqrt{x}-5}-\frac{10\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-\frac{5}{\sqrt{x}+5}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-\frac{10\sqrt{x}}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-\frac{5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+5\right)-10\sqrt{x}-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

\(=\frac{x+5\sqrt{x}-10\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

\(=\frac{x-10\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

\(=\frac{\left(\sqrt{x}-5\right)^2}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{\sqrt{x}-5}{\sqrt{x}+5}\)

b, Thay x = 9 vào A, ta được:

\(A=\frac{\sqrt{9}-5}{\sqrt{9}+5}=\frac{3-5}{3+5}=\frac{-2}{8}=-\frac{1}{4}\)

c, \(A< \frac{1}{3}\Leftrightarrow\frac{\sqrt{x}-5}{\sqrt{x}+5}< \frac{1}{3}\Leftrightarrow\frac{\sqrt{x}-5}{\sqrt{x}+5}-\frac{1}{3}< 0\)

\(\Leftrightarrow\frac{3\left(\sqrt{x}-5\right)}{3\left(\sqrt{x}+5\right)}-\frac{\sqrt{x}+5}{3\left(\sqrt{x}+5\right)}< 0\)

\(\Leftrightarrow\frac{3\sqrt{x}-15-\sqrt{x}-5}{3\left(\sqrt{x}+5\right)}< 0\)

\(\Leftrightarrow\frac{2\sqrt{x}-20}{3\left(\sqrt{x}+5\right)}< 0\) 

\(\Rightarrow2\sqrt{x}-20< 0\) (vì \(3\left(\sqrt{x}+5\right)>0\) )

\(\Leftrightarrow2\sqrt{x}< 20\)

\(\Leftrightarrow\sqrt{x}< 10\)

\(\Leftrightarrow x< 100\)

Vậy \(0\le x< 100\)và \(x\ne25\) là giá trị cần tìm.

a, Ta có : \(x=\sqrt{3+2\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)

\(=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}=4\)

Thay x = 4 => \(\sqrt{x}=2\) vào B ta được : 

\(B=\frac{2+5}{2-3}=-7\)

b, Ta có : Với \(x\ge0;x\ne9\)

\(A=\frac{4}{\sqrt{x}+3}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{\sqrt{x}}{\sqrt{x}-3}\)

\(=\frac{4\left(\sqrt{x}-3\right)+2x-\sqrt{x}-13-\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}\)

\(=\frac{4\sqrt{x}-12+2x-\sqrt{x}-13-x-3\sqrt{x}}{x-9}=\frac{x-25}{x-9}\)

Lại có \(P=\frac{A}{B}\Rightarrow P=\frac{\frac{x-25}{x-9}}{\frac{\sqrt{x}+5}{\sqrt{x}-3}}=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)