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

a: Thay x=9 vào A, ta được:

\(A=\dfrac{3+2}{3-5}=\dfrac{5}{-2}=\dfrac{-5}{2}\)

\(B=\dfrac{3\sqrt{x}-15+20-2\sqrt{x}}{x-25}=\dfrac{\sqrt{x}+5}{x-25}=\dfrac{1}{\sqrt{x}-5}\)

b: Để \(A=B\cdot\left|x-4\right|\) thì \(\left|x-4\right|=\dfrac{A}{B}=\dfrac{\sqrt{x}+2}{\sqrt{x}-5}:\dfrac{1}{\sqrt{x}-5}=\sqrt{x}+2\)

\(\Leftrightarrow x-4=\sqrt{x}+2\)

\(\Leftrightarrow x-\sqrt{x}-6=0\)

=>x=9

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

Bài 1 : Với : \(x>0;x\ne1\)

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

Thay vào ta được : \(P=x=25\)

Bài 2 : 

a, Với \(x\ge0;x\ne1\)

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

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

Thay x = 9 vào A ta được : \(\frac{3}{3+1}=\frac{3}{4}\)

Bài 3 : \(x\ge0;x\ne1\)

\(P=\left(\frac{3}{x-1}+\frac{1}{\sqrt{x}+1}\right):\frac{1}{\sqrt{x}+1}\)

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

b, Ta có : \(P=\frac{\sqrt{x}+2}{\sqrt{x}-1}=\frac{5}{4}\Rightarrow4\sqrt{x}+8=5\sqrt{x}-5\)

\(\Leftrightarrow\sqrt{x}=13\Leftrightarrow x=169\)(tmđk )

Bài 1:

Thay x=9 vào biểu thức \(A=\frac{2\sqrt{x}+1}{\sqrt{x}+2}\), ta được:

\(\frac{2\cdot\sqrt{9}+1}{\sqrt{9}+2}=\frac{2\cdot3+1}{3+2}=\frac{7}{5}\)

Vậy: \(\frac{7}{5}\) là giá trị của biểu thức \(A=\frac{2\sqrt{x}+1}{\sqrt{x}+2}\) tại x=9

Bài 2:

a) Ta có: \(B=\left(\frac{x+14\sqrt{x}-5}{x-25}+\frac{\sqrt{x}}{\sqrt{x}+5}\right):\frac{\sqrt{x}+2}{\sqrt{x}-5}\)

\(=\left(\frac{x+14\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}+\frac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\right)\cdot\frac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(=\frac{2x+9\sqrt{x}-5}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-5\right)}\cdot\frac{\sqrt{x}-5}{\sqrt{x}+2}\)

\(=\frac{2x+10\sqrt{x}-\sqrt{x}-5}{\sqrt{x}+5}\cdot\frac{1}{\sqrt{x}+2}\)

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