a: Thay x=36 vào A, ta được:

\(A=\frac{\sqrt{36}+4}{\sqrt{36}+2}=\frac{6+4}{6+2}=\frac{10}{8}=\frac54\)

b: \(B=\left(\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right):\frac{x+16}{\sqrt{x}+2}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}-4\right)+4\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}\cdot\frac{\sqrt{x}+2}{x+16}\)

\(=\frac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}\cdot\frac{\sqrt{x}+2}{x+16}=\frac{\sqrt{x}+2}{x-16}\)

c: Đặt P=B(A-1)

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

\(=\frac{\sqrt{x}+2}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}\cdot\frac{2}{\sqrt{x}+2}=\frac{2}{x-16}\)

Để P là số nguyên thì 2⋮x-16

=>x-16∈{1;-1;2;-2}

=>x∈{17;15;18;14}

tìm cả đk giúp mik vs

ĐKXĐ: \(x>0;x\ne1\)

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

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

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

b.

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

\(\Rightarrow A=\dfrac{4+2\sqrt{3}}{\sqrt{3}+1-1}=\dfrac{4+2\sqrt{3}}{\sqrt{3}}=\dfrac{6+4\sqrt{3}}{3}\)

c.

Để \(\sqrt{A}\) xác định \(\Rightarrow\sqrt{x}-1>0\Rightarrow x>1\)

Ta có:

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

Dấu "=" xảy ra khi \(\sqrt{x}-2=0\Rightarrow x=4\)

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

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

\(Q=x+1\)

Không thể tìm được GTLN hay GTNN của Q.

b)

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

Để \(\frac{3Q}{\sqrt{x}}\) nguyên thì \(\frac{3}{\sqrt{x}}\)nguyên hay \(\sqrt{x}\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

Vì \(\sqrt{x}\)dương nên \(\sqrt{x}\in\left\{1;3\right\}\)

Vậy x=1, x=9 là các giá trị cần tìm

a:

ĐKXĐ: x>=0; x<>1

Ta có: \(\frac{2}{\sqrt{x}-1}-\frac{5}{x+\sqrt{x}-2}\)

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

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

Ta có: \(1+\frac{3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)+3-x}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+2\right)}\)

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

Ta có: \(P=\left(\frac{2}{\sqrt{x}-1}-\frac{5}{x+\sqrt{x}-2}\right):\left(1+\frac{3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)\)

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

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

b: Thay \(x=6-2\sqrt5=\left(\sqrt5-1\right)^2\) vào P, ta được:

\(P=\frac{2\cdot\sqrt{\left(\sqrt5-1\right)^2}-1}{\sqrt{\left(\sqrt5-1\right)^2}+1}\)

\(=\frac{2\left(\sqrt5-1\right)-1}{\sqrt5-1+1}=\frac{2\sqrt5-3}{\sqrt5}=2-\frac{3}{\sqrt5}=2-\frac{3\sqrt5}{5}=\frac{10-3\sqrt5}{5}\)

c: \(P=\frac{1}{\sqrt{x}}\)

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

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

=>\(2x-2\sqrt{x}-1=0\)

=>\(x-\sqrt{x}-\frac12=0\)

=>\(x-\sqrt{x}+\frac14-\frac34=0\)

=>\(\left(\sqrt{x}-\frac12\right)^2=\frac34\)

=>\(\left[\begin{array}{l}\sqrt{x}-\frac12=\frac{\sqrt3}{2}\\ \sqrt{x}-\frac12=-\frac{\sqrt3}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}\sqrt{x}=\frac{\sqrt3+1}{2}\\ \sqrt{x}=\frac{-\sqrt3+1}{2}\left(loại\right)\end{array}\right.\)

=>\(\sqrt{x}=\frac{\sqrt3+1}{2}\)

=>\(x=\left(\frac{\sqrt3+1}{2}\right)^2=\frac{4+2\sqrt3}{4}=\frac{2+\sqrt3}{2}\)

d: Để P là số nguyên thì \(2\sqrt{x}-1\) ⋮\(\sqrt{x}+1\)

=>\(2\sqrt{x}+2-3\) ⋮\(\sqrt{x}+1\)

=>-3⋮\(\sqrt{x}+1\)

=>\(\sqrt{x}+1\in\left\lbrace1;3\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace0;2\right\rbrace\)

=>x∈{0;4}

e: \(P<1-\sqrt{x}\)

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

=>\(2\sqrt{x}-1<\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)=-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=-\left(x-1\right)=-x+1\)

=>\(2\sqrt{x}-1+x-1<0\)

=>\(x+2\sqrt{x}+1-3<0\)

=>\(\left(\sqrt{x}+1\right)^2<3\)

=>\(\sqrt{x}+1<\sqrt3\)

=>\(\sqrt{x}<\sqrt3-1\)

=>\(x<4-2\sqrt3\)

Kết hợp ĐKXĐ, ta được: 0<=x<\(4-2\sqrt3\)

a:

ĐKXĐ: x>=0; x<>1

Ta có: \(\frac{2}{\sqrt{x}-1}-\frac{5}{x+\sqrt{x}-2}\)

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

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

Ta có: \(1+\frac{3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

\(=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)+3-x}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+2\right)}\)

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

Ta có: \(P=\left(\frac{2}{\sqrt{x}-1}-\frac{5}{x+\sqrt{x}-2}\right):\left(1+\frac{3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)\)

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

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

b: Thay \(x=6-2\sqrt5=\left(\sqrt5-1\right)^2\) vào P, ta được:

\(P=\frac{2\cdot\sqrt{\left(\sqrt5-1\right)^2}-1}{\sqrt{\left(\sqrt5-1\right)^2}+1}\)

\(=\frac{2\left(\sqrt5-1\right)-1}{\sqrt5-1+1}=\frac{2\sqrt5-3}{\sqrt5}=2-\frac{3}{\sqrt5}=2-\frac{3\sqrt5}{5}=\frac{10-3\sqrt5}{5}\)

c: \(P=\frac{1}{\sqrt{x}}\)

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

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

=>\(2x-2\sqrt{x}-1=0\)

=>\(x-\sqrt{x}-\frac12=0\)

=>\(x-\sqrt{x}+\frac14-\frac34=0\)

=>\(\left(\sqrt{x}-\frac12\right)^2=\frac34\)

=>\(\left[\begin{array}{l}\sqrt{x}-\frac12=\frac{\sqrt3}{2}\\ \sqrt{x}-\frac12=-\frac{\sqrt3}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}\sqrt{x}=\frac{\sqrt3+1}{2}\\ \sqrt{x}=\frac{-\sqrt3+1}{2}\left(loại\right)\end{array}\right.\)

=>\(\sqrt{x}=\frac{\sqrt3+1}{2}\)

=>\(x=\left(\frac{\sqrt3+1}{2}\right)^2=\frac{4+2\sqrt3}{4}=\frac{2+\sqrt3}{2}\)

d: Để P là số nguyên thì \(2\sqrt{x}-1\) ⋮\(\sqrt{x}+1\)

=>\(2\sqrt{x}+2-3\) ⋮\(\sqrt{x}+1\)

=>-3⋮\(\sqrt{x}+1\)

=>\(\sqrt{x}+1\in\left\lbrace1;3\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace0;2\right\rbrace\)

=>x∈{0;4}

e: \(P<1-\sqrt{x}\)

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

=>\(2\sqrt{x}-1<\left(1-\sqrt{x}\right)\left(\sqrt{x}+1\right)=-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=-\left(x-1\right)=-x+1\)

=>\(2\sqrt{x}-1+x-1<0\)

=>\(x+2\sqrt{x}+1-3<0\)

=>\(\left(\sqrt{x}+1\right)^2<3\)

=>\(\sqrt{x}+1<\sqrt3\)

=>\(\sqrt{x}<\sqrt3-1\)

=>\(x<4-2\sqrt3\)

Kết hợp ĐKXĐ, ta được: 0<=x<\(4-2\sqrt3\)

a) \(B=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+2}{x-4}\left(đk:x\ge0,x\ne4\right)\)

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

c) \(C=A\left(B-2\right)=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\left(\dfrac{2\sqrt{x}+2}{\sqrt{x}+2}-2\right)\)

\(=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}.\dfrac{-2}{\sqrt{x}+2}=\dfrac{-2}{\sqrt{x}-2}\in Z\)

\(\Rightarrow\sqrt{x}-2\inƯ\left(2\right)=\left\{1;-1;2-2\right\}\)

\(\Rightarrow\sqrt{x}\in\left\{3;1;4;0\right\}\)

\(\Rightarrow x\in\left\{0;1;9;16\right\}\)

a) Tại x=16 thì A = \(\dfrac{\sqrt{16}-1}{\sqrt{16}+2}=\dfrac{4-1}{4+2}=\dfrac{1}{2}\)

b) B = \(\dfrac{\sqrt{x}+1+\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\div\dfrac{\sqrt{x}}{x+\sqrt{x}}\)

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

        = \(\dfrac{x+1}{\sqrt{x}}\)

B = \(\dfrac{x+1}{\sqrt{x}}\)= 2

   ⇒ x + 1 = 2\(\sqrt{x}\) 

   ⇒ x - \(2\sqrt{x}\) +1 = 0

   ⇒ \(\left(\sqrt{x}-1\right)^2\) = 0

   ⇒ \(\sqrt{x}-1=0\)

⇒  x = 1 

a, ĐK: \(x\ge0;x\ne9\)

\(P=\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{3x+9}{9-x}\)

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

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

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

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

b, \(P>0\Leftrightarrow-\dfrac{3}{\sqrt{x}-3}>0\)

\(\Leftrightarrow\sqrt{x}-3>0\)

\(\Leftrightarrow x>9\)

c, \(P=-\dfrac{3}{\sqrt{x}-3}\in Z\)

\(\Leftrightarrow\sqrt{x}-3\inƯ_3=\left\{\pm1;\pm3\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{0;2;4;6\right\}\)

\(\Leftrightarrow x\in\left\{0;4;16;36\right\}\)