a: Ta có: \(A=\frac{-x}{4-x}+\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\)

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

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

b: Thay x=36 vào A, ta được:

\(A=\frac{6}{6-2}=\frac64=\frac32\)

c: \(A=-\frac13\)

=>\(\frac{\sqrt{x}}{\sqrt{x}-2}=-\frac13\)

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

=>\(4\sqrt{x}=2\)

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

=>\(x=\frac14\left(nhận\right)\)

e: A:B=-2

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

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

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

=>\(3\sqrt{x}=-2\) (vô lý)

=>x∈∅

a:

ĐKXĐ của B là x>=0; x<>9

Thay x=36 vào B, ta được:

\(B=\frac{\sqrt{36}}{\sqrt{36}-3}=\frac{6}{6-3}=\frac63=2\)

b: \(B<\frac12\)

=>\(\frac{\sqrt{x}}{\sqrt{x}-3}<\frac12\)

=>\(\frac{\sqrt{x}}{\sqrt{x}-3}-\frac12<0\)

=>\(\frac{2\sqrt{x}-\sqrt{x}+3}{2\cdot\left(\sqrt{x}-3\right)}<0\)

=>\(\frac{\sqrt{x}+3}{2\left(\sqrt{x}-3\right)}<0\)

=>\(2\left(\sqrt{x}-3\right)<0\)

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

=>\(\sqrt{x}<3\)

=>0<=x<9

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

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

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

d: \(P=A\cdot B=\frac{\sqrt{x}+2}{\sqrt{x}}\cdot\frac{\sqrt{x}}{\sqrt{x}-3}=\frac{\sqrt{x}+2}{\sqrt{x}-3}\)

Để P là số nguyên thì \(\sqrt{x}+2\) ⋮\(\sqrt{x}-3\)

=>\(\sqrt{x}-3+5\) ⋮\(\sqrt{x}-3\)

=>5⋮\(\sqrt{x}-3\)

=>\(\sqrt{x}-3\in\left\lbrace1;-1;5;-5\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace4;2;8;-2\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace2;4;8\right\rbrace\)

=>x∈{4;16;64}

\(a)A=(\frac{x}{(x+6)(x+6)}-\frac{x-6}{x(x+6)})\cdot\frac{x(x+6)}{2x-6}+\frac{x}{x-6}\)

\(A=\frac{x^2-(x-6)^2}{x(x+6)(x-6)}\cdot\frac{x(x+6)}{2x-6}-\frac{x}{x-6}=\frac{(x-x+6)(x+x-6)}{(x-6)(2x-6)}-\frac{x}{x-6}\)

\(=\frac{6(2x-6)}{(x-6)(2x-6)}-\frac{x}{x-6}=\frac{6}{(x-6)}-\frac{x}{x-6}\cdot\frac{6-x}{x-6}=-1\)

\(b)\text{A luôn = -1 với mọi x}\)

bạn viết thế này khó nhìn quá

nhìn hơi đau mắt nhá bạn hoa mắt quá

\(a,\) ta có : 

\(\Leftrightarrow\left\{{}\begin{matrix}A=\sqrt{3}+\sqrt{2^2.3}-\sqrt{3^2.3}-\sqrt{6^2}\\A=\sqrt{3}+2\sqrt{3}-3\sqrt{3}-6\\A=\sqrt{3}.\left(1+2-3\right)-6\\A=-6\end{matrix}\right.\)

\(\Rightarrow A=-6\) . vậy \(A=9\sqrt{5}\)

__________________________________________________________

\(b,\) với \(x>0\) và \(x\ne1\) . ta có :

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

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

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

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

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

\(\Leftrightarrow B=\dfrac{4}{\sqrt{x}}\)

vậy với \(x>0\) \(;\) \(x\ne1\) thì \(B=\dfrac{4}{\sqrt{x}}\)

để \(B=2\) thì \(\dfrac{4}{\sqrt{x}}=2\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)

vậy để \(B=2\) thì \(x=4\)

a: Thay x=49 vào A, ta được:

\(A=\dfrac{2\cdot7+1}{7-3}=\dfrac{14+1}{4}=\dfrac{15}{4}\)

b: \(B=\dfrac{2x+36}{x-9}-\dfrac{9}{\sqrt{x}-3}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

\(=\dfrac{2x+36}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{9}{\sqrt{x}-3}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)

\(=\dfrac{2x+36-9\left(\sqrt{x}+3\right)-\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{2x+36-9\sqrt{x}-27-x+3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

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

c: \(P=A\cdot B=\dfrac{\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{2\sqrt{x}+1}{\sqrt{x}+3}\)

P>1 khi P-1>0

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

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

=>\(\sqrt{x}>2\)

=>x>4

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x>4\\x\ne9\end{matrix}\right.\)

\(=2\sqrt{3}-4\sqrt{3}+5\sqrt{3}=3\sqrt{3}\)

Bài 5: 

\(\widehat{B}=60^0\)

\(AB=8\sqrt{3}\left(cm\right)\)

\(BC=16\sqrt{3}\left(cm\right)\)

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}

Bài 1 :

a) \(ĐKXĐ:x\ne1\)

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

\(\Leftrightarrow A=\frac{3+x-1}{\left(x-1\right)\left(x+1\right)}\cdot\left(x+1\right)\)

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

b) Thay x = \(\frac{2}{5}\)vào A ta được :

\(A=\frac{\frac{2}{5}+2}{\frac{2}{5}-1}=\frac{\frac{12}{5}}{-\frac{3}{5}}=-4\)

c) Để \(A=\frac{5}{4}\)

\(\Leftrightarrow\frac{x+2}{x-1}=\frac{5}{4}\)

\(\Leftrightarrow4x+8=5x-5\)

\(\Leftrightarrow x=13\)

d) Để \(A>\frac{1}{2}\)

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

\(\Leftrightarrow\frac{x+2}{x-1}-\frac{1}{2}>0\)

\(\Leftrightarrow2x+4-x+1>0\)

\(\Leftrightarrow x+5>0\)

\(\Leftrightarrow x>-5\)

Bài 2 :

a) \(ĐKXĐ:\hept{\begin{cases}x\ne-1\\x\ne0\end{cases}}\)

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

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

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

b) Để \(A=1\)

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

\(\Leftrightarrow2x-1=x+1\)

\(\Leftrightarrow x=2\)

b) Để \(A< 2\)

\(\Leftrightarrow\frac{2x-1}{x+1}< 2\)

\(\Leftrightarrow\frac{2x-1}{x+1}-2< 0\)

\(\Leftrightarrow2x-1-2x-1< 0\)

\(\Leftrightarrow-2< 0\)(luôn đúng)

Vậy A < 2 <=> mọi x

Bài 1 : 

a, \(A=\frac{2x^2-4x+8}{x^3+8}=\frac{2\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\frac{2}{x+2}\)

b, Ta có : \(\left|x\right|=2\Rightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

TH1 : Thay x = 2 vào biểu thức trên ta được : 

\(\frac{2}{2+2}=\frac{2}{4}=\frac{1}{2}\)

TH2 : Thay x = -2 vào biểu thức trên ta được : 

\(\frac{2}{-2+2}=\frac{2}{0}\)vô lí 

c, ta có A = 2 hay \(\frac{2}{x+2}=2\)ĐK : \(x\ne-2\)

\(\Rightarrow2x+4=2\Leftrightarrow2x=-2\Leftrightarrow x=-1\)

Vậy với x = -1 thì A = 2 

d, Ta có A < 0 hay \(\frac{2}{x+2}< 0\)

\(\Rightarrow x+2< 0\)do 2 > 0 

\(\Leftrightarrow x< -2\)

Vậy với A < 0 thì x < -2 

e, Để A nhận giá trị nguyên khi \(x+2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

2.

ĐKXĐ : \(x\ne\pm2\)

a. \(B=\frac{x^2-4x+4}{x^2-4}=\frac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}=\frac{x-2}{x+2}\)

b. | x - 1 | = 2 <=>\(\hept{\begin{cases}x-1=2\\x-1=-2\end{cases}}\)<=>\(\hept{\begin{cases}x=3\\x=-1\end{cases}}\)

Với x = 3 thì \(B=\frac{3-2}{3+2}=\frac{1}{5}\)

Với x = - 1 thì \(B=\frac{-1-2}{-1+2}=-3\)

Vậy với | x - 1 | = 2 thì B đạt được 2 giá trị là B = 1/5 hoặc B = - 3

c. \(B=\frac{x-2}{x+2}=-1\)<=>\(-\left(x-2\right)=x+2\)

<=> \(-x+2=x+2\)<=>\(-x=x\)<=>\(x=0\)

d. \(B=\frac{x-2}{x+2}< 1\)<=>\(x-2< x+2\)luôn đúng \(\forall\)x\(\ne\pm2\)

e. \(B=\frac{x-2}{x+2}=\frac{x+2-4}{x+2}=1-\frac{4}{x+2}\)

Để B nguyên thì 4/x+2 nguyên => x + 2\(\in\){ - 4 ; - 2 ; - 1 ; 1 ; 2 ; 4 }

=> x \(\in\){ - 6 ; - 4 ; - 3 ; - 1 ; 0 ; 2 }