\(a,A=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\\ b,x=36\Leftrightarrow A=\dfrac{6}{6-2}=\dfrac{6}{4}=\dfrac{3}{2}\\ c,A=-\dfrac{1}{3}\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-2}=-\dfrac{1}{3}\Leftrightarrow3\sqrt{x}=2-\sqrt{x}\\ \Leftrightarrow\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\\ d,A\in Z\Leftrightarrow1+\dfrac{2}{\sqrt{x}-2}\in Z\\ \Leftrightarrow\sqrt{x}-2\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\\ \Leftrightarrow\sqrt{x}\in\left\{0;1;3;4\right\}\\ \Leftrightarrow x\in\left\{0;1;9;16\right\}\)

\(e,A:B=\dfrac{\sqrt{x}}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{\sqrt{x}}{\sqrt{x}+1}=-2\\ \Leftrightarrow\sqrt{x}=-2\sqrt{x}-2\\ \Leftrightarrow\sqrt{x}=-\dfrac{2}{3}\left(ktm\right)\\ \Leftrightarrow x\in\varnothing\)

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

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

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

\(a,A=4\sqrt{3}-5\sqrt{3}+2-\sqrt{3}=2-2\sqrt{3}\\ B=\dfrac{x+2\sqrt{x}+8+2\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\dfrac{\sqrt{x}}{\sqrt{x}-4}\\ b,B-\dfrac{1}{2}A=\dfrac{\sqrt{x}}{\sqrt{x}-4}-\dfrac{1}{2}\left(2-2\sqrt{3}\right)=0\\ \Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-4}=1+\sqrt{3}\\ \Leftrightarrow\sqrt{x}=\left(1+\sqrt{3}\right)\left(\sqrt{x}-4\right)\Leftrightarrow\sqrt{x}=\sqrt{x}-4\sqrt{3}+\sqrt{3x}-4\\ \Leftrightarrow\sqrt{3x}=4\sqrt{3}+4\\ \Leftrightarrow\sqrt{x}=\dfrac{4\sqrt{3}+4}{\sqrt{3}}\\ \Leftrightarrow\sqrt{x}=\dfrac{12+4\sqrt{3}}{3}\\ \Leftrightarrow x=\dfrac{192+96\sqrt{3}}{9}=\dfrac{64+32\sqrt{3}}{3}\)

Để \(A=\frac{5}{\sqrt{x}-1}\)nguyên thì \(5⋮\sqrt{x}-1\)

\(\Rightarrow\sqrt{x}-1\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

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

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

Để \(B=\frac{\sqrt{x}+1}{\sqrt{x}-3}\)nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)

\(\Leftrightarrow\sqrt{x}-3+4⋮\sqrt{x}-3\)

Vì \(\sqrt{x}-3⋮\sqrt{x}-3\)

\(\Rightarrow4⋮\sqrt{x}-3\)

\(\Rightarrow\sqrt{x}-3\inƯ\left(4\right)=\left\{\pm1;\pm2;\pm4\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{4;2;5;1;7;-1\right\}\)

\(\Leftrightarrow x\in\left\{16;4;25;1;49\right\}\)

Máy kia tương tự đi ăn cơm đây :>

Mk lm nốt câu C

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

Để C nguyên\(\Leftrightarrow\sqrt{x}-1\inƯ_{\left(3\right)}=\left\{\pm3;\pm1\right\}\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1=3\left(tm\right)\\\sqrt{x}-1=-3\left(l\right)\\\sqrt{x}-1=-1\left(tm\right)\\\sqrt{x}-1=1\left(tm\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=16\\x=0\\x=4\end{matrix}\right.\)

câu này hay thế!

a: 

Sửa đề: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{3x+3}{9-x}\right)\cdot\left(\dfrac{\sqrt{x}-7}{\sqrt{x}+1}+1\right)\)

\(P=\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\right)\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)

\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)

\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{2}{\sqrt{x}+1}=\dfrac{-6}{\sqrt{x}+3}\)

b: P>=1/2

=>P-1/2>=0

=>\(\dfrac{-6}{\sqrt{x}+3}-\dfrac{1}{2}>=0\)

=>\(\dfrac{-12-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>=0\)

=>\(-\sqrt{x}-15>=0\)

=>\(-\sqrt{x}>=15\)

=>căn x<=-15

=>\(x\in\varnothing\)

c: căn x+3>=3

=>6/căn x+3<=6/3=2

=>P>=-2

Dấu = xảy ra khi x=0

Bài 1: 

a: \(A=\left(\sqrt{x}+\sqrt{y}-\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)

\(=\dfrac{x+2\sqrt{xy}+y-x-\sqrt{xy}-y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{x}+\sqrt{y}}{x-\sqrt{xy}+y}\)

\(=\dfrac{\sqrt{xy}}{x-\sqrt{xy}+y}\)

b: \(\sqrt{xy}>=0;x-\sqrt{xy}+y>0\)

Do đó: A>=0

a/ \(A=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\)

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

\(=\dfrac{-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+2}{6}=\dfrac{-1}{\sqrt{x}-2}\)

b/ \(A>0\Leftrightarrow\dfrac{-1}{\sqrt{x}-2}>0\)

Ta thấy: - 1 < 0 nên để A > 0 thì:

\(\sqrt{x}-2< 0\)\(\Leftrightarrow\sqrt{x}< 2\Leftrightarrow x< 4\)

kết hợp với đkxđ: => \(0\le x< 4\)

a) Ta có: \(A=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\cdot\dfrac{\left(1-x\right)^2}{2}\)

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

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

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

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

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

\(=-x+1\)

b) Để A=-6 thì -x+1=-6

\(\Leftrightarrow-x=-6-1=-7\)

hay x=7(thỏa ĐK)

Vậy: Để A=-6 thì x=7

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

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

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

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

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

\(A=\dfrac{2}{\sqrt{x}+2}\)

b, \(A=\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{1}{4}\)

\(\Rightarrow\sqrt{x}+2=8\)

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

\(\Leftrightarrow x=36\left(tm\right)\)

Vậy x = 36 thì \(A=\dfrac{1}{4}\)

Bạn hình như ghi sai đề rồi: \(4-x\) chứ hình như không phải 4-\(\sqrt{x}\)