ĐKXĐ: \(x\notin\left\{-1;-\dfrac{1}{2}\right\}\)

a) Ta có: \(P=\left(\dfrac{2x}{x^3+x^2+x+1}+\dfrac{1}{x+1}\right):\left(1+\dfrac{x}{x+1}\right)\)

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

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

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

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

b) Vì \(x=\dfrac{1}{4}\) thỏa mãn ĐKXĐ

nên Thay \(x=\dfrac{1}{4}\) vào biểu thức \(P=\dfrac{x^2+2x+1}{\left(2x+1\right)\left(x^2+1\right)}\), ta được:

\(P=\left[\left(\dfrac{1}{4}\right)^2+2\cdot\dfrac{1}{4}+1\right]:\left[\left(2\cdot\dfrac{1}{4}+1\right)\left(\dfrac{1}{16}+1\right)\right]\)

\(=\left(\dfrac{1}{16}+\dfrac{1}{2}+1\right):\left[\left(\dfrac{1}{2}+1\right)\left(\dfrac{1}{16}+1\right)\right]\)

\(=\dfrac{25}{16}:\dfrac{51}{32}=\dfrac{25}{16}\cdot\dfrac{32}{51}=\dfrac{50}{51}\)

Vậy: Khi \(x=\dfrac{1}{4}\) thì \(P=\dfrac{50}{51}\)

a) Ta có: \(A=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

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

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

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

b) Để \(A< -\dfrac{1}{3}\) thì \(A+\dfrac{1}{3}< 0\)

\(\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}+\dfrac{1}{3}< 0\)

\(\Leftrightarrow\dfrac{-9+\sqrt{x}+3}{3\left(\sqrt{x}+3\right)}< 0\)

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

\(\Leftrightarrow x< 36\)

Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 36\\x\ne9\end{matrix}\right.\)

a: ĐKXĐ: x>0; x<>1

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

c: A nguyên

=>x-1 thuộc {1;-1;2;-2}

=>x thuộc {2;3}

a) ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne4\end{matrix}\right.\)

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

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

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

d) Để A>0 thì \(\sqrt{x}-2>0\)

hay x>4

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 \(\sqrt{x}-2=0\Rightarrow x=4\)

Bài 1:

\(a,ĐK:x\ne\pm5\\ b,P=\dfrac{x-5+2x+10-2x-10}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}=\dfrac{1}{x+5}\\ c,P=-3\Leftrightarrow x+5=-\dfrac{1}{3}\Leftrightarrow x=-\dfrac{16}{3}\\ d,P\in Z\Leftrightarrow x+5\inƯ\left(1\right)=\left\{-1;1\right\}\\ \Leftrightarrow x\in\left\{-6;-4\right\}\)

Bài 2:

\(a,\Leftrightarrow\dfrac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3}{x-2}=0\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow\dfrac{x\left(2-x\right)}{\left(x-2\right)\left(x+2\right)}=0\Leftrightarrow\dfrac{-x}{x+2}=0\Leftrightarrow x=0\)