a)\(A=\frac{x^2}{5x+25}+\frac{2x-10}{x}+\frac{50+5x}{x^2+5x}\left(ĐK:x\ne0;-5\right)\)

\(\Leftrightarrow A=\frac{x^2}{5\left(x+5\right)}+\frac{2\left(x-5\right)}{x}+\frac{5\left(x+10\right)}{x\left(x+5\right)}\)

\(\Leftrightarrow A=\frac{x^3+10\left(x^2-25\right)+25x+250}{5x\left(x+5\right)}\)

\(\Leftrightarrow A=\frac{x^3+10x^2+25x}{5x\left(x+5\right)}\)

\(\Leftrightarrow A=\frac{x\left(x+5\right)^2}{5x\left(x+5\right)}\)

\(\Leftrightarrow A=\frac{x+5}{5}\)

b)Để A=-4 \(\Leftrightarrow\frac{x+5}{5}=-4\)

                  \(\Leftrightarrow x+5=-20\)

                   \(\Leftrightarrow x=-25\)

a).....

\(=\frac{x^2}{5\left(x+5\right)}+\frac{2x-10}{x}+\frac{50+5x}{x\left(x+5\right)}\)                                MTC= 5x (x+5)                 ĐK\(\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)

\(=\frac{x^2.x}{5x\left(x+5\right)}+\frac{5.\left(2x-10\right).\left(x+5\right)}{5x\left(x+5\right)}+\frac{5.\left(50+5x\right)}{5x\left(x+5\right)}\)

\(=\frac{x^3+\left(10x-50\right).\left(x+5\right)+250+25x}{5x\left(x+5\right)}\)

\(=\frac{x^3+10x^2+50x-50x-250+250+25x}{5x\left(x+5\right)}\)

\(=\frac{x^3+10x^2+25x}{5x\left(x+5\right)}\)

\(=\frac{x\left(x^2+10x+25\right)}{5x\left(x+5\right)}\)

\(=\frac{x\left(x+5\right)^2}{5x\left(x+5\right)}=\frac{x+5}{5}\)

b) A=-4

=>\(\frac{x+5}{5}=-4\)

=> x = -25

c)

d) Để A đạt gt nguyên thì 5\(⋮\)x+5

=> \(\left(x+5\right)\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)

*x+5=1 => x=-4 \(\in Z\)

*x+5=-1 => x=-6\(\in Z\)

*x+5=5  => x=0\(\in Z\)

*x+5=-5  => x=-10\(\in Z\)

Vậy...........

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

\(\Leftrightarrow x-2< 0\)  ( vì \(-1< 0\))

\(\Leftrightarrow x< 2\)

\(A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)

\(A=\)\(\left[\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)

  \(:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)

\(A=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}:\left[\frac{x^2-4+10-x^2}{x+2}\right]\)

\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)

\(A=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}\)

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

theo câu a) \(A=\frac{-1}{x-2}\)  với ĐKXĐ: \(x\ne\pm2\)

b) \(\left|2x-1\right|=3\)

\(\Rightarrow\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\Rightarrow\orbr{\begin{cases}2x=4\\2x=-2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)   \(\Rightarrow x=-1\)  ( vì \(x=2\)  ko TM ĐKXĐ )

+) khi \(x=-1\)thì \(A=\frac{-1}{-1-2}=\frac{-1}{-3}=\frac{1}{3}\)

vậy khi \(x=-1\)  thì \(A=\frac{1}{3}\)

\(ĐKXĐ:x\ne\pm1\)

a) \(A=\left(\frac{1}{1-x}+\frac{2}{1+x}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}\)

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

\(=\frac{1+x+2-2x-5+x}{1-x^2}:\frac{2x-1}{1-x^2}\)

\(=\frac{8}{1-x^2}.\frac{1-x^2}{2x-1}=\frac{8}{2x-1}\)

b) Để A nguyên thì \(\frac{8}{2x-1}\inℤ\)

\(\Leftrightarrow8⋮2x-1\Rightarrow2x-1\inƯ\left(8\right)=\left\{\pm1;\pm2;\pm4;\pm8\right\}\)

Mà dễ thấy 2x - 1 lẻ nên\(2x-1\in\left\{\pm1\right\}\)

+) \(2x-1=1\Rightarrow x=1\left(ktmđkxđ\right)\)

+) \(2x-1=-1\Rightarrow x=0\left(tmđkxđ\right)\)

Vậy x nguyên bằng 0 thì A nguyên

c) \(\left|A\right|=A\Leftrightarrow A\ge0\)

\(\Rightarrow\frac{8}{2x-1}\ge0\Rightarrow2x-1>0\Leftrightarrow x>\frac{1}{2}\)

Vậy \(x>\frac{1}{2}\)thì |A| = A

a, \(A=\left(\frac{1}{1-x}+\frac{2}{1+x}-\frac{5-x}{1-x^2}\right):\frac{1-2x}{x^2-1}\left(x\ne\frac{1}{2};x\ne\pm1\right)\)

\(\Leftrightarrow A=\left(\frac{1+x}{\left(1-x\right)\left(1+x\right)}+\frac{2-2x}{\left(1-x\right)\left(1+x\right)}-\frac{5-x}{\left(1-x\right)\left(1+x\right)}\right):\frac{\left(x+1\right)\left(x-1\right)}{2x-1}\)

\(\Leftrightarrow A=\frac{1+x+2-2x-5+x}{\left(1-x\right)\left(1+x\right)}\cdot\frac{\left(x-1\right)\left(x+1\right)}{2x-1}\)

\(\Leftrightarrow A=\frac{-2\left(1-x^2\right)}{\left(1-x^2\right)\left(2x-1\right)}=\frac{2}{2x-1}\)

Vậy \(A=\frac{2}{2x-1}\left(x\ne\frac{1}{2};x\ne\pm1\right)\)

b) \(A=\frac{2}{2x-1}\left(x\ne\frac{1}{2};x\ne\pm1\right)\)

Để A nhận giá trị nguyên thì 2 chia hết cho 2x-1

Mà x nguyên => 2x-1 nguyên

=> 2x-1 thuộc Ư (2)={-2;-1;1;2}
Ta có bảng

Đối chiếu điều kiện

=> x=0

a, ĐKXĐ: \(\hept{\begin{cases}5x+25\ne0\\x\ne0\\x^2+5x\ne0\end{cases}\Rightarrow\hept{\begin{cases}5\left(x+5\right)\ne0\\x\ne0\\x\left(x+5\right)\ne0\end{cases}\Rightarrow}}\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)

b, \(P=\frac{x^2}{5x+25}+\frac{2x-10}{x}+\frac{50+5x}{x^2+5x}\)

\(=\frac{x^3}{5x\left(x+5\right)}+\frac{5\left(2x-10\right)\left(x+5\right)}{5x\left(x+5\right)}+\frac{\left(50+5x\right).5}{5x\left(x+5\right)}\)

\(=\frac{x^3+10\left(x-5\right)\left(x+5\right)+250+25x}{5x\left(x+5\right)}\)

\(=\frac{x^3+10x^2+25x}{5x\left(x+5\right)}=\frac{x\left(x+5\right)^2}{5x\left(x+5\right)}=\frac{x+5}{5}\)

c, \(P=-4\Rightarrow\frac{x+5}{5}=-4\Rightarrow x+5=-20\Rightarrow x=-25\)

d, \(\frac{1}{P}\in Z\Rightarrow\frac{5}{x+5}\in Z\Rightarrow5⋮\left(x+5\right)\Rightarrow x+5\inƯ\left(5\right)=\left\{-5;-1;1;5\right\}\Rightarrow x\in\left\{-10;-6;-4;0\right\}\)

Mà x khác 0 (ĐKXĐ của P) nên \(x\in\left\{-10;-6;-4\right\}\)

a) \(ĐKXĐ:\hept{\begin{cases}5x+25\ne0\\x\ne0\\x^2+5x\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)

b) \(P=\frac{x^2}{5x+25}+\frac{2x-10}{x}+\frac{50+5x}{x^2+5x}\)

\(P=\frac{x^3}{5x\left(x+5\right)}+\frac{10x^2-250}{5x\left(x+5\right)}+\frac{250+25x}{5x\left(x+5\right)}\)

\(P=\frac{x^3+10x^2+25x}{5x\left(x+5\right)}=\frac{x\left(x+5\right)^2}{5x\left(x+5\right)}=\frac{x+5}{5}\)

c) \(P=4\Leftrightarrow\frac{x+5}{5}=4\Leftrightarrow x+5=20\Leftrightarrow x=15\)

d) \(\frac{1}{P}=\frac{5}{x+5}\in Z\Leftrightarrow5⋮x+5\)

\(\Leftrightarrow x+5\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

Lập bảng nhé

e) \(Q=P+\frac{x+25}{x+5}=\frac{x+30}{x+5}=1+\frac{25}{x+5}\)

\(Q_{min}\Leftrightarrow\frac{25}{x+5}_{min}\)