a, \(ĐKXĐ:x\ne\pm\frac{1}{5},x\ne\frac{3}{2}\)

\(\Rightarrow P=\frac{\left(5x+1\right)\left(x+2\right)}{\left(2x-3\right)\left(5x-1\right)\left(5x+1\right)}-\frac{\left(8-3x\right)\left(5x+1\right)}{\left(5x-1\right)\left(5x+1\right)\left(2x-3\right)}\)

\(=\frac{x+2}{\left(2x-3\right)\left(5x-1\right)}-\frac{8-3x}{\left(5x-1\right)\left(2x-3\right)}\)

\(=\frac{2\left(2x-3\right)}{\left(2x-3\right)\left(5x-1\right)}=\frac{2}{5x-1}\)

b, Để P có giá trị nguyên thì  \(2⋮5x-1\)

\(\Rightarrow5x-1\in\left\{1,2,-1,-2\right\}\)

=> x=..............

ĐKXĐ : x \(\ne\frac{3}{2}\) ; \(x\ne\frac{1}{5};x\ne-\frac{1}{5}\) 

P= \(\frac{5x+1}{2x-3}.\left(\frac{x+2}{25x^2-1}-\frac{8-3x}{25x^2-1}\right)\) 

P= \(\frac{5x-1}{2x-3}.\left(\frac{4x-6}{\left(5x+1\right).\left(5x-1\right)}\right)\)

P= \(\frac{5x-1}{2x-3}.\frac{2\left(2x-3\right)}{\left(5x-1\right)\left(5x+1\right)}\) 

P= \(\frac{2}{5x-1}\) 

KL

b,

Ta có \(x\inℤ\Rightarrow\frac{2}{5x-1}\inℤ\) \(\Leftrightarrow5x-1\inƯ(5)\) mà Ư(5)={+-1;+-5}

a) x=1/-3/3/6

b)x=?

có ai giúp em bài này với khó quá

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

mà \(x\inℤ\Rightarrow x+4\inƯ\left(5\right)=\left\{-5,-1,1,5\right\}\)

\(\Leftrightarrow x\in\left\{-9,-5,-3,1\right\}\)

b) \(\frac{11-2x}{x-5}=\frac{1+10-2x}{x-5}=\frac{1}{x-5}-2\inℤ\Leftrightarrow\frac{1}{x-5}\inℤ\)

mà \(x\inℤ\Rightarrow x-5\inƯ\left(1\right)=\left\{-1,1\right\}\Leftrightarrow x\in\left\{4,6\right\}\)

c) \(\frac{x+1}{2x+1}\inℤ\Rightarrow\frac{2\left(x+1\right)}{2x+1}=\frac{2x+1+1}{2x+1}=1+\frac{1}{2x+1}\inℤ\Leftrightarrow\frac{1}{2x+1}\inℤ\)

mà \(x\inℤ\Rightarrow2x+1\inƯ\left(1\right)=\left\{-1,1\right\}\Leftrightarrow x\in\left\{-1,0\right\}\).

Thử lại đều thỏa mãn.