Ta có : \(\frac{2x-1}{2X+3}\)=\(\frac{2x+3-3-1}{2x+3}\)

=1-\(\frac{4}{2x+3}\)

Muốn \(\frac{2x-1}{2x+3}\)

là số nguyên thì suy ra \(\frac{4}{2x+3}\) nguyên 

suy ra 2 x+3 thuộc ư(4)= +-1 , +- 2 , +-4

Nên: 2 x+3=1 suy ra x=-1

        2 x+3 = - 1 suy ra x = -2

         2x+3= 2   suy ra x   =\(\frac{-1}{5}\)

       2x+3= -2    suy ra x =\(\frac{-5}{2}\)

      2x+3= -4   suy ra x= \(\frac{-4}{2}\)

  2x+3  = 4  suy ra     x= \(\frac{1}{2}\)

Vậy   : x = -1 và x = -2

de phan so nay la so nguyen thi 

            2x-1 chia het cho 2x+3

ta co 2x-1= 2x+3-4

dan den 4 chia het cho 2x+3

ma uoc cua 4 la 1;4;2;-1;-2;-4

khi 2x+3=1 thi x=-1              khi 2x+3 =-1 thi x=-2

khi 2x+3=4 thi x loai           khi 2x+3=-4 thi x loai 

khi 2x+3 =-2thi xloai              khi 2x+3 =2 thi xloai

vay x=-1 hoac =-2

a:

ĐKXĐ: x<>2

|2x-3|=1

=>\(\left[{}\begin{matrix}2x-3=1\\2x-3=-1\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=2\left(loại\right)\\x=1\left(nhận\right)\end{matrix}\right.\)

Thay x=1 vào A, ta được:

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

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

\(B=\dfrac{2x}{x+1}+\dfrac{3}{x-2}-\dfrac{2x^2+1}{x^2-x-2}\)

\(=\dfrac{2x}{x+1}+\dfrac{3}{x-2}-\dfrac{2x^2+1}{\left(x-2\right)\left(x+1\right)}\)

\(=\dfrac{2x\left(x-2\right)+3\left(x+1\right)-2x^2-1}{\left(x+1\right)\left(x-2\right)}\)

\(=\dfrac{2x^2-4x+3x+3-2x^2-1}{\left(x+1\right)\left(x-2\right)}\)

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

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

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

Để P lớn nhất thì \(\dfrac{2}{x-2}\) max

=>x-2=1

=>x=3(nhận)