\(=\left[\dfrac{2x-3}{\left(2x-5\right)\left(2x-1\right)}-\dfrac{3}{2x-1}-\dfrac{2\left(x-4\right)}{\left(x-4\right)\left(2x-5\right)}\right].\dfrac{2x\left(2x+3\right)-\left(2x+3\right)}{-2x\left(4x-7\right)-3\left(4x-7\right)}+1\)

\(=\left[\dfrac{2x-3-6x+15-4x+2}{\left(2x-5\right)}\right].\dfrac{2\left(x+\dfrac{3}{2}\right)}{\left(-2x-3\right)\left(4x-7\right)}+1\)

\(=\dfrac{-2\left(4x-7\right)}{2x-5}.\dfrac{2\left(x+\dfrac{3}{2}\right)}{\left(-2x-3\right)\left(4x-7\right)}+1\)

\(=\dfrac{1}{2x-5}.2+1\)

\(=\dfrac{2+2x-5}{2x-5}\)

\(=\dfrac{-3+2x}{2x-5}\)

Đặt bthuc = A nhé

ĐKXĐ : \(2x\ne3y\)

\(A=\left[\dfrac{2x\left(4x^2+6xy+9y^2\right)}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}-\dfrac{27y^3+36xy^2}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}-\dfrac{24xy\left(2x-3y\right)}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\right]\left[\dfrac{2x\left(2x-3y\right)}{\left(2x-3y\right)}+\dfrac{9y^2+12xy}{\left(2x-3y\right)}\right]\)\(=\left[\dfrac{8x^3+12x^2y+18xy^2-27y^3-36xy^2-48x^2y+72xy^2}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\right]\left[\dfrac{4x^2-6xy+9y^2+12xy}{\left(2x-3y\right)}\right]\)

\(=\dfrac{8x^3-36x^2y+36xy^2-27y^3}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\cdot\dfrac{4x^2+6xy+9y^2}{2x-3y}\)

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

Với x = 1/3 ; y = -2 (tmđk) thay vào A ta được : A = 2.1/3 - 3.(-2) = 20/3

ĐKXĐ: \(x\ne\left\{\dfrac{-3}{2};\dfrac{1}{2};\dfrac{7}{4};\dfrac{5}{2};4;\right\}\)

\(P=\left(\dfrac{2x-3}{\left(2x-1\right)\left(2x-5\right)}-\dfrac{3}{2x-1}-\dfrac{2\left(x-4\right)}{\left(2x-5\right)\left(x-4\right)}\right)\div\dfrac{\left(7-4x\right)\left(2x+3\right)}{\left(2x-1\right)\left(2x+3\right)}+1\)

\(P=\left(\dfrac{2x-3-3\left(2x-5\right)-2\left(2x-1\right)}{\left(2x-1\right)\left(2x-5\right)}\right)\dfrac{2x-1}{7-4x}+1\)

\(P=\dfrac{-8x+14}{\left(2x-5\right)\left(7-4x\right)}+1=\dfrac{2}{2x-5}+1\)

b/ \(\left|x\right|=\dfrac{1}{2}\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{-1}{2}\end{matrix}\right.\)

Với \(x=\dfrac{1}{2}\Rightarrow P=\dfrac{2}{2.\dfrac{1}{2}-5}+1=\dfrac{1}{2}\)

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

c/ Để P nguyên \(\Rightarrow\dfrac{2}{2x-5}\) nguyên \(\Rightarrow2⋮\left(2x-5\right)\Rightarrow2x-5=Ư\left(2\right)=\left\{-2;-1;1;2\right\}\)

\(2x-5=-2\Rightarrow x=\dfrac{3}{2}\left(l\right)\)

\(2x-5=-1\Rightarrow x=2\)

\(2x-5=1\Rightarrow x=3\)

\(2x-5=2\Rightarrow x=\dfrac{7}{2}\left(l\right)\)

Vậy \(x=\left\{2;3\right\}\) thì P nguyên

d/ \(P>0\Rightarrow\dfrac{2}{2x-5}+1>0\Rightarrow\dfrac{2x-3}{2x-5}>0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-3>0\\2x-5>0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-3< 0\\2x-5< 0\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x< \dfrac{5}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x>\dfrac{5}{2}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x< \dfrac{3}{2}\\x>\dfrac{5}{2}\end{matrix}\right.\)

a: \(=\left(\dfrac{2x-3}{\left(2x-5\right)\left(2x-1\right)}-\dfrac{2x-8}{\left(2x-5\right)\left(x-4\right)}-\dfrac{3}{2x-1}\right)\cdot\dfrac{4x^2+4x-3}{-8x^2+2x+21}+1\)

\(=\dfrac{2x-3-2\left(2x-1\right)-3\left(x-4\right)}{\left(2x-1\right)\left(2x-5\right)}\cdot\dfrac{\left(2x+3\right)\left(2x-1\right)}{\left(4x-7\right)\cdot\left(-2x-3\right)}+1\)

\(=\dfrac{2x-3-4x+2-3x+12}{1}\cdot\dfrac{-1}{4x-7}+1\)

\(=\dfrac{5x-11}{4x-7}+1=\dfrac{9x-4}{4x-7}\)

b: |x|=1/2

=>x=1/2(loại) hoặc x=-1/2(nhận)

Khi x=-1/2 thì \(P=\dfrac{\dfrac{-9}{2}-4}{-2-7}=-\dfrac{17}{2}:\left(-9\right)=\dfrac{17}{18}\)

c: Để P là số nguyên thì \(36x-16⋮4x-7\)

\(\Leftrightarrow4x-7\in\left\{1;-1;47;-47\right\}\)

hay \(x\in\left\{2;\dfrac{3}{2};\dfrac{27}{2};-10\right\}\)

\(A=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}\\ A=\dfrac{2\left(\dfrac{1}{2}-2\right)}{\dfrac{1}{2}+2}=\dfrac{2\left(-\dfrac{3}{2}\right)}{\dfrac{5}{2}}=\left(-3\right)\cdot\dfrac{2}{5}=-\dfrac{6}{5}\)

\(B=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}=\dfrac{-5}{-5+10}=\dfrac{-5}{5}=-1\)