Đề sai rồi bạn

a: \(A=\dfrac{-\left(x+2\right)^2-2x\left(x-2\right)-4x^2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)\left(x-3\right)}{\left(x-3\right)^2}\)

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

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

b: Khi x=1/3 thì \(A=\dfrac{7\cdot\dfrac{1}{9}+4}{\left(\dfrac{1}{3}-2\right)\left(\dfrac{1}{3}-3\right)}=\dfrac{43}{40}\)

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

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

b: A nguyên

=>x^2-1+3 chia hết cho x-1

=>\(x-1\in\left\{1;-1;3;-3\right\}\)

=>\(x\in\left\{2;0;4;-2\right\}\)

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

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

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

b: |x-3|=2

=>x-3=2 hoặc x-3=-2

=>x=5(nhận) hoặc x=1(loại)

Khi x=5 thì \(E=\dfrac{5^2}{5-1}=\dfrac{25}{4}\)

c: Để E=1/2 thì \(\dfrac{x^2}{x-1}=\dfrac{1}{2}\)

\(\Leftrightarrow2x^2-x+1=0\)

hay \(x\in\varnothing\)

f) \(A=\dfrac{x^2}{x-1}=\dfrac{x^2-x+x-1+1}{x-1}=\dfrac{x\left(x-1\right)+x-1+1}{x-1}=x+1+\dfrac{1}{x-1}=x-1+\dfrac{1}{x-1}+2\ge2\sqrt{\left(x-1\right).\dfrac{1}{x-1}}+2=4\)\(A=4\Leftrightarrow x=2\)

-Vậy \(A_{min}=4\)

\(a,A=\dfrac{2x\left(x-3\right)+8\left(x+3\right)-2x-12}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x^2+6}\\ A=\dfrac{2x^2-6x+8x+24-2x-12}{\left(x-3\right)}\cdot\dfrac{1}{x^2+6}\\ A=\dfrac{2x^2+12}{\left(x-3\right)\left(x^2+6\right)}=\dfrac{2\left(x^2+6\right)}{\left(x-3\right)\left(x^2+6\right)}=\dfrac{2}{x-3}\)

\(b,A=5\Leftrightarrow\dfrac{2}{x-3}=5\Leftrightarrow5x-15=2\Leftrightarrow x=\dfrac{17}{5}\)