a: \(A=\dfrac{x^2+2x}{x^2-4x+4}:\left(\dfrac{x+2}{x}-\dfrac{1}{2-x}+\dfrac{6-x^2}{x^2-2x}\right)\)
\(=\dfrac{x\left(x+2\right)}{\left(x-2\right)^2}:\left(\dfrac{x+2}{x}+\dfrac{1}{x-2}+\dfrac{6-x^2}{x\left(x-2\right)}\right)\)
\(=\dfrac{x\left(x+2\right)}{\left(x-2\right)^2}:\dfrac{\left(x+2\right)\left(x-2\right)+x+6-x^2}{x\left(x-2\right)}\)
\(=\dfrac{x\left(x+2\right)}{\left(x-2\right)^2}\cdot\dfrac{x\left(x-2\right)}{x^2-4+x+6-x^2}\)
\(=\dfrac{x^2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x^2}{x-2}\)
b: |2x+1|=3
=>\(\left[{}\begin{matrix}2x+1=3\\2x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-2\left(loại\right)\end{matrix}\right.\)
Khi x=1 thì \(A=\dfrac{1^2}{1-2}=-1\)
c: A<0
=>\(\dfrac{x^2}{x-2}< 0\)
mà \(x^2>0\forall x\) thỏa mãn ĐKXĐ
nên x-2<0
=>x<2
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}x< 2\\x\notin\left\{0;-2\right\}\end{matrix}\right.\)
d: Để A là số nguyên thì \(x^2⋮x-2\)
=>\(x^2-4+4⋮x-2\)
=>\(4⋮x-2\)
=>\(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)
=>\(x\in\left\{3;1;4;0;6;-2\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{3;1;4;6\right\}\)
e)
\(A=\dfrac{x^2}{x-2}=\dfrac{x^2-4x+4+4x-4}{x-2}\\ =\dfrac{\left(x-2\right)^2}{x-2}+\dfrac{4x-4}{x-2}\\ =x-2+\dfrac{4x-8+4}{x-2}\\ =x-2+4+\dfrac{4}{x-2}\\ =x+2+\dfrac{4}{x-2}\\ =x-2+\dfrac{4}{x-2}+4\)
Vì \(x>2\Rightarrow\left\{{}\begin{matrix}x-2>0\\\dfrac{4}{x-2}>0\end{matrix}\right.\)
Áp dụng bất đẳng thức cô-si ta có:
\(A\ge2\sqrt{\left(x-2\right)\cdot\dfrac{4}{x-2}}+4=2\cdot2+4=8\)
Dấu "=" xảy ra khi: \(x-2=\dfrac{4}{x-2}\Leftrightarrow\left(x-2\right)^2=4\Leftrightarrow\left[{}\begin{matrix}x=6\left(tm\right)\\x=0\left(ktm\right)\end{matrix}\right.\)
Vậy: ...
