a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\\x< >4\end{matrix}\right.\)

b: \(A=\dfrac{\left(\sqrt{x}-3\right)}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{x-3\sqrt{x}+2}\)

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

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

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

c: A nguyên khi \(\sqrt{x}-2\in\left\{1;-1\right\}\)

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

Kết hợp ĐKXĐ, ta được: \(\sqrt{x}=3\)

=>x=9

\(x^2-2>-\sqrt{3}\)

=>\(x^2>2-\sqrt{3}\)

=>\(\left[{}\begin{matrix}x>\sqrt{2-\sqrt{3}}=\dfrac{1}{\sqrt{2}}\cdot\sqrt{4-2\sqrt{3}}=\dfrac{\sqrt{2}}{2}\left(\sqrt{3}-1\right)\\x< -\sqrt{2-\sqrt{3}}=-\dfrac{\sqrt{2}}{2}\left(\sqrt{3}-1\right)\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x>\dfrac{\sqrt{6}-\sqrt{2}}{2}\\x< \dfrac{-\sqrt{6}+\sqrt{2}}{2}\end{matrix}\right.\)

x² - 2 > -√3

⇔ x² > 2 - √3

⇔ x > √(2 - √3) hoặc x < -√(2-√3)

a: \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{x+4}{x-4}\right):\dfrac{2\sqrt{x}-1-\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

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

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

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

b: P>-1/2

=>\(P+\dfrac{1}{2}>0\)

=>\(-\dfrac{2\sqrt{x}}{\sqrt{x}+1}+\dfrac{1}{2}>0\)

=>\(\dfrac{-4\sqrt{x}+\sqrt{x}+1}{2\left(\sqrt{x}+1\right)}>0\)

=>\(-3\sqrt{x}+1>0\)

=>\(-3\sqrt{x}>-1\)

=>\(\sqrt{x}< \dfrac{1}{3}\)

=>\(0< x< \dfrac{1}{9}\)

\(\left\{{}\begin{matrix}mx-y=1\\x+my=m+6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}m^2x-my=m\\x+my=m+6\end{matrix}\right.\) \(\left(m\ne0\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(m^2+1\right)x=2m+6\\mx-y=1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2m+6}{m^2+1}\\y=mx-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2m+6}{m^2+1}\\y=m.\dfrac{2m+6}{m^2+1}-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2m+6}{m^2+1}\\y=\dfrac{m^2+6m-1}{m^2+1}\end{matrix}\right.\)

Theo đề bài :

\(3x-y=1\)

\(\Leftrightarrow3.\dfrac{2m+6}{m^2+1}-\dfrac{m^2+6m-1}{m^2+1}=1\)

\(\Leftrightarrow\dfrac{19-m^2}{m^2+1}=1\)

\(\Leftrightarrow19-m^2=m^2+1\)

\(\Leftrightarrow2m^2=18\)

\(\Leftrightarrow m^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}m=3\\m=-3\end{matrix}\right.\) thỏa mãn yêu cầu của đề bài

`1/(x-2) -x/(x^2-4)`

`= 1/(x-2) - x/((x-2)(x+2))`

`= (x+2)/((x-2)(x+2))  - x/((x-2)(x+2))`

`= (x+2-x)/((x-2)(x+2)) `

`= 2/(x-4)`

\(\dfrac{1}{x^2-3x+2}-\dfrac{1}{x-2}=2\left(đkxđ:x\ne1;x\ne2\right)\\ \Leftrightarrow\dfrac{1}{\left(x-2\right)\left(x-1\right)}-\dfrac{1}{x-2}=2\\ \Leftrightarrow\dfrac{1}{\left(x-2\right)\left(x-1\right)}-\dfrac{x-1}{\left(x-2\right)\left(x-1\right)}=\dfrac{2\left(x-2\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}\\ \Rightarrow1-x+1=\left(2x-4\right)\left(x-1\right)\\ \Leftrightarrow2-x=2x^2-2x-4x+4\\ \Leftrightarrow-x-2x^2+2x+4x=4-2\\ \Leftrightarrow-2x^2+5x-2=0\\ \Leftrightarrow-\left(2x^2-x-4x+2\right)=0\\ \Leftrightarrow-\left[\left(2x^2-x\right)-\left(4x-2\right)\right]=0\)

\(\Leftrightarrow-\left[x\left(2x-1\right)-2\left(2x-1\right)\right]=0\\ \Leftrightarrow-\left(2x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-2x+1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-2x=-1\\x=2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=2\end{matrix}\right.\)

\(\dfrac{1}{x-2}+\dfrac{1}{x^2+x-6}=\dfrac{1}{x+3}-1\left(đkxđ:x\ne2;x\ne-3\right)\\ \Leftrightarrow\dfrac{1}{x-2}+\dfrac{1}{\left(x-2\right)\left(x+3\right)}=\dfrac{1}{x+3}-1\\ \Leftrightarrow\dfrac{x+3}{\left(x-2\right)\left(x+3\right)}+\dfrac{1}{\left(x-2\right)\left(x+3\right)}=\dfrac{x-2}{\left(x-2\right)\left(x+3\right)}-\dfrac{\left(x-2\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\\ \Rightarrow x+3+1=x-2-\left(x^2+3x-2x-6\right)\\ \Leftrightarrow x+4=x-2-x^2-3x+2x+6\\ \Leftrightarrow x-x+x^2+3x-2x=-2+6-4\\ \Leftrightarrow x^2+x=0\)

\(\Leftrightarrow x\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)

\(\sqrt{3+\sqrt{5}}\cdot\sqrt{2}\)

\(=\sqrt{2\cdot\left(3+\sqrt{5}\right)}\)

\(=\sqrt{6+2\sqrt{5}}\)

\(=\sqrt{\left(\sqrt{5}\right)^2+2\cdot\sqrt{5}\cdot1+1^2}\)

\(=\sqrt{\left(\sqrt{5}+1\right)^2}\)

\(=\left|\sqrt{5}+1\right|\)

\(=\sqrt{5}+1\)

Xét tứ giác ADHE có

\(\widehat{ADH}=\widehat{AEH}=\widehat{DAE}=90^0\)

=>ADHE là hình chữ nhật

ΔHDB vuông tại D có DK là trung tuyến

nên KH=KB=KD

ΔHEC vuông tại E có EI là trung tuyến

nên EI=IH=IC

\(\widehat{IED}=\widehat{IEH}+\widehat{DEH}\)

\(=\widehat{IHE}+\widehat{DAH}\)

\(=\widehat{HAB}+\widehat{HBA}=90^0\)

=>IE vuông góc ED(1)

\(\widehat{KDE}=\widehat{KDH}+\widehat{EDH}\)

\(=\widehat{KHD}+\widehat{EAH}=\widehat{HAC}+\widehat{HCA}=90^0\)

=>KD vuông góc DE(2)

Từ (1), (2) suy ra DKIE là hình thang vuông

\(S_{DKIE}=\dfrac{1}{2}\left(DK+EI\right)\cdot ED\)

\(=\dfrac{1}{2}\cdot AH\cdot\left(\dfrac{1}{2}HC+\dfrac{1}{2}HB\right)\)

\(=\dfrac{1}{4}\cdot AH\cdot BC\)

=>\(\dfrac{S_{DKIE}}{S_{ABC}}=\dfrac{1}{4}:\dfrac{1}{2}=\dfrac{1}{2}\)