a: Xét tứ giác OBAC có 

\(\widehat{OBA}+\widehat{OCA}=180^0\)

Do đó: OBAC là tứ giác nội tiếp

\(\dfrac{x}{a}=\dfrac{m-\dfrac{x}{2}}{m}\) 

\(\Rightarrow xm=a\left(m-\dfrac{x}{2}\right)\)

\(\Rightarrow xm=am-\dfrac{ax}{2}\)

\(\Rightarrow2xm=2am-ax\)

\(\Rightarrow2xm+ax=2am\)

\(\Rightarrow x\left(2m+a\right)=2am\)

\(\Rightarrow x=\dfrac{2am}{a+2m}\)

vâng em cảm ơn ạ

\(b,\Leftrightarrow\left\{{}\begin{matrix}m+2=1\\m\ne2\end{matrix}\right.\Leftrightarrow m=-1\\ c,\text{PT giao Ox: }y=0\Leftrightarrow\left(m+2\right)x-m=0\\ \text{Thay }x=2\Leftrightarrow2m+4-m=0\\ \Leftrightarrow m=-4\\ d,\text{PT giao Ox và Oy: }\\ y=0\Leftrightarrow x=\dfrac{m}{m+2}\Leftrightarrow A\left(\dfrac{m}{m+2};0\right)\Leftrightarrow OA=\left|\dfrac{m}{m+2}\right|\\ x=0\Leftrightarrow y=-m\Leftrightarrow B\left(0;-m\right)\Leftrightarrow OB=\left|m\right|\\ \Delta OAB\text{ cân }\Leftrightarrow OA=OB\Leftrightarrow\left|\dfrac{m}{m+2}\right|=\left|m\right|\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{m}{m+2}=m\\\dfrac{m}{m+2}=-m\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m\left(m+1\right)=0\\m\left(m+3\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}m=0\\m=-1\\m=-3\end{matrix}\right.\)

a: ĐKXĐ: x<>-1; y<>3

\(\begin{cases}\frac{5}{x+1}+\frac{y}{y-3}=-18\\ \frac{2x}{x+1}-\frac{3}{y-3}=9\end{cases}\Rightarrow\begin{cases}\frac{5}{x+1}+\frac{y-3+3}{y-3}=-18\\ \frac{2x+2-2}{x+1}-\frac{3}{y-3}=9\end{cases}\)

=>\(\begin{cases}\frac{5}{x+1}+1+\frac{3}{y-3}=-18\\ 2-\frac{2}{x+1}-\frac{3}{y-3}=9\end{cases}\Rightarrow\begin{cases}\frac{5}{x+1}+\frac{3}{y-3}=-19\\ \frac{2}{x+1}+\frac{3}{y-3}=-7\end{cases}\)

=>\(\begin{cases}\frac{5}{x+1}+\frac{3}{y-3}-\frac{2}{x+1}-\frac{3}{y-3}=-19+7\\ \frac{2}{x+1}+\frac{3}{y-3}=-7\end{cases}\Rightarrow\begin{cases}\frac{3}{x+1}=-12\\ \frac{3}{y-3}=-7-\frac{2}{x+1}\end{cases}\)

=>\(\begin{cases}x+1=-\frac14\\ \frac{3}{y-3}=-7-2:\frac{-1}{4}=-7-2\cdot\left(-4\right)=-7+8=1\end{cases}\Rightarrow\begin{cases}x=-\frac54\\ y-3=3\end{cases}\)

=>\(\begin{cases}x=-\frac54\\ y=3+3=6\end{cases}\) (nhận)

b: ĐKXĐ: x<>-1; y<>-1

\(\begin{cases}\frac{2}{x+1}+\frac{3}{y+1}=2\\ \frac{5}{x+1}+\frac{3y}{y+1}=5\end{cases}\Rightarrow\begin{cases}\frac{2}{x+1}+\frac{3}{y+1}=2\\ \frac{5}{x+1}+\frac{3y+3-3}{y+1}=5\end{cases}\)

=>\(\begin{cases}\frac{2}{x+1}+\frac{3}{y+1}=2\\ \frac{5}{x+1}+3-\frac{3}{y+1}=5\end{cases}\Rightarrow\begin{cases}\frac{2}{x+1}+\frac{3}{y+1}=2\\ \frac{5}{x+1}-\frac{3}{y+1}=2\end{cases}\)

=>\(\begin{cases}\frac{2}{x+1}+\frac{3}{y+1}+\frac{5}{x+1}-\frac{3}{y+1}=2+2\\ \frac{2}{x+1}+\frac{3}{y+1}=2\end{cases}\Rightarrow\begin{cases}\frac{7}{x+1}=4\\ \frac{3}{y+1}=2-\frac{2}{x+1}\end{cases}\)

=>\(\begin{cases}x+1=\frac74\\ \frac{3}{y+1}=2-2:\frac74=2-2\cdot\frac47=2-\frac87=\frac67\end{cases}\)

=>\(\begin{cases}x+1=\frac74\\ y+1=3\cdot\frac76=1\cdot\frac72=\frac72\end{cases}\Longrightarrow\begin{cases}x=\frac74-1=\frac34\\ y=\frac72-1=\frac52\end{cases}\) (nhận)

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

?

\(\sqrt{\dfrac{x^2+2x+1}{16x^2}}=\sqrt{\dfrac{\left(x+1\right)^2}{16x^2}}=\dfrac{\left|x+1\right|}{4\left|x\right|}=\dfrac{1-x}{-4x}=\dfrac{x-1}{4x}\left(do.x\le-1\right)\)

=> 2(2x +1) = 3(x-5) 

=> 4x + 2 = 3x - 15 

=> 4x - 3x = -15 - 2 

=> x = -17

\(\dfrac{2x+1}{3}=\dfrac{x-5}{2}\)

`=> 2(2x+1)=3(x-5)`

`=> 4x +2=3x-15`

`=> 4x-3x=-15-2`

`=> x= -17`

Vậy `x=-17`

`@ ` \(\text{Mạc Nhược Hàn}\)

\(\Leftrightarrow4x^2-20x-4x^2+3x+12x-3=5\)

\(\Leftrightarrow-5x=8\)

hay \(x=-\dfrac{8}{5}\)