\(a,\left\{{}\begin{matrix}AM=DM\\BM=MC\\\widehat{AMB}=\widehat{DMC}\end{matrix}\right.\Rightarrow\Delta ABM=\Delta DCM\left(c.g.c\right)\\ b,\Delta ABM=\Delta DCM\Rightarrow\widehat{B}=\widehat{MCD}\)

Mà 2 góc này ở vị trí so le trong

\(\Rightarrow AB\text{//}CD\\ c,\left\{{}\begin{matrix}AB=AC\\BM=MC\\AM\text{ chung}\end{matrix}\right.\Rightarrow\Delta AMB=\Delta AMC\left(c.c.c\right)\\ \Rightarrow\widehat{BAM}=\widehat{CAM}\\ \Rightarrow AM\text{ là p/g }\widehat{A}\\ d,\Delta AMB=\Delta AMC\Rightarrow\widehat{AMB}=\widehat{AMC}\\ \text{Mà }\widehat{AMB}+\widehat{AMC}=180^0\\ \Rightarrow\widehat{AMB}=90^0\\ \Rightarrow AM\bot BC\)

Mà M là trung điểm BC nên AM là trung trực BC

\(b,\Leftrightarrow\left\{{}\begin{matrix}m+1=3\\m-3\ne-3\end{matrix}\right.\Leftrightarrow m=2\\ c,\text{PT giao Ox tại hoành độ 3: }\\ x=-3;y=0\Leftrightarrow\left(m+1\right)\left(-3\right)+m-3=0\\ \Leftrightarrow-2m-6=0\Leftrightarrow m=-3\)

\(A=\left(x^2+2\cdot\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{5}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4}\\ A_{min}=-\dfrac{5}{4}\Leftrightarrow x=-\dfrac{3}{2}\\ B=\left(x^2+2xy+y^2\right)+\left(x^2+6x+9\right)+3\\ B=\left(x+y\right)^2+\left(x+3\right)^2+3\ge3\\ B_{min}=3\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-3\end{matrix}\right.\\ C=-\left(x^2-2x+1\right)+1=-\left(x-1\right)^2+1\le1\\ C_{max}=1\Leftrightarrow x=1\)

\(1,A=\dfrac{5+1}{25-9}=\dfrac{6}{16}=\dfrac{3}{8}\\ 2,B=\dfrac{x-\sqrt{x}-12-x+3\sqrt{x}+18}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{2\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{2}{\sqrt{x}-3}\\ 3,P=2A:B=\dfrac{2\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{2}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\\ P=1-\dfrac{2}{\sqrt{x}+3}\)

P đạt min \(\Leftrightarrow\dfrac{2}{\sqrt{x}+3}\) đạt max \(\Leftrightarrow\sqrt{x}+3\) đạt min \(\Leftrightarrow\sqrt{x}+3=1+3=4\left(x_{min}\text{ nguyên dương}\right)\)

Vậy \(P_{min}=\dfrac{1+1}{1+3}=\dfrac{1}{2}\Leftrightarrow x=1\)

Nguyễn Minh Sơn Nguyễn Minh Trẩu Sơn :v

Đệ nhất trẩu tre

Nguyễn Minh Sơn kệ nó :))

Nguyễn Minh Sơn ko phải chữ t :v