Bài 2:

\(a,m=2\Leftrightarrow\left(d_1\right):y=x+3\\ \text{PT hoành độ giao điểm: }x+3=-x+1\Leftrightarrow x=-1\Leftrightarrow y=2\Leftrightarrow A\left(1;2\right)\\ b,\Leftrightarrow x=3;y=4\Leftrightarrow3\left(m-1\right)+m+1=4\\ \Leftrightarrow m=\dfrac{3}{2}\\ c,\Leftrightarrow\left\{{}\begin{matrix}m-1\ne-1\\m+1\ne1\end{matrix}\right.\Leftrightarrow m\ne0\)

\(\Leftrightarrow\left(x^2-2x+2-1\right)^2=0\\ \Leftrightarrow\left[\left(x-1\right)^2\right]^2=0\\ \Leftrightarrow x-1=0\Leftrightarrow x=1\)

\(a,ĐK:x\le-2;x\ge2\\ PT\Leftrightarrow\sqrt{x+2}\left(\sqrt{x-2}+\sqrt{x+2}\right)=0\\ \Leftrightarrow\sqrt{x+2}=0\left(\sqrt{x-2}+\sqrt{x+2}>0\right)\\ \Leftrightarrow x=-2\left(tm\right)\)

Từ GT, suy ra \(\dfrac{b+c-2a}{a}+3=\dfrac{a+c-2b}{b}+3=\dfrac{a+b-2c}{c}+3\)

\(\Rightarrow\dfrac{a+b+c}{a}=\dfrac{a+b+c}{b}=\dfrac{a+b+c}{c}\)

Nếu \(a+b+c=0\Rightarrow A=\dfrac{2a+2b}{a}\cdot\dfrac{2c+2a}{c}\cdot\dfrac{2b+2c}{b}\)

\(\Rightarrow A=\dfrac{8\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{8\left(-c\right)\left(-b\right)\left(-a\right)}{abc}=-8\)

Nếu \(a+b+c\ne0\Rightarrow a=b=c\Rightarrow A=\dfrac{8\cdot2a\cdot2a\cdot2a}{a\cdot a\cdot a}=64\)

\(a,\left\{{}\begin{matrix}MB=MC\\AB=AC\\AM\text{ chung}\end{matrix}\right.\Rightarrow\Delta ABM=\Delta ACM\left(c.c.c\right)\\ b,\left\{{}\begin{matrix}AM=ME\\BM=MB\\\widehat{AMB}=\widehat{EMC}\left(đđ\right)\end{matrix}\right.\Rightarrow\Delta AMB=\Delta EMC\left(c.g.c\right)\\ \Rightarrow\widehat{ABC}=\widehat{BME}\\ \text{Mà 2 góc này ở vị trí slt nên }AB\text{//}EC\\ c,\Delta AMB=\Delta AMC\\ \Rightarrow\widehat{AMB}=\widehat{AMC}\widehat{AMC}\\ \text{Mà }\widehat{AMC}+\widehat{AMB}=180^0\\ \Rightarrow\widehat{AMC}=\widehat{AMB}=90^0\\ \Rightarrow AM\bot BC\)