a: =>2mx-x+my-2y=2
=>m(2x+y)-x-2y-2=0
Tọa độ mà (d) luôn đi qua là: 2x+y=0 và x+2y=2
=>x=-2/3; y=4/3
b: \(d\left(O;d\right)=\dfrac{\left|0\cdot\left(2m-2\right)+0\cdot\left(m-2\right)-2\right|}{\sqrt{\left(2m-2\right)^2+\left(m-2\right)^2}}=\dfrac{2}{\sqrt{\left(2m-2\right)^2+\left(m-2\right)^2}}\)
Để d lớn nhất thì \(\sqrt{\left(2m-2\right)^2+\left(m-2\right)^2}_{MIN}\)
=>\(\sqrt{4m^2-8m+4+m^2-4m+4}_{MIN}\)
=>\(\sqrt{5m^2-12m+8}_{MIN}\)
=>\(\sqrt{5\left(m^2-\dfrac{12}{5}m+\dfrac{8}{5}\right)}_{MIN}\)
=>\(m^2-\dfrac{12}{5}m+\dfrac{8}{5}_{MIN}\)
=>\(m^2-2\cdot m\cdot\dfrac{6}{5}+\dfrac{36}{25}+\dfrac{4}{25}=\left(m-\dfrac{6}{5}\right)^2+\dfrac{4}{25}>=\dfrac{4}{25}\)
Dấu = xảy ra khi m=6/5
