`(2m-10)^2/((m+5)^2+1)`
`=(2m-10)^2/(m^2+10m+26)-404+404`
`=(4m^2-40m+100)/(m^2+10m+26)-404+404`
`=(4m^2-40m+100-404m^2-4040m-10504)/(404[(m+5)^2+1])+404`
`=(-400m^2-4080m-10404)/(404[(m+5)^2+1])+404`
`=(-(400m^2+4080m+10404))/(404[(m+5)^2+1])+404`
`=(-(20m+102)^2)/(404[(m+5)^2+1])+404<=404`
Dấu "=" xảy ra khi `20m+102=0<=>m=(-51)/10`
Bài này giải kiểu lớp 8 thì nó cực kì vô duyên:
\(P=\dfrac{4m^2-40m+100}{m^2+10m+26}=\dfrac{404\left(m^2+10m+26\right)-4\left(100m^2+1020m+2601\right)}{m^2+10m+26}\)
\(P=404-\dfrac{4\left(10m+51\right)^2}{\left(m+5\right)^2+1}\le404\)
\(P_{max}=404\) khi \(m=-\dfrac{51}{10}\)