Question

cho x,y thỏa: (x+y)^2 +7(x+y)+y^2+10=0 . Tìm Max và Min của T=x+y+1

Answer 1

Ta có:

\(\left(x+y\right)^2+7\left(x+y\right)+y^2+10=0\)

\(x^2+y^2+2xy+7x+7y+y^2+10=0\)

\(x^2+y^2+1+2xy+2x+2y+5x+5y+5+4=0\)

\(\left(x+y+1\right)^2+5\left(x+y+1\right)+4=0\)

\(\left(x+y+1\right)^2+\left(x+y+1\right)+4\left(x+y+1\right)+4=0\)

\(\left(x+y+1\right)\left(x+y+2\right)+4\left(x+y+1\right)=0\)

\(\left(x+y+1\right)\left(x+y+6\right)=0\)

\(x+y=-1\)\(x+y=-6\)​

Max T=x+y+1=-6+1=-5 <=> x+y=-6

Min T=x+y+1=-1+1=0 <=> x+y=-1