2X+4Y=1 比较X平方加Y平方与1/20的大小
热心网友
∵2x+4y=1 ∴y=(1-2x)/4x^2+y^2=x^2+[(1-2x)/4]^2=(5/4)x^2-(x/4)+(1/16)令x^2+y^2=(5/4)x^2-(x/4)+(1/16)=m使得(5/4)x^2-(x/4)+(1/16)=m有解，Δ=(1/4)^2-4*(5/4)*[(1/16)-m]=(1/4)(20m-1)≥0∴m≥1/20即x^2+y^2=(5/4)x^2-(x/4)+(1/16)≥1/20∴x^2+y^2≥1/20
热心网友
2x+4y=1---y=(1-2x)/4---x^2+y^2=x^2+(1/2*x^2-x/4+1/16=5/4*x^2-x/4+1/16=5/4*(x-1/10)^2+1/20=1/20---x^2+y^2=1/20