证明不等式(x2+y2)2>=xy(x+y)2
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/07/17 09:04:48
证明不等式(x2+y2)2>=xy(x+y)2
2都代表平方
2都代表平方
证明:(x^2+y^2)^2-xy(x+y)^2
=x^4+2x^2 y^2+y^4-xy(x^2+2xy+y^2)
=x^4-x^3 y+y^4-xy^3
=x^3(x-y)-y^3 (x-y)
=(x-y)^2 (x^2+xy+y^2)
=(x-y)^2 [(x+y/2)^2+3y^2/4]
显然在实数范围内上式≥0,故(x^2+y^2)^2≥xy(x+y)^2.
=x^4+2x^2 y^2+y^4-xy(x^2+2xy+y^2)
=x^4-x^3 y+y^4-xy^3
=x^3(x-y)-y^3 (x-y)
=(x-y)^2 (x^2+xy+y^2)
=(x-y)^2 [(x+y/2)^2+3y^2/4]
显然在实数范围内上式≥0,故(x^2+y^2)^2≥xy(x+y)^2.
已知2x=3y,求xy/(x2+y2)-y2/(x2-y2)的值
已知:x=3,y=-5,x2-2xy+y2/x2-y2的值是
若X2+Y2-2X-6Y+10=0 ,求(x2-y2)/xy的值
已知x2+4y2+x2y2-6xy+1=0,求 x4-y4/2x-y 乘 2xy-y2/xy-y2 除以(x2+y2/x
正数xy满足x2-y2=2xy,求x+y分之x-y的值
若x<y<0,则x2−2xy+y2+x2+2xy+y2=( )
已知X2-2x+y2+6y+10=0,求(x2-2xy)/(xy+y2)的值
正整数x,y满足x2-y2=2xy,求x-y/x+y的值
整数x,y满足x2-y2=2xy,求x-y/x+y的值
已知x2+y2=20,x2+xy-2y2=2(x+2y),求x+y的值
已知x-y+1,X2+Y2=25 求(x+y)2和x2-xy+y2的值
已知X2+Y2+8X+6Y+25=0 求代数式X2++XY+4Y2分之X2-4Y2 减X+2Y分之X的值