已知x y>1,x一定比y大吗?举例说明
来源:学生作业帮助网 编辑:作业帮 时间:2024/09/29 21:34:55
原式=(y-x)/xy=-xy/xy=-1
移项,x=y+xy:同除y得x/y=1+x;同除X得1/y=1/x+1;移项1/x-1/y=-1
x+y=3,xy=14x+3xy+4y=4(x+y)+3xy=4x3+3x1=12+3=15祝学习进步,望采纳,不懂的欢迎追问.再问:怎么又是你啊。。。再问一道题:代数式x²+x+3的值为7
再问:该方法此处计算是错的,应该为,接下来的都不对了再答:那就从那步开始吧x+y=xy-8若x,y大于0xy-8=x+y≥2√xyxy-8≥2√xyxy-2√xy-8≥0(√xy-4)(√xy+2)≥
=-x-(2y-2+3x)+2(x+4)=-x-2y+2-3x+2x+8=-4x-2y+10
绝对值大于等于0,相加等于0,若有一个大于0,则另一个小于0,不成立所以两个都等于0所以x+y+1=0xy+3=0xy=-3x+y=-1两边平方x^2+2xy+y^2=(-1)^2x^2+y^2=1-
x+15=y3:x=5:(x+15)x=22.5y=37.5
(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)=-6xy+3x-3y=-6×(-2)+3×1=15
答:x+y=-1,xy=-2-5(x+y)+(x-y)+x(xy+y)=-5x-5y+x-y+xy(x+1)=-4x-6y+(-2)(x+1)=-4x-6y-2x-2=-6x-6y-2=-6(x+y)
x^2y+xy^2=xy(x+y)=1/5
设y/x=k,即有y=kx代入方程中有:x^2+k^2x^2-4x+1=0(1+k^2)x^2-4x+1=0判别式=16-4(1+k^2)>=01+k^2
x²-5xy+6y²=0(x-2y)(x-3y)=0x=2y或x=3y∴x:y=2:1或x:y=3:1
(1)∵xy+x=-1①,xy-y=-2②,∴①-②得x+y=1;(2)先把xy+x=-1,xy-y=-2的值代入代数式,得原式=-x-[2y-1+3x]+2[x+4]=-x-2y+1-3x+2x+8
3/(x-y)=1/xyx-y=3xyy-z=-3xy原式=[(y-x)-2xy]/[2(x-y)+3xy]=[(-3xy)-2xy]/[2(3xy)+3xy]=-5xy/9xy=-5/9
xy+1/xy+y/x+x/y=[(xy)^2+1+x^2+y^2]/(xy)=[(xy)^2-2xy+1+x^2-2xy+y^2+4xy]/(xy)=[(xy-1)^2+(x-y)^2+4xy]/(
xy+1/xy>=2√(xy*1/xy)=2(当xy=1/xy即xy=1时取等号)x/y+y/x>=2√(x/y*y/x)=2(当x/y=y/x即x=y取等号)当x=y=1时可以同时满足两项的等号要求
因为xy/(x+y)=1/2所以x+y=2xy原式=3(x+y)-5xy/(-x-y+3xy)=3*2xy-5xy/(-2xy+3xy)=xy/xy=1
∵xy≤(x+y 2)2=14,设xy=t,令f(t)=t+1t,因其f′(t)=1-1t2,当0<t≤14时,f′(t)<0,故函数f(t)在(0,14]上是减函数,∴t+1t≥14+4=
Bxyy那么x为正数,因为负数a为任意有理数a^2等于0所以选B
(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)=-2xy+2x+3y-3xy-2y+2x-x-4y-xy=-6xy+3x-3y=-6*(-2)+3*1=15不懂可追问,有帮助请采