x²+2xy-1=0.求x+y的取值范围
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xy/(x+y)=1,取倒数(x+y)/xy=1x/xy+y/xy=11/y+1/x=1.1yz/(y+z)=2,取倒数(y+z)/yz=1/2y/yz+z/yz=1/21/z+1/y=1/2.2xz
=-x-(2y-2+3x)+2(x+4)=-x-2y+2-3x+2x+8=-4x-2y+10
即x=y=1xy=1对不对?如果对的话x^2+2y^2-2xy-2y+1=0化简为你做的很对.就是这样解的,没有其他更好的方法了.这里用到的是数学里的
(x+y-2xy)(x+y-2)+(1-xy)^2=(x+y)^2-2(1+xy)(x+y)+4xy+1-2xy+x^2y^2=(x+y)^2-2(1+xy)(x+y)+(1+xy)^2=(x+y)^
(-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)
∵1/x-1/y=3∴y-x=3xy2x+3xy-2y/x-y=2(x-y)+3xy/x-y=-6xy+3xy/-3xy=1
原式=3x-3y-6xy=3-12=-9
x(x+y)(x-y)-x(x+y)2=x(x+y)[(x-y)-(x+y)]=x(x+y)(-2y)=-2xy(x+y)=-2×(-1/2)×1=1再问:18p3q3-2pq再答:7(x-1)3-1
x(x-1)-(x^2-y)=-3x²-x-x²+y=-3y-x=-3(y-x+2xy)/2-xy=(y-x)/2+xy-xy=(y-x)/2=-3/2
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
(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)=-2xy+2x+3y-3xy-2y+2x-x-4y-xy=-6xy+3x-3y=-6xy+3*(x-y)当时,原式=-6*1+3*
-5(x+y)+(x-y)+2(xy+y)=-5x-5y+x-y+2xy+2y=-4x-4y+2xy=-4(x+y)+2xy=-4×(-1)+2×(-2)=4+(-4)=0你有问题也可以在这里向我提问
这道题目还是在考察韦达定理的运用用伟大定理求出xy的值再代入代数式否则是求不出来的(x+y)^2=x^2+y^2+2xy=1x^2+y^2=5(x-y)^2=5-2(-2)=9下面分两种情况讨论1x-
x{x-1}-{x^2-y}=-3即x^2-x-x^2+y=-3-x+y=-3x-y=3而x^2+y^2-2xy=(x-y)^2=3^2=9
(-2xy+2x+3y)-(3xy+2y-2x)-(x+4y+xy)=(-2xy-3xy-xy)+(2x+2x-x)+(3y-2y-4y)=-6xy+3x-3y=-6+3*3=3
x(x-1)-(x^2-y)=5化简得y-x=5(y-x)^2=25y^2+x^2-2xy=25
x(x+y)(x-y)-x(x+y)2=x(x+y)[(x-y)-(x+y)]=-2xy(x+y).当x+y=1,xy=-12时,原式=-2×(-12)×1=1.
(-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