解方程组x y=10,7x 4y=3
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第一题设x+y=axy=b则x^2+y^2=a^2-2b原方程变为a^2-2b+b=13①b+a=7②这就变成了简单的二元二次方程由②得b=7-a带入①得a^2+a-20=0(a-4)(a+5)=0得
x:y=3:2①y:z=5:4②x+y+z=66③,由①得x=32y④,由②得z=45y⑤,把④⑤代入③得32y+y+45y=66,解得y=20,把y=20代入④得x=32×20=30,把y=20代入
x+y=7-xyx^2+y^2=10即(x+y)^2=(7-xy)^2=x^2+y^2+2xy=10+2xy即(7-xy)^2=10+2xy即x^2y^2+39-16xy=0解得xy=3代入x+xy+
∵2xy-5√(xy+1)=10==>2xy+2-5√(xy+1)=12==>2(xy+1)-5√(xy+1)-12=0==>[2√(xy+1)+3][√(xy+1)-4]=0(用十字相乘法分解)==
x+y=1①xy=−2②,∵由①得:x=1-y,③把③代入②得:(1-y)y=-2,即y2-y-2=0,(y-2((y+1)=0,解得:y1=-1,y2=2,代入③得:x1=2,x2=-1,故答案为:
将x+y=2两边分别平方,得x2+2xy+y2=4(1)把方程xy-z2=1两边都乘以2得2xy-2z2=2(2)(1)-(2)得:x2+y2+2z2=2(3)由x+y=2得2x+2y=4(4)(3)
(1)×7+(2)×328y+27y=55xy所以y=0或x=1分别代入所以x=0.y=0x=1,y=1/2
xy/3x+2y=1/8,xy/2x+3y=1/73x+2y/xy=8,2x+3y/xy=73/y+2/x=8,2/y+3/x=73/y+2/x=8,2/y+3/x=71/x=1,1/y=2,x=1,
两式相减x²-10x-y²+10y=0x²-y²-10(x-y)=0(x+y)(x-y)-10(x-y)=0(x+y-10)(x-y)=0x+y=10x-y=0
x+y=7x=7-y(7-y)y=127y-y^2=12y^2-7y+12=0(y-3)(y-4)=0y1=3y2=4x1=4x2=3
啊,由于我只是初二,能力有限,所以我只知道猜的这种方法由x+y=10得y=10-x把y=10-x代入XY=24,得(10-x)x=24由(10-x)x=24得x²-10x+24=0用十字相乘
∵x^2+y^2=10,x+xy+y=7∴7-xy=x+y,且2xy≤x^2+y^2=10,∴(7-xy)^2=(x+y)^2=x^2+y^2+2xy=10+2xy,且xy≤5∴49-14xy+(xy
答:x²+y²=10xy=3,y=3/x代入上式得:x²+(3/x)²=10整理得:(x²)²-10*x²+9=0(x²
x^2+xy=12xy+y^2=4因式分解下,得x(x+y)=12.y(x+y)=4两个方程相加,得(x+y)^2=16所以x+y=±4当x+y=4时,代入x(x+y)=12.y(x+y)=4解得x=
令x+y=c,x-y=d,则方程组3(x+y)-a(x-y)=16,2(x+y)+b(x-y)=15等价3c-ad=16,2c+bd=15与条件所给方程组相同,所以c=x+y=7,d=x-y=1,解得
构造一元二次方程m^2-7m+10=0,则1/x,1/y是它的两个根很容易解得m=2,m=5因此x=1/2,y=1/5,或x=1/5,y=1/2
因为:xy=(x-10)(y+1)→xy=xy+x-10y-10①xy=(x-15)(y+3)→xy=+3x-15y-45②由①、②得:x=10y+10③x=5y+15④将③‐④得:10y+
∵x^2+y^2=10,x+xy+y=7∴7-xy=x+y,且2xy≤x^2+y^2=10,∴(7-xy)^2=(x+y)^2=x^2+y^2+2xy=10+2xy,且xy≤5∴49-14xy+(xy
由xy+x=16,得x=16/(y+1)代入xy-x=8,得16y/(y+1)-16/(y+1)=8=>16(y-1)/(y+1)=8=>(y-1)/(y+1)=1/2移项,通分得y-3/2(y+1)
令1/x=m,1/y=nm+n=5m*n=6m=3,n=2或m=2,n=3x=1/3,y=1/2或,x=1/2,y=1/3