二次方程组2x y=14 ,-3x 2y=21,则x和y等于多少
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3x^2-y^2=8(1)x^2+xy+y^2=4(2)将2*(2)有2x^2+2xy+2y^2=8,与(1)连等2x^2+2xy+2y^2=3x^2-y^2,移项有x^2-2xy-3y^2=0用十字
即(x+y)²=1(x-3y)²=4x+y=±1,x-3y=±2所以x+y=1,x-3y=2x+y=1,x-3y=-2x+y=-1,x-3y=2x+y=-1,x-3y=-2
x^2-2xy-3y^2=0(x-y)^2-4y^2=0(x-y)^2=(2y)^2x-y=±2yx-y=2y或x-y=-2yx1=3yx2=-y将x1,x2分别代入②,得:(3y)^2-y*3y+y
xy/(2x+y)-xy/(2x-y)=3xy/(2x-y)+xy/(2x+y)=4;一、两公式对加得2xy/(2x+y)=7xy=(2x+y)*7/2二、把xy代入原公式求解就行了:(2x+y)*7
∵x-3y=0∴x=3y∴(x²+xy+2y²)÷(x²-xy+y²)=[(3y)²+(3y)y+2y²]÷[(3y)²-(3y)
x^2+xy+y^2+x+5y=0,-----------(1)x+2y=0,x=-2y.代入方程(1)整理后得:y(3y-3)=0,y1=0,x1=0.y2=1,x2=-2.
mx+4y=83x+2y=6得:x=4/(6-m)y=(14-3m)/(6-m)x的二次方=y的二次方[4/(6-m)]^2=[(14-3m)/(6-m)]^2m=10/3或6(舍去)
由y-x=1得y=x+1代入到2x^2-xy-2=0中得x²-x-2=0所以x=-1或x=2因此当x=-1时y=0当x=2时y=3所以{x=-1,y=0{x=2,y=3
x+y=-3x=-3-y带入xy=2(-3-y)y=2-y^2-3y-2=0y^2+3y+2=0(y+1)(y+2)=0y1=-1y2=-2x1=-2x2=-1
你的解法不对着...由(2)得:(x-2y)y=1(3)化解得:x=1/y+2y(4)把(4)代入(1)得到:(1/y+2y)方+(1/y+2y)-12=0化解得到:1/(y方)+6(y方)=7(5)
仔细观察题目后会发现,等式的右边是不为零的整数,这样无法判断XYZ的值所以用加减消元法,将这几个等式变形,变为右边=0的另外几个等式,然后再因式分解.这样为从新列出关XYZ的三元一次方程组吧.然后解出
x=2,y=0.5
x2+y2+2xy=9(x+y)²=9x+y=3或-3(x-y)2-3x+3y+2=0(x-y)²-3(x-y)+2=0(x-y-1)(x-y-2)=0x-y=1或2所以组成4个方
因为X^2-Y^2=(X+Y)(X-Y)x^2+y^2+2xy=(X+Y)^2这个题目可以因式分解成1,(X+Y)(X-Y)=32,(X+Y)^2+(X+Y)=12设X+Y=AX-Y=B那么方程变成A
(x+1)²+|y-1|=0两个非负数的和为0,这两个非负数都为0x+1=0,y-1=0x=-1,y=12(xy-5xy²)-(3xy²-xy)=2xy-10xy
因为是二元二次方程组,所以m-1和n+1=0或者2且不能都为0所以m-i=2,n+1=2,m+n=4m-1=0,n+1=2m+n=2m-1=2,n+1=0m+n=2故m+n=2或者4再问:是不是应该还
由(x-2)(y+3)=xy(x-5)(y+2)=xy得3x-2y-6=02x-5y-10=0上面方程化为6x-4y-12=06x-15y-30=0两式相减得11y=-18,y=-18/11,代入任意
x^2-xy+y^2=3,2x^2-xy-y^2=55x^2-5xy+5y^2=156x^2-3xy-3y^2=15两式子相减x^2+2xy-8y^2=0(x+4y)(x-2y)=0x=2y或者x=-
x+y=11x=11-y将x=11-y代入xy=24得:(11-y)*y=2411y-y^2=24y^2-11y+24=0(y-8)(y-3)=0y=8或y=3则x=3或x=8