已知圆(x-3)² (y 4)²=25 则在该圆内的点
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![已知圆(x-3)² (y 4)²=25 则在该圆内的点](/uploads/image/f/4254946-34-6.jpg?t=%E5%B7%B2%E7%9F%A5%E5%9C%86%28x-3%29%C2%B2+%28y+4%29%C2%B2%3D25+%E5%88%99%E5%9C%A8%E8%AF%A5%E5%9C%86%E5%86%85%E7%9A%84%E7%82%B9)
x2y+xy2=xy(x+y)=66,设xy=m,x+y=n,由xy+x+y=17,得到m+n=17,由xy(x+y)=66,得到mn=66,∴m=6,n=11或m=11,n=6(舍去),∴xy=m=
原式分解因式得x^3y^3(2x-y)=(xy)^3(2x-y)=8/3.(x^3表示x的3次方)
把原式两边对x求导得:x^2+12y^3*dy/dx+1+2dy/dx=0合并同类项移项得:dy/dx=-(1+2x)/(12y^3+2)
原式=(x4-xy3)+(y4-x3y)+(3xy2-3x2y)=x(x3-y3)+y(y3-x3)+3xy(y-x)=(x3-y3)(x-y)-3xy(x-y)=(x-y)(x3-y3-3xy)=(
∵x+y=6,xy=4,∴(1)x2+y2=(x+y)2-2xy,=62-2×4,=28;(2)(x-y)2=x2+y2-2xy,=28-2×4,=20;(3)x4+y4=(x2+y2)2-2x2y2
(x+y+z)²-(x²+y²+z²)=2(xy+yz+zx)=-1,xy+yz+zx=-1/2x3+y3+z3=3xyz+(x+y+z)(x²+y&
(x+y+z)^2=[(x+y)+z]^2=(x^2+2xy+y^2)+z^2+2zx+2zy=x^2+y^2+z^2+2xy+2xz+2yz=x^2+y^2+z^2+2(xy+xz+yz)=0x+y
∵x+y=a∴x2+y2+2xy=a2又∵x2+y2=b2∴2xy=a2-b2x4+y4=(x2+y2)2-2x2y2=(x2+y2)2-(2xy)22=b4−(a2−b2)22=-12a4+a2b2
将各个点坐标代入反比例函数中,可求得:Y1=½,Y2=1, Y3=-1, Y4=-½∴Y2>Y1>Y4>Y3
(1)因为y=1/xk>0y的值随x值的增大而减小因为2>1>-1>-4所以y4>y3>y2>y1(2)1因为y=1/xk>0x1>x2y的值随x值的增大而减小所以y2>y1
x2+y2=(x+y)2-2xy=14x3+y3=(x2+y2)×(x+y)-xy2-yx2=14×4-xy(x+y)=52……剩下的就是这么个算法,手机党,求个最佳哈
方程ax^2+bx+c=0,判断这个方程有没有实数根,有几个实数根,就要用ΔΔ=b^2-4ac若Δ<0,则方程没有实数根Δ=0,则方程有两个相等实数根,也即只有一个实数根Δ>0,则方程有两个不相等的实
(x2+z2)(x2+y2)(y2+z2)=(x+y)2-2xy×(x+z)2-2xz×(y+z)2-2yz--之后不清楚了
y^2+3y-1=0把y=0代入-1=0,不成立所以y不等于0两边除以yy+3-1/y=0y-1/y=-3平方y^2-2+1/y^2=9y^2+1/y^2=11平方y^4+2+1/y^4=121y^4
根据题意得:x3−y4=33x+2y=78,整理得:4x−3y=36①3x+2y=78②,①×2+②×3得:17x=306,解得:x=18,将x=18代入①得:y=12,则方程组的解为x=18y=12
根据题意,得x+y2=3x−2yx+y2=10+6x+y4,整理得x−y=0(1)4x−y=−10(2),由(1)-(2),并解得x=-103(3).把(3)代入(1),解得y=-103,所以原方程组
x4-xy3-x3y-3x2y+3xy2+y4=(x4-xy3)+(y4-x3y)+(3xy2-3x2y)=x(x3-y3)+y(y3-x3)+3xy(y-x)=(x3-y3)(x-y)-3xy(x-
(x^4-y^4)÷(x^2+y^2)/(x+y)=(x^2-y^2)(x^2+y^2)÷(x^2+y^2)/(x+y)=(x^2-y^2)(x^2+y^2)*(x+y)/(x^2+y^2)=(x^2
因为x²+4y²+x²y²-6xy+1=0(x²-4xy+4y²)+(x²y²-2xy+1)=0(x-2y)²