设实数x y满足x方加y方减4x加3等于0
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k=(y+2)/(x+1)所以k就是过点(-1,-2)的直线的斜率x,y满足x^2+y^2=1所以就是求过点(-1,-2)的直线和单位圆有公共点是斜率的取值范围显然相切时有最值y+2=kx+kkx-y
X²+4X+Y²-6Y+13=0X²+4X+4+Y²-6Y+9=0(X+2)²+(Y-3)²=0X=-2,Y=3
很高兴喂你解答!原式=4√[(x^2+xy+y^2)/(x-y)*1/2√[(x^2-xy+y^2)/(x+y)*3√(x^3+y^3)=6√[(x^2+xy+y^2)/(x-y)*√[(x^2-xy
再问:图片看不全再答: 能看到了吧?
把原方程看做是关于x的一元二次方程,整理得:x^2-4(y+1)x+(6y^2-4y+6)=0这个方程需有正整数根,因此其判别式:△=16(y+1)^2-4(6y^2-4y+6)=-8(y^2-6y+
2x²=x²+x²所以(x²-2xy+y²)+(x²+2x+1)=0(x-y)²+(x+1)²=0平方大于等于0,相加等
1/(x^4)+4/(y^4)xy=1,所以y=1/x即y^4=1/(x^4)代入原式,得到1/(x^4)+4(x^4)再由重要不等式,得到1/(x^4)+4(x^4)≥2(根号)1/(x^4)·4(
首先x3-y3=(x-y)(x2+xy+y2)其次x3-y3-3xy=1所以x3-y3=1+3xy=(x-y)(x2+xy+y2)x-y=(1+3xy)/(x2+xy+y2)=(1+3xy)/[(x-
x^2+y^2=4,设x=2cosa,y=2sina,则x^2-2√3xy-y^2=4(cosa)^2-8√3cosasina-4(sina)^2=4cos2a-4√3sin2a=8cos(2a+60
(x+y)/(x)=31+(y/x)=3得:y/x=2则:(x²+y²)/(x²-xy+y²)【分子分母同除以x²】=[1+(y/x)²]/
(X-1/2Y)方
x²+2xy-3y²=0(x+3y)(x-y)=0解得x=y或x=-3y又x,y为正数所以y=x;x=-3y舍(x²+xy+y²)/(x²-xy+y&
(x^2+xy+y^2)^2-4xy(x^2+y^2)=(x^2+y^2)^2+2xy(x^2+y^2)+x^2y^2-4xy(x^2+y^2)=(x^2+y^2)^2-2xy(x^2+y^2)+x^
第一题:设3^x=y,这里3^x表示3的x次方的意思那么原式变为y^2-8y-9=0,这个就简单了,解得y=-9或者y=1由于y=3^x,只能是3的整数次幂,所以y不能是-9,所以y=1那么3^x=1
x³-y³-x²y+xy²=(x³-y³)-(x²y-xy²)=(x-y)(x²+xy+y²)-xy
X方加2XY加2Y方减4Y加5=(X²+2XY+Y²)+(Y²-4Y+4)+1=(X+Y)²+(Y-2)²+1当Y=2,X=-2时,原式的值最小,为1
2X方-4XY+4Y方-6X+9=0(x^2-4xy+4y^2)+(x^2-6x+9)=0(x-2y)^2+(x-3)^2=0所以有x-2y=0,x-3=0即有x=3,y=x/2=3/2x*根号8Y=
原式=2x²-4xy+2y²-7y²=(√2x-√2y)²-7y²=(√2x-√2y+√7y)(√2x-√2y-√7y)
2x^2+xy-y^2-4x+5y-6双十字相乘法,2x^2+xy-y^2-4x+5y-62x-y2xy-3(2x-y+2)(x+y-3)或者待定系数法.有问题追问再问:谢谢谢谢谢谢谢你再答:不用谢