x^2 y^2-4x 1=0,求y x,y-x,x^2 y^2的最值问题

(2x-y)/(x+3y)=2所以原式=2(2x-y)/(x+3y)-3(x+4y)/(2x-y)=2(2x-y)/(x+3y)-3/[(2x-y)/(x+3y)]=2×2-3/2=5/2

原式=16x²-24xy+9y²-25x²-10xy-y²+9x²-6xy-8y²=-40xy=-40*1/3*6=-80

绝对值项恒非负,两绝对值项之和=0,两绝对值项分别=0x-2=0x=2y-4=0y=4y=x+21/(xy)+1/[(x+2)(y+2)]+1/[(x+4)(y+4)]+...+1/[(x+1994)

直线Y=KX(K≠0与双曲线Y=-4/X在区间(-∞,0)∪(0,+∞)上均为奇函数,所以：X₁=-X₂；Y₁=-Y₂由题知：X₁YS

kx=4/xx=±√(4/k)x1=√(4/k)y1=√(4/k)*kx2=-√(4/k)y2=-√(4/k)*k2x2y1=2*(-√(4/k))*√(4/k)*k=-8

(x*x+y*y)(x*x+y*y)-4x*x*y*y=(x^4-2x^2y^2+y^4)=(x^2-y^2)^2=0x^2=y^2x/y=±1(x*x+5xy+y*y)/(x*x+2xy+y*y)=

x1=2cosay1=4sina设那点是Q则A(2cosa+4sina,2cosa-4sina)x=2cosa+4sinay=2cosa-4sina所以x+y=4cosax-y=8sinasin&su

即(x-2y)²=0x-2y=0所以x=2y所以原式=(2x²+2xy-xy-y²)/(4x²-4xy+y²)=(2x²+xy-y²

2x^2-4mx+(2m^2-4m-3)=0y=x1^2+x2^2=(x1+x2)^2-2x1x2=4m^2-(2m^2-4m-3)=2m^2+4m+3m的取值就是判别式>=0即16m^2-8(2m^

已知x^2+4y^2-4x+4y+5=0求(（y^4-x^4）/(y-2x)(x+y))*((2x-y)/(xy-y^2))/((x^2+y^2)/y)的值答案：x²+4y²-4x

用维达定理(X2)+(X1)=(-a分之b)=(-1分之-2)=2(X1)*(X2)=(a分之c)=(-1分之m-3)所以(X2)+(X1)最小是2

答案是-4

这就是韦达定理对一元二次方程ax²+bx+c=0的两根为x1x2,则x1+x2=-a/bx1x2=a/c令y=a（x-x1）（x-x2）=0得x=x1x=x2即图像与x轴的交点也就是a（x-

1x1(1,+无穷)2.x^2-4x+5=(x-2)^2+1>=1值域为(-无穷,0]3.y=0(x>=0)值域恒为0y=x(x

y1=x1+2,y2=x2+2,(y1-y2)=(x1-x2)√[(x1—x2)^2+(y1—y2)^2]=√2|x1-x2|连立代入有:x^2-2x+3=x+2x^2-3x+1=0|x1-x2|=√

kx=4/xx²=4/kx1=2√k/k,x2=-2√k/ky1=2√k,y2=-2√k2x1y2-7x2y1=-8+28=20

首先考虑固定一点(x1,y1),求(x2,y2)使|x1-x2|+|y1-y2|最小.代入y2=6-2x2得|x1-x2|+|y1-6+2x2|=|x1-x2|+2|(y1/2-3)+x2|≥|x1-

∵|x+2y-1|+y²+4y+4=0∴|x+2y-1|+(y+2)²=0∴x=5,y=-2(2x-y)²-2(2x-y)(x+2y)+(x+2y)²=[(2x

f[(x1+x2)/2]=2^[(x1+x2)/2][f(x1)+f(x2)]/2=(2^x1+2^x2)/2由基本不等式(2^x1+2^x2)/2≧√[(2^x1)(2^x2)]=2^[(x1+x2

什么意思?再问：tangram_guid_1358503626031再答：1：2x+3y=7,3x-5y=1;x=2y=12:3x+5y=5,3x-4y=23;x=5y=-23:3x+5y=5,3x-