(x+√x∧2-2010)(y+√y^2-2010)-2010=0,求x,y
来源:学生作业帮助网 编辑:作业帮 时间:2024/11/05 20:41:16
(x+√x∧2-2010)(y+√y^2-2010)-2010=0,求x,y
(x+√x∧2-2010)(y+√y^2-2010)-2010=0,求x,y
(x+√x∧2-2010)(y+√y^2-2010)-2010=0,求x,y
由题意知:X2-2010≥0 并且Y2-2010≥0
解得X≥√2010或X≤-√2010,Y≥√2010或Y≤-√2010
X=√2010,Y=√2010,原式=(√2010+0)(√2010+0)-2010=0成立
X=-√2010,Y=-√2010,原式=(-√2010+0)(-√2010+0)-2010=0成立
所以,本题有两组X=√2010,Y=√2010,X=-√2010,Y=-√2010
设x,y是有理数,且x,y满足等式x+2y-y√2=17+4√2,求(x+y)∧2010的值
(x+√x∧2-2010)(y+√y^2-2010)-2010=0,求x,y
先化简 再求值 (2x-y)(y+x)-(x-2y)(2y+x)-(-3y+x)^2其(√x+1)+y^2+4=-4y
x-y/x-x+y/y-(x+y)(x-y)/y² y/x=2
化简y(x+y)+(x+y)(x-y)-x^2
√X/Y+Y/X+2-√X/Y-√Y/X(注√X/Y+Y/X+2在一个根号2中)
计算:(x√y/x-2y√x/y)(x√y/x+2y√x/y)求你们帮帮忙啊
(x+y)(x-y)-(x-y)^2如题 (x+y)(x-y)-(x-y)^2
设X,Y是有理数,且X,Y满足X+2Y-(√2)Y=17+4√2试求(√X+Y)^2010的值.
【2(x-y)(x-y)(x-y)-8(x-y)(x-y)(x+y)+6y(x-y)(x-y)]/2(x-y)(x-y)
因式分解x(x+y)(x-y)-x(x+y)^2
若x>0,y>0,且√x(√x+√y)=3√y(√x+5√y),求2x+2√x*√y+3y/x-√x*√y+y
y^2-2(√x+1/√x)y+3
y=arcsin(x/√1+x^2),求y'
y=ln(x+√x^2+1),求y
RT,若(√2x-y)+y^2+4y+4=0,求[(x-y)^2+(x+y)(x-y)]÷2x的值
y=2x+√(1-x)
求一阶微分方程y'=(y√y)/(2x√y-x^2)的通解