100/(x-y)+100/(x+y)=100/y 怎么解
来源:百度知道 编辑:UC知道 时间:2024/09/12 17:06:49
怎么解出 x=(1+根号2)y ?
100/(x-y)+100/(x+y)=100/y
1/(x-y)+1/(x+y)=1/y
通分
(x+y+x-y)/(x+y)(x-y)=1/y
2x/(x+y)(x-y)=1/y
对角相乘
2xy=(x+y)(x-y)=x^2-y^2
x^2-2xy-y^2=0
把y看成已知
则x^2-2xy+y^2=2y^2
(x-y)^2=2y^2=(±√2y)^2
x-y=±√2y
x=y±√2y=(1±√2)y
约分得1/(X-Y)+1/(X+Y)=1/Y
同时乘以(X-Y)(X+Y)Y得
Y(X+Y)+Y(X-Y)=(X+Y)(X-Y)
化简 2XY=X^2-Y^2
遗式 X^2-2XY-Y^2+(Y^2-Y^2)=0
(X+Y)^2=2Y^2
X+Y=√(2Y^2)=±Y√2
X=±Y√2-Y=±Y(√2-1)
解:由100/(x-y)+100/(x+y)=100/y
得100/(x+y)+100/(x-y)=100/(x+y)(x-y)y
所以 100/y=100/(x+y)(x-y)y
所以(x+y)(x-y)=1
所以X=根号下(1+y^2)
你给的答案不对
100/(x-y)+100/(x+y)=100/y
先约分,得1/(x-y)+1/(x=y)=1/y
再通分得:2x/(x^2-y^2)=1/y
去分母得:x^2-2xy-y^2=0
将这个方程看成关于x的一元二次方程,运用求根公式,解得x=(1+根号2)y
此题得解
给我分,他分太多,这是我原创,没抄袭他!
100/(x-y)+100/(x+y)=100/y
200x/x^2-y^2=100/y
x^2-y^2=2xy
x^2-2xy+y^2-2y^2=0
(x-y)^2=2y^2
x-y=±√2y