已知,y=(x^2+2x+1/x^2-1)÷(x+1/x^2-x)-x+1,是说明不论x为何值时,y的值不变
来源:学生作业帮助网 编辑:作业帮 时间:2024/06/30 20:38:40
![已知,y=(x^2+2x+1/x^2-1)÷(x+1/x^2-x)-x+1,是说明不论x为何值时,y的值不变](/uploads/image/z/14404153-49-3.jpg?t=%E5%B7%B2%E7%9F%A5%2Cy%3D%28x%5E2%2B2x%2B1%2Fx%5E2-1%29%C3%B7%28x%2B1%2Fx%5E2-x%29-x%2B1%2C%E6%98%AF%E8%AF%B4%E6%98%8E%E4%B8%8D%E8%AE%BAx%E4%B8%BA%E4%BD%95%E5%80%BC%E6%97%B6%2Cy%E7%9A%84%E5%80%BC%E4%B8%8D%E5%8F%98)
x){}Ku*m5*⌴*
]C5`
M] Sٌ/oy6ɎvU<ٱީOL/g/ 3+mkk@,==Z!P"g5*@ʵ+
P{dR03_6Y:O{7]F 1>6`PP @POӃ@) -
已知,y=(x^2+2x+1/x^2-1)÷(x+1/x^2-x)-x+1,是说明不论x为何值时,y的值不变
已知,y=(x^2+2x+1/x^2-1)÷(x+1/x^2-x)-x+1,是说明不论x为何值时,y的值不变
已知,y=(x^2+2x+1/x^2-1)÷(x+1/x^2-x)-x+1,是说明不论x为何值时,y的值不变
因为
y=[(x^2+2x+1)/(x^2-1)]÷[(x+1)/(x^2-x)]-x+1
=[(x+1)^2/(x+1)(x-1)]*[x(x-1)/(x+1)]-x+1
=x-x+1=1,
所以
不论x为何值时,y的值都是:1,即y的值不变.
y=[(x^2+2x+1)/(x^2-1)]÷[(x+1)/(x^2-x)]-x+1
=[(x+1)^2/(x+1)(x-1)]*[x(x-1)/(x+1)]-x+1
=x-x+1=1,
不论x为何值时,y的值都是:1,即y的值不变。
已知x+y=a,2x-y=-2a,求[(x/y-y/x)/(x+y)-x(1/x-1/y)]/[(x+1)/y]的值
已知x,y满足约束条件:x-y+1>=0,x+y-2>=0,x
mathematica软件,已知y[x],Y=f1(x),X(x)=f2(x),如何plot Y[X]?y[x_] := x + 9.81/2*x*xk=1X[x] = x - k*y'[x]/(1 + (y'[x])^2)^0.5Y[x] = y[x] + k/(1 + (y'[x])^2)^0.5Plot[{y[x],Y[X],},{x,0,2}]
已知x(x-1)-(x^2-y)=-3.求x^2+y^2-2xy
已知x^2+y^2-8x+12y+52=0 ,求1/2x-1/x-y(x-y/2x-x^2+y^2)
已知x^2+y^2-8x+12y+52=0 ,求1/2x-1/x-y(x-y/2x-x^2+y^2)
已知实数x,y满足x-ay-1>=0,2x+y>=0,x
已知x-y=1,求[(x+2y)^2+(2x+y)(x-4y)-3(x+y)(x-y)]除以y的值大神们帮帮忙
已知x(x+1)-(x²+y)=3求2/x²+y²-xy
已知:【(x²+y²)-(x-y)²+2y(x-y)】/4y=1,求4x/4x²-y²-1/2x+y
已知:2x+y=6,x-3y=1 求7y(x-3y)(x-3y)-2(3y-x)(3y-x)
已知x-y=1,求[(x+2y)²+(2x+y)(x-4y)-3(x+y)(x-y)]÷y的值
1.(x-2y)(x+2y)-(x平方+y)(x平方-y) 2.(2x-y)(y+2x)-(2y+x)(2y-x)3.已知:x+y=6,x平方-y平方=48,求x,y的值4.x平方-2x分之x+2 -- x平方-4x+4分之x-1
已知x=2+根号3,y=2-根号3,计算代数式(x+y/x-y-x-y/x+y)乘以(1/x^2-1/y^2)
已知x/y=1/2,求2x/(x²-2xy+y² )* (x²-y²/x+y)+2y/x-y
已知x/y=1/2,求2x/(x²-2xy+y² )* (x²-y²/x+y)+2y/x-y
已知y = |2x-6| + |x-1| - 4|x+1| ,求y已知y = |2x-6| + |x-1| - 4|x+1| ,求y的最大值
已知 x-2y=-2 求(2y-x)平方-2x+4y-1