分式数学问题1.已知x+1/y=z+1/x=1,求y+1/z的值2.解方程:(x-4)/(x-5) - (x-5)/(x
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/10/06 08:47:37
分式数学问题
1.
已知x+1/y=z+1/x=1,求y+1/z的值
2.
解方程:(x-4)/(x-5) - (x-5)/(x-6) = (x-7)/(x-8) - (x-8)(x-9)
1.
已知x+1/y=z+1/x=1,求y+1/z的值
2.
解方程:(x-4)/(x-5) - (x-5)/(x-6) = (x-7)/(x-8) - (x-8)(x-9)
x + 1/y = 1 则:y = 1/(1-x)
z + 1/x = 1 则:z = 1- 1/x = (x-1)/x ,所以,1/z = x/(x-1) = 1+ 1/(x-1) = 1-1/(1-x)
所以 y + 1/z = 1
(x-4)/(x-5) - (x-5)/(x-6) = (x-7)/(x-8) - (x-8)(x-9)
即:1+1/(x-5) - 1 - 1/(x-6) = 1+1/(x-8) - 1 -1/(x-9)
1/(x-5) - 1/(x-6) = 1/(x-8) -1/(x-9)
-1/[(x-5)(x-6)] = -1/[(x-8)(x-9)]
(x-8)(x-9) = (x-5)(x-6)
x²-17x+72 = x²-11x+30
解得:x = 7
z + 1/x = 1 则:z = 1- 1/x = (x-1)/x ,所以,1/z = x/(x-1) = 1+ 1/(x-1) = 1-1/(1-x)
所以 y + 1/z = 1
(x-4)/(x-5) - (x-5)/(x-6) = (x-7)/(x-8) - (x-8)(x-9)
即:1+1/(x-5) - 1 - 1/(x-6) = 1+1/(x-8) - 1 -1/(x-9)
1/(x-5) - 1/(x-6) = 1/(x-8) -1/(x-9)
-1/[(x-5)(x-6)] = -1/[(x-8)(x-9)]
(x-8)(x-9) = (x-5)(x-6)
x²-17x+72 = x²-11x+30
解得:x = 7
一条分式数学题已知x y z满足x/x+y + y/z+x + z/x+y =1,求x²/x+y + y&su
已知x::y:z=3:4:5,(1)求x+y分之z的值;(2)若x+y+z=6,求x,y,z.
已知X,Y,Z是三个非零数,且X:Y:Z=4:3:1,求分式2X-3Y+5Z除于X+y+Z的值
已知X:Y=1:3,Y:Z=5:4 (1)求X:Y:Z (2)如果X+Y+Z=96,求X,Y,Z的值 3元1次方程
已知方程2x+2y+3z=10且x+3y+5z=1,求2z+y-x的值
已知X、Y、Z满足2X-Y+4Z=8(1);X-2Y-Z=7(2)求X-Y-Z的值
1 已知想x,y,z满足2x-y+4z=8,x-2y-z=7,求x-y+z的值
已知x/2=y/3=z/-1≠0,求分式2x+6y-3z/x+y+2z的值
1.已知x,y,z满足2│x-y│+(根号2y-z)+z平方-z+(1/4)=0,求x,y,z值.
已知x,y,z满足x/(y+z)+y/(z+x)+z/(x+y)=1,求代数式x2/(y+z)+y2/(x+z)+z2/
已知实数x,y,z满足x/(y+z)+y/(z+x)+z/(x+y)=1,求x2/(y+z)+y2/(z+x)+z2/(
已知{x:y:z=1:2:3,x+y+z=12,求x、y、z的值