这几道数学题咋做啊 1.已知x+y+z=0,则1/(y²+z²-x²)+1/(z²
来源:学生作业帮 编辑:作业帮 分类:数学作业 时间:2024/08/23 12:47:45
这几道数学题咋做啊
1.已知x+y+z=0,则1/(y²+z²-x²)+1/(z²+x²-y²)+1/(x²+y²-z²)= 2.若 3x-4y-z=0,2x+y-8z=0,则(x+y-z)/(x-y+z)= 3.已知a+b+c=0,且a,b,c均不为0,则a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=
1.已知x+y+z=0,则1/(y²+z²-x²)+1/(z²+x²-y²)+1/(x²+y²-z²)= 2.若 3x-4y-z=0,2x+y-8z=0,则(x+y-z)/(x-y+z)= 3.已知a+b+c=0,且a,b,c均不为0,则a(1/b+1/c)+b(1/c+1/a)+c(1/a+1/b)=
1.∵x+y+z=0
∴y²+z²-x²=y²+z²-(y+z)²= -2yz
同理z²+x²-y²= -2zx x²+y²-z²= -2xy
原式=1/-2yz+ 1/-2zx+ 1/ -2xy=(通分)-(x+y+z)/ -2xyz =o (分子x+y+z=0)
2.3x-4y-z=0 (1) 2x+y-8z=0(2)
(1)+(2)×4得 11x-33z=0 所以x=3z z=1/3 x
(1)×8 - (2)得 22x-33y=0 ∴2x=3y y=2/3x
带入原式
原式=(x+2/3x—1/3 x)/(x -2/3x+1/3 x)=2
3.原式=a(1/b+1/c+1/a)+b(1/c+1/a+1/b)+c(1/a+1/b+1/c)-3 (每项加一个1)
=(1/b+1/c+1/a)(a+b+c) -3 (合并同类项 将三项合并)
= -3
做完啊...详细吧.
∴y²+z²-x²=y²+z²-(y+z)²= -2yz
同理z²+x²-y²= -2zx x²+y²-z²= -2xy
原式=1/-2yz+ 1/-2zx+ 1/ -2xy=(通分)-(x+y+z)/ -2xyz =o (分子x+y+z=0)
2.3x-4y-z=0 (1) 2x+y-8z=0(2)
(1)+(2)×4得 11x-33z=0 所以x=3z z=1/3 x
(1)×8 - (2)得 22x-33y=0 ∴2x=3y y=2/3x
带入原式
原式=(x+2/3x—1/3 x)/(x -2/3x+1/3 x)=2
3.原式=a(1/b+1/c+1/a)+b(1/c+1/a+1/b)+c(1/a+1/b+1/c)-3 (每项加一个1)
=(1/b+1/c+1/a)(a+b+c) -3 (合并同类项 将三项合并)
= -3
做完啊...详细吧.
这几道数学题咋做啊 1.已知x+y+z=0,则1/(y²+z²-x²)+1/(z²
已知x+y+z=0,求(y²+z²-x²)分之1+(z²+x²-y
数学题已知xyz≠0,且x+y+z=1,x²+y²+z²=1,求1/x+1/y+1/z的值
一道根式数学题已知x>0,y>0,z>0,x,y,z满足2006x²=2007y²=2008z
已知x,y,z满足x+y+2z=1,x²+y²+6z+1.5=0,求x,y,z的值
已知x²+y²+z²-2x+4y-6z+14=0,求x+y-z/x+y+z的值
初一 下 数学 1、已知3x-4y-z=0,2x+y-8z=0,求 x²+y²-z²÷xy
已知丨x+1丨+(y+2)²+丨z-1丨=0,求(x+y)²-(y+z)²-(z+x)
已知实数x、y、z满足:2x+3y+z=1,则x²+y²+z²的最小值为
如题 已知x²+y²+z²+4x+4z+4y+1=0求x+y+z
代数式的恒等变形(1).已知x²+y²+z²-2x+4y-6z+14=0,求(x-y-z)
已知x-y+z=0,求(x²-y²-z²)-4y²z²等于多少?