2008x-2007y=2009且2007x-2008y=2006则y和x于

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怀旧地小青年儿共回答了4个问题 | 采纳率: 2008x-2007y=2009
+ + +
2007x-2008y=2006
4015x - 4015y =4015
x - y =1 x=y+1
2008x-2007y=2009
- - -
2007x-2008y=2006
x+y =3 x=2 y=1; 1年前

1+tanα／1-tanα=2006则1／cosα+tan2α=? wangpeng19861年前1: 妍欢儿共回答了16个问题 | 采纳率87.5%

(1+tanα)／(1-tanα)=2006
1+tanα=2006(1-tanα)
tanα=2005/2007
cos2α=(1-tan^2α)/(1+tan^2α)
=(1-2005^2/2007^2)/(1+2005^2/2007^2)
tan2α=2tanα/(1-tan^2α)
=(2*2005/2007)/(1-2005^2/2007^2)
1／cos2α+tan2α
=1/[(1-2005^2/2007^2)/(1+2005^2/2007^2)]+(2*2005/2007)/(1-2005^2/2007^2)
=(1+2005^2/2007^2)/(1-2005^2/2007^2)+(2*2005/2007)/(1-2005^2/2007^2)
=(1+2005^2/2007^2+2*2005/2007)/(1-2005^2/2007^2)
=(1+2005/2007)^2/[(1-2005/2007)(1+2005/2007)]
=(1+2005/2007)/(1-2005/2007)
=(4012/2007)/(2/2007)
=4012/2
=2006
1／cos2α+tan2α
=1/[(1-tan^2α)/(1+tan^2α)]+2tanα/(1-tan^2α)
=(1+tan^2α)/(1-tan^2α)+2tanα/(1-tan^2α)
=(1+tan^2α+2tanα)/(1-tan^2α)
=(1+tanα)^2/(1-tan^2α)
=(1+tanα)^2/[(1-tanα)(1+tanα)]
=(1+tanα)/(1-tanα)
=(1+tanα)/(1-tanα)
=2006

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一道有点难度的题,尽快解决已知X=2005A+2004 Y=2005A+2005 Z=2005A+2006则X^2+Y^ 一道有点难度的题,尽快解决 已知X=2005A+2004 Y=2005A+2005 Z=2005A+2006 则X^2+Y^2+Z^2-XY-YZ-XZ的值是? 这个题是擂台赛的.回答的好追加分,谢谢 感觉回忆V1年前1: apperever 共回答了18个问题 | 采纳率88.9%

X=2005A+2004
Y=2005A+2005
Z=2005A+2006
x-y=-1
y-z=-1
x-z=-2
X^2+Y^2+Z^2-XY-YZ-XZ
=(x^2-2xy+y^2+x^2-2xz+z^2+y^2-2yz+z^2)/2
=[(x-y)^2+(x-z)^2+(y-z)^2]/2
=(1+1+4)/2
=3

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