_____已知|ax-2y-4|√2x+4 互为相反数,且x与y互为相反数,求a的值._____ _____ _____若x.y满足y=√ 1-2x+√(x-1)²+√2x-1,则x+y的算术平方根是多少?_____ _____若√ a-1 +√ ab-2=0.求1 1 1 1_____+_______________+________
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![_____已知|ax-2y-4|√2x+4 互为相反数,且x与y互为相反数,求a的值._____ _____ _____若x.y满足y=√ 1-2x+√(x-1)²+√2x-1,则x+y的算术平方根是多少?_____ _____若√ a-1 +√ ab-2=0.求1 1 1 1_____+_______________+________](/uploads/image/z/1665854-62-4.jpg?t=_____%E5%B7%B2%E7%9F%A5%7Cax-2y-4%7C%E2%88%9A2x%2B4+%E4%BA%92%E4%B8%BA%E7%9B%B8%E5%8F%8D%E6%95%B0%2C%E4%B8%94x%E4%B8%8Ey%E4%BA%92%E4%B8%BA%E7%9B%B8%E5%8F%8D%E6%95%B0%2C%E6%B1%82a%E7%9A%84%E5%80%BC._____+_____+_____%E8%8B%A5x.y%E6%BB%A1%E8%B6%B3y%3D%E2%88%9A+1-2x%2B%E2%88%9A%28x-1%29%26sup2%3B%2B%E2%88%9A2x-1%2C%E5%88%99x%2By%E7%9A%84%E7%AE%97%E6%9C%AF%E5%B9%B3%E6%96%B9%E6%A0%B9%E6%98%AF%E5%A4%9A%E5%B0%91%3F_____+_____%E8%8B%A5%E2%88%9A+a-1+%2B%E2%88%9A+ab-2%3D0.%E6%B1%821+1+1+1_____%2B_______________%2B________)
_____已知|ax-2y-4|√2x+4 互为相反数,且x与y互为相反数,求a的值._____ _____ _____若x.y满足y=√ 1-2x+√(x-1)²+√2x-1,则x+y的算术平方根是多少?_____ _____若√ a-1 +√ ab-2=0.求1 1 1 1_____+_______________+________
_____
已知|ax-2y-4|√2x+4 互为相反数,且x与y互为相反数,求a的值.
_____ _____ _____
若x.y满足y=√ 1-2x+√(x-1)²+√2x-1,则x+y的算术平方根是多少?
_____ _____
若√ a-1 +√ ab-2=0.求
1 1 1 1
_____+_______________+_______________+.+____________________的
ab (a+1)(b+1) (a+2)(b+2) (a+2009)(b+2009)
值
_____已知|ax-2y-4|√2x+4 互为相反数,且x与y互为相反数,求a的值._____ _____ _____若x.y满足y=√ 1-2x+√(x-1)²+√2x-1,则x+y的算术平方根是多少?_____ _____若√ a-1 +√ ab-2=0.求1 1 1 1_____+_______________+________
首先要明确一点,根号内的数不是正数就是零,所以遇见√A+√B=0时,A和B都为零.(因为若A为正数,那么√A就是正数,要满足√A+√B=0,√B就必须为负数,而√B不可能为负数,所以A和B都是零)
1.∵(ax-2y-4)的绝对值与根号(2x+4)互为相反数
∴(ax-2y-4)的绝对值+根号(2x+4)=0
∵(ax-2y-4)的绝对值≥0,且根号(2x+4) ≥0
∴(ax-2y-4)的绝对值=0,根号(2x+4)=0
X=-2
又因为x与y互为相反数
∴y=2
∴a*-2-2*4=0
∴a=-4
2.y=√ 1-2x+√(x-1)²+√2x-1,
可以注意到√ 1-2x和√2x-1,
√2x-1=√ -(1-2x)
所以√ 1-2x和√2x-1互为相反数.
(相反数不是一正一负,就是零,而这两者不可能为一正一负,所以两者都为零)
所以1-2x=0,X=1/2
将X=1/2代入原式,可得Y=1/2
所以X+Y的算术平方根=1
3.
a-1=0,ab-2=0
a=1,ab=2
所以b=2 ,a=2
则原式=1/1*2+1/2*3+1/3*4+……+1/2010*2011
=1-1/2+1/2-1/3+1/3-1/4+……+1/2010-1/2011
=1-1/2011
=2010/2011
累死我了.