乙题: 解:(1)因为反比例函数的图象经过点有.········································································——青夏教育精英家教网—

乙题: 解:(1)因为反比例函数的图象经过点有.················································································································ 2分．····················································································································· 3分所以反比例函数的解析式为.············································································· 4分 (2)当为一.三象限角平分线与反比例函数图像的交点时. 线段最短．············································································································ 5分将代入.解得.即.．····················· 6分 .··········································································································· 7分则.··········································································································· 8分又为反比例函数图像上的任意两点. 由图象特点知.线段无最大值.即．·················································· 9分【查看更多】

题目列表(包括答案和解析)

（2012•峨边县模拟）甲题：关于x的一元二次方程x²+（2m+1）x+m²=0的实数解为x₁和x₂．
（1）求m的取值范围．
（2）当

=0时，求m的值．
乙题：如图，在△ABC中，AC=AB，以AB为直径的半⊙O交AC于点E交BC于点D，连AD、BE．
（1）求证：△BEC∽△ADC；
（2）BC²=2AB•CE．

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小明学习了“第八章幂的运算”后做这样一道题：若（2x-3）^x+3=1，求x的值，他解出来的结果为x=1，老师说小明考虑问题不全面，聪明的你能帮助小明解决这个问题吗？
小明解答过程如下：
解：因为1的任何次幂为1，所以2x-3=1，x=2．且2+3=5
故（2x-3）^x+3=（2×2-3）²⁺³=1⁵=1，所以x=2
你的解答是：

解：①∵1的任何次幂为1，所以2x-3=1，x=2．且2+3=5，
∴（2x-3）^x+3=（2×2-3）²⁺³=1⁵=1，
∴x=2；
②∵-1的任何偶次幂也都是1，
∴2x-3=-1，且x+3为偶数，
∴x=1，
当x=1时，x+3=4是偶数，
∴x=1；
③∵任何不是0的数的0次幂也是1，
∴x+3=0，2x-3≠0，
解的：x=-3，
综上：x=2或3或1．

．

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甲、乙两人解方程组

ax+by=2

cx-7y=8

，甲正确地解得

x=3

y=-2

，乙因为把c看错，误认为d，解得

x=-2

y=2

，求a、b、c、d．