求助几道七年级数学难题!!!

2024-11-15 15:07:58
推荐回答(2个)
回答1:

1,√(a-1993)有意义,则a大于1992
原式可化为:a-1992+√(a-1993)=a
不难得到,a-1992^2=1993

2, 原式4m^2+9n^2-4m+6n+2=0可化为
4m^2-4m+1+9n^2+6n+1=0
即(2m-1)^2+(3n+1)^2=0
(2m-1)^2大于等于0
3n+1)^2大于等于0
要使得两者相加等于0,那么他们各自都为0
即m=1/2,n=1/3
带入(18n^2+24n+4)/(4m^2+4m-1),
(18n^2+24n+4)/(4m^2+4m-1)=3.5

3.设此数为A,循环节是B,有n个,A=1.2345…B=0.345
n可以是任意大的整数
1*10^n*A=B*10^n+A
(10^n-1)*A=B*10^n
A=B*10^n/(10^n-1)
于是就化成了两个整数之比

4. 证明:根号5是无理数。
证明:可以用‘反证法’来证明:
假设√5是有理数,那么它一定可以用一个最简的既约分数a/b表示,
√5=a/b
两边同时平方,得
5=a^2/b^2
得:a^2=5b^2,
由此可见,a是5的倍数,于是设a=5k,则有
(5k)^2=5b^2
25k^2=5b^2
得:b^2=5k^2,
也就是说b也是5的倍数,
综上,a、b都是5的倍数,那么a/b就不是最简分数了,与假设矛盾,
因此,根号5不是有理数,必定是无理数。

5. √(a^2+b^2)=√{(a+b)^2-2ab}
=2^2-2*3
=-2
(估计是你的题目打错了,但是做法基本上就是这样了~~~)

回答2:

1,√(a-1993)有意义,则a大于1992
原式可化为:a-1992+√(a-1993)=a
不难得到,a-1992^2=1993 2, 原式4m^2+9n^2-4m+6n+2=0可化为
4m^2-4m+1+9n^2+6n+1=0
即(2m-1)^2+(3n+1)^2=0
(2m-1)^2大于等于0
3n+1)^2大于等于0
要使得两者相加等于0,那么他们各自都为0
即m=1/2,n=1/3
带入(18n^2+24n+4)/(4m^2+4m-1),
(18n^2+24n+4)/(4m^2+4m-1)=3.5 . 证明:根号5是无理数。
证明:可以用‘反证法’来证明:
假设√5是有理数,那么它一定可以用一个最简的既约分数a/b表示,
√5=a/b
两边同时平方,得
5=a^2/b^2
得:a^2=5b^2,
由此可见,a是5的倍数,于是设a=5k,则有
(5k)^2=5b^2
25k^2=5b^2
得:b^2=5k^2,
也就是说b也是5的倍数,
综上,a、b都是5的倍数,那么a/b就不是最简分数了,与假设矛盾,
因此,根号5不是有理数,必定是无理数。a+b=√(2^2),ab=3,a^2+b^2+2ab=4,得到a^2+b^2=-2,是不可能的事情,你题目是不是错了。

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