llllllllllllllllllllllllllllllllllllllllllllllllll (29 8:41:

byef2022-10-04 11:39:545条回答

llllllllllllllllllllllllllllllllllllllllllllllllll (29 8:41:46)
1.已知a≥1,比较M=√（a+1）-√a与N=√a-√（a-1）的大小.

2.比较2a2+3和4a的大小.

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SCDMA_BJ 共回答了18个问题 | 采纳率83.3%: 以下以（）^0.5代替开方,以（）^2代替平方.
第一道题：首先,M>0,N>0;
其次M^2-N^2=(a+1+a-2*(a*(a+1))^0.5)-(a+1+a-2*(a*(a-1))^0.5)
=2*(a*(a-1))^0.5-2*(a*(a+1))^0.5,显然小于零,故而M0
所以应该是2a^2+3>4a; 1年前

甘蔗渣共回答了53个问题 | 采纳率: M2=2a+1-2√a(a+1)
N2=2a+1-2√a(a+1)
M>0 N>0 2√a(a+1)>2√a(a+1)
M
1年前

1; 1年前

becky_2020 共回答了137个问题 | 采纳率: 有难度，怪不得没人回答。
第二问：2a2+3-4a=2(a-1)方+1，因为a≥1，所以（a-1）方大于等于0，所以2a2+3-4a=2(a-1)方+1大于等于1，所以2a2+3大于4a; 1年前

付体昌共回答了26个问题 | 采纳率: M>0,N>0;
其次M^2-N^2=(a+1+a-2*(a*(a+1))^0.5)-(a+1+a-2*(a*(a-1))^0.5)
=2*(a*(a-1))^0.5-2*(a*(a+1))^0.5,显然小于零，故而M第二道题：2*a^2+3-4*a=(2*a^2-4a+2)+1=2*(a-1)^2+1>0
所以应该是2a^2+3>4a; 1年前

hongfengling 共回答了227个问题 | 采纳率: 1. M-N=√（a+1）+√（a-1）>0所以 M>N
2.2a2+3-4a
=a2-2a+1+a2-2a+1+1
=2（a-1）2+1>0 所以2a2+3>4a; 1年前

llllllllllllllllllllllllllllllllllllllllllllllllll (18 11:40 llllllllllllllllllllllllllllllllllllllllllllllllll (18 11:40:56) 已知函数f（X）=ax+b,若f(0)=1,且f(x+1)=f(x+1)=f(x)+3,求f(x). 爱卫会1年前2: 1212ws1a2w 共回答了20个问题 | 采纳率95%

将f（0）带进去得到b=1 ,则f（x+1）=a（x+1）+1 ,f（x）+3=ax+4
又f(x+1)=f(x)+3,即a（x+1）+1 =ax+4 化简得到a=3
所以f(x）=3x+1

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sin(π-a）-cos（π+a）=√2/3
由诱导公式,sina+cosa=√2/3

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