If n is positive, is ~$\sqrt{n}$~ > 100 ?
(1) ~$\sqrt{n-1}$~ > 99
(2) ~$\sqrt{n+1}$~ > 101
Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are NOT sufficient.
题目分析:
要判断~$\sqrt{n}$~>100,即判断是否~$n>100^2$~。
(1)~$\sqrt{n-1}>99$~ ,两边平方得~$n-1>99^2$~,即~$n>(99+1)^2-99*2$~,即~$n>100^2-198$~ 不满足~$n>100^2$~,无法判断题目。
(2)~$\sqrt{n+1}>101$~ ,两边平方得~$n+1>101^2$~,即~$n>(101+1)(101-1)$~,满足~$n>100^2$~,即满足~$\sqrt{n}>100$~。
因此,本题答案为(B),仅条件2可以解题,条件1不行。
99^2 +1 = (99+1)^2-99*2 < 100^2
将三个式子都进行平方,之后在进行比较
这题怎么想呢?
解题错误