If ~$x\neq -y$~, is ~$\frac{x-y}{x+y}> 1$~ ?
(1) x > 0
(2) y < 0
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.
注意,只知道分母不等于0,不知道分母的正负,所以不能判断啊!
分母的符号决定等式两边同乘分母后的大于小于符号
never multiply an inequality by variable (or expression with variable) unless you know the sign of variable (or expression with variable)
一定要过脑子……先同分
可不可以这样做呢:令x-y-(x+y)=x-y-x-y=-2y,由条件(2)y<0,那-2y>0,所以x-y>x+y,分子大于分母,不就式子大于1了吗?为啥条件2不对呢?
不要想当然地做题
这个题虽然是大于1还是小于1,但是这完全是在混淆视听
转化成(x-y)(x+y)=》x^2-y^2=》这个值大于0还是小于0==》如果极可能大于0 也可能小于0 那么大于还是小于1也就无法判断了
x小于y与x大于y结果不同
不等式两边不能同时乘以x+y,因为不知道x+y是正还是负;
(1)+(2),x+y可以是正值,也可以是负值