Chứng minh đẳng thức sai

E

endinovodich12

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kakashi_hatake

2x42 \le x \le 4
1x2+1<1<2 lex\dfrac{1}{\sqrt{x-2}+1} <1<2 \ le x
14x+1>0\dfrac{1}{\sqrt{4-x}+1}>0
-> 1x2+114x+1(x+12)<0\dfrac{1}{\sqrt{x-2}+1} - \dfrac{1}{\sqrt{4-x}+1}-(x+\dfrac{1}{2}) <0
đpcm
 
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