giải BPT: $3x^3+6x^2+5x+4 \le (x^2+3x+4)\sqrt{x+2}$
giải BPT: $3x^3+6x^2+5x+4 \le (x^2+3x+4)\sqrt{x+2}$
Started By chieckhantiennu, 13-02-2016 - 22:43
#1
Posted 13-02-2016 - 22:43
#2
Posted 16-02-2016 - 16:48
$(x^2+3x+4)(x+1-\sqrt{x+2})+2x(x^2+x-1)\leq 0 \Leftrightarrow (x^2+x-1)(\frac{x^2+3x+4}{x+1+\sqrt{x+2}}+2x)\leq 0 \Leftrightarrow (x^2+x-1)(3x^2+5x+4+2x\sqrt{x+2}) \leq 0 \Leftrightarrow (x^2+x-1)[2(x+1)^2+(x+\sqrt{x+2})^2]\leq 0 \Rightarrow x^2+x-1\leq 0 \Leftrightarrow ......$
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