Toán chứng minh x.( x mũ 2 – 4x + 5). (2x – x mũ 2 + 3) < 0 04/08/2021 By Savannah chứng minh x.( x mũ 2 – 4x + 5). (2x – x mũ 2 + 3) < 0
Giải thích các bước giải: $x(x^2-4x+5)(2x-x^2+3)$ $=-x((x-2)^2+1)(x^2-2x-3)$ $=-x((x-2)^2+1)((x-1)^2-4)$ $=-x((x-2)^2+1)(x-1-2)(x-1+2)$ $=-x((x-2)^2+1)(x-3)(x+1)$ $=\dfrac{1}{2}((x-2)^2+1)(-2x.(x-3)(x+1))$ Ta có: $-2x.(x-3)(x+1)\le \dfrac{(-2x+x-3+x+1)^3}{27}=\dfrac{-8}{27}<0$ (cosi 3 số) Mà $(x-2)^2+1>0\quad \forall x$ $\rightarrow (-2x(x-3)(x+1)).((x-2)^2+1)<0$ $\rightarrow x(x^2-4x+5)(2x-x^2+3)<0\rightarrow đpcm$ Trả lời
Giải thích các bước giải:
$x(x^2-4x+5)(2x-x^2+3)$
$=-x((x-2)^2+1)(x^2-2x-3)$
$=-x((x-2)^2+1)((x-1)^2-4)$
$=-x((x-2)^2+1)(x-1-2)(x-1+2)$
$=-x((x-2)^2+1)(x-3)(x+1)$
$=\dfrac{1}{2}((x-2)^2+1)(-2x.(x-3)(x+1))$
Ta có: $-2x.(x-3)(x+1)\le \dfrac{(-2x+x-3+x+1)^3}{27}=\dfrac{-8}{27}<0$ (cosi 3 số)
Mà $(x-2)^2+1>0\quad \forall x$
$\rightarrow (-2x(x-3)(x+1)).((x-2)^2+1)<0$
$\rightarrow x(x^2-4x+5)(2x-x^2+3)<0\rightarrow đpcm$