Chủ đề này có 1 trả lời

[Rias Gremory]

Cho $x\neq 0$ . Tìm Min $T=\frac{(x^{2}+16\left | x \right |+48)(x^{2}+12\left | x \right |+27)}{x^{2}}$

[angleofdarkness]

Cho $x\neq 0$ . Tìm Min $T=\frac{(x^{2}+16\left | x \right |+48)(x^{2}+12\left | x \right |+27)}{x^{2}}$

 

 

Biến đổi: $T=\frac{(x^2+16\left | x \right |+48)(x^2+12\left | x \right |+27)}{x^2} \\ =\frac{(\left | x \right |+4)(\left | x \right |+12)(\left | x \right |+3)(\left | x \right |+9)}{\left | x \right | ^2} \\  =\frac{(\left | x \right | ^2+13\left | x \right |+36)(\left | x \right | ^2+15\left | x \right |+36)}{\left | x \right | ^2} \\  =\Big( \left | x \right |+\frac{36}{\left | x \right |}+13 \Big) \Big( \left | x \right |+\frac{36}{\left | x \right |}+15 \Big)$

 

Có $\left | x \right | + \frac{36}{\left | x \right |} \geq 12$ (Cauchy) nên $T \geq (12+13)(12+15)=675$

 

Dấu = khi $\left | x \right |=\frac{36}{\left | x \right |}$ tức x = -6; 6.