$\begin{array}{l}B=\dfrac{3x}{2}+\dfrac{2}{x-1}=\dfrac{3x-3+3}{2}+\dfrac{2}{x-1}=\dfrac{3x-3}{2}+\dfrac{2}{x-1}+\dfrac{3}{2}\\\text{Áp dụng BĐT Cô-si cho hai số không âm, ta có:}\\\dfrac{3x-3}{2}+\dfrac{2}{x-1}\ge 2\sqrt{\dfrac{3x-3}{2} . \dfrac{2}{x-1}}=2\sqrt{\dfrac{3(x-1)}{2} . \dfrac{2}{x-1}}=2\sqrt3\\\to B\ge 2\sqrt3+\dfrac{3}{2}=\dfrac{3+4\sqrt3}{2}\\\text{Dấu = xảy ra}\\\leftrightarrow \dfrac{3x-3}{2}=\dfrac{2}{x-1}\\\to 3(x-1)^2=4\\\to (x-1)^2=\dfrac{4}{3}\\\to \left[\begin{array}{l}x-1=\dfrac{2}{\sqrt3}\\x-1=-\dfrac{2}{\sqrt3}\end{array}\right.\\\to \left[\begin{array}{l}x=\dfrac{3+2\sqrt3}{3}(\text{thõa mãn})\\x=\dfrac{3-2\sqrt3}{3}(\text{loại})\end{array}\right.\\\text{Vậy $Min_B=\dfrac{3+4\sqrt3}{2}$ đạt được khi $x=\dfrac{3+2\sqrt3}{3}$}\end{array}$
Đáp án:
$\min B = 2\sqrt3 +\dfrac32\Leftrightarrow x = 1 +\dfrac{2}{\sqrt3}$
Giải thích các bước giải:
$B =\dfrac{3x}{2} +\dfrac{2}{x-1}\qquad (x > 1)$
$\to B =\dfrac{3(x-1) + 3}{2} +\dfrac{2}{x-1}$
$\to B =\dfrac{3(x-1)}{2} +\dfrac{2}{x-1} +\dfrac{3}{2}$
Áp dụng bất đẳng thức $AM-GM$ ta được:
$\dfrac{3(x-1)}{2} +\dfrac{2}{x-1}\geq 2\sqrt{\dfrac{3(x-1)}{2}\cdot\dfrac{2}{x-1}}$
$\to \dfrac{3(x-1)}{2} +\dfrac{2}{x-1}\geq 2\sqrt3$
$\to \dfrac{3(x-1)}{2} +\dfrac{2}{x-1} +\dfrac{3}{2}\geq 2\sqrt3 +\dfrac32$
$\to B \geq 2\sqrt3 +\dfrac32$
Dấu $=$ xảy ra $\Leftrightarrow \dfrac{3(x-1)}{2} =\dfrac{2}{x-1}$
$\Leftrightarrow (x-1)^2 =\dfrac43$
$\Leftrightarrow x = 1 +\dfrac{2}{\sqrt3}\quad (Do\,\,x > 1)$
Vậy $\min B = 2\sqrt3 +\dfrac32\Leftrightarrow x = 1 +\dfrac{2}{\sqrt3}$
Đáp án:
$\underline{\text{Bạn tham khảo !!}}$
Giải thích các bước giải:
$\begin{array}{l}B=\dfrac{3x}{2}+\dfrac{2}{x-1}=\dfrac{3x-3+3}{2}+\dfrac{2}{x-1}=\dfrac{3x-3}{2}+\dfrac{2}{x-1}+\dfrac{3}{2}\\\text{Áp dụng BĐT Cô-si cho hai số không âm, ta có:}\\\dfrac{3x-3}{2}+\dfrac{2}{x-1}\ge 2\sqrt{\dfrac{3x-3}{2} . \dfrac{2}{x-1}}=2\sqrt{\dfrac{3(x-1)}{2} . \dfrac{2}{x-1}}=2\sqrt3\\\to B\ge 2\sqrt3+\dfrac{3}{2}=\dfrac{3+4\sqrt3}{2}\\\text{Dấu = xảy ra}\\\leftrightarrow \dfrac{3x-3}{2}=\dfrac{2}{x-1}\\\to 3(x-1)^2=4\\\to (x-1)^2=\dfrac{4}{3}\\\to \left[\begin{array}{l}x-1=\dfrac{2}{\sqrt3}\\x-1=-\dfrac{2}{\sqrt3}\end{array}\right.\\\to \left[\begin{array}{l}x=\dfrac{3+2\sqrt3}{3}(\text{thõa mãn})\\x=\dfrac{3-2\sqrt3}{3}(\text{loại})\end{array}\right.\\\text{Vậy $Min_B=\dfrac{3+4\sqrt3}{2}$ đạt được khi $x=\dfrac{3+2\sqrt3}{3}$}\end{array}$