Giải thích các bước giải:
a.Ta có:
$y=3x+\dfrac{x^2+x+3}{x-2}$
$\to y=3x+\dfrac{(x^2-2x)+(3x-6)+9}{x-2}$
$\to y=3x+\dfrac{x(x-2)+3(x-2)+9}{x-2}$
$\to y=3x+x+3+\dfrac{9}{x-2}$
$\to y=4x+3+\dfrac{9}{x-2}$
$\to y=4(x-2)+11+\dfrac{9}{x-2}$
$\to y=4(x-2)+\dfrac{9}{x-2}+11$
$\to y\ge 2\sqrt{4(x-2)\cdot \dfrac{9}{x-2}}+11$
$\to y\ge 23$
Dấu = xảy ra khi $4(x-2)=\dfrac9{x-2}\to x=\dfrac72$ vì $x>2$
b.Ta có:
$y=\dfrac1x+\dfrac1{1-x}\ge\dfrac4{x+1-x}=4$
Dấu = xảy ra khi $\dfrac1x=\dfrac1{1-x}\to x=\dfrac12$
