Đáp án: $3m+n=57$
Giải thích các bước giải:
Ta có:
$\lim_{x\to3}\dfrac{\sqrt{6x-9}-\sqrt[3]{27x-54}}{(x-3)(x^2+3x-18)}$
$=\lim_{x\to3}\dfrac{\sqrt{6x-9}-\sqrt[3]{27x-54}}{(x-3)(x-3)(x+6)}$
$=\lim_{x\to3}\dfrac{(\sqrt{6x-9}-x)+(x-\sqrt[3]{27x-54})}{(x-3)^2(x+6)}$
$=\lim_{x\to3}\dfrac{\dfrac{6x-9-x^2}{\sqrt{6x-9}+x}+\dfrac{x^3-(27x-54)}{x^2+x\sqrt[3]{27x-54}+(\sqrt[3]{27x-54})^2}}{(x-3)^2(x+6)}$
$=\lim_{x\to3}\dfrac{\dfrac{-(x-3)^2}{\sqrt{6x-9}+x}+\dfrac{(x-3)^2(x+6)}{x^2+x\sqrt[3]{27x-54}+(\sqrt[3]{27x-54})^2}}{(x-3)^2(x+6)}$
$=\lim_{x\to3}\dfrac{\dfrac{-1}{\sqrt{6x-9}+x}+\dfrac{x+6}{x^2+x\sqrt[3]{27x-54}+(\sqrt[3]{27x-54})^2}}{x+6}$
$=\dfrac{\dfrac{-1}{\sqrt{6\cdot 3-9}+3}+\dfrac{3+6}{3^2+3\sqrt[3]{27\cdot 3-54}+(\sqrt[3]{27\cdot 3-54})^2}}{3+6}$
$=\dfrac1{54}$
$\to m=1, n=54$
$\to 3m+n=57$
