\(\dfrac{6^6+6^3.3^3+3^6}{-73}\)

\(\dfrac{3^6.2^6+3^6.2^3+3^6.1}{-73}\)

\(\dfrac{3^6.\left(2^6+2^3+1\right)}{-73}\)

\(=\dfrac{3^6.73}{-73}=-729\)

\(N=\dfrac{2^6\cdot3^6+2^3\cdot3^3\cdot3^3+3^6}{-73}=\dfrac{3^6\left(2^6+2^3+1\right)}{-73}=\dfrac{3^6\cdot73}{-73}=-3^6=-729\)

1.

Điều kiện xác định của căn thức: \(\left[{}\begin{matrix}x\ge3\\x\le-3\end{matrix}\right.\)

\(\lim\limits_{x\rightarrow+\infty}\dfrac{\sqrt{x^2+1}-x}{\sqrt{x^2-9}-4}=\dfrac{1-1}{1}=0\Rightarrow y=0\) là 1 TCN

\(\lim\limits_{x\rightarrow-\infty}\dfrac{\sqrt{x^2+1}-x}{\sqrt{x^2-9}-4}=\dfrac{-1-1}{-1}=2\Rightarrow y=2\) là 1 TCN

\(\lim\limits_{x\rightarrow-5}\dfrac{\sqrt{x^2+1}-x}{\sqrt{x^2-9}-4}=\dfrac{\sqrt{26}+5}{0}=+\infty\Rightarrow x=-5\) là 1 TCĐ

\(\lim\limits_{x\rightarrow5}\dfrac{\sqrt{x^2+1}-x}{\sqrt{x^2-9}-4}=\dfrac{\sqrt{26}-5}{0}=+\infty\Rightarrow x=5\) là 1 TCĐ

Hàm có 4 tiệm cận

2.

Căn thức của hàm luôn xác định

Ta có:

\(\lim\limits_{x\rightarrow2}\dfrac{2x-1-\sqrt{x^2+x+3}}{x^2-5x+6}=\lim\limits_{x\rightarrow2}\dfrac{\left(2x-1\right)^2-\left(x^2+x+3\right)}{\left(x-2\right)\left(x-3\right)\left(2x-1+\sqrt{x^2+x+3}\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{\left(x-2\right)\left(3x+1\right)}{\left(x-2\right)\left(x-3\right)\left(2x-1+\sqrt{x^2+x+3}\right)}\)

\(=\lim\limits_{x\rightarrow2}\dfrac{3x+1}{\left(x-3\right)\left(2x-1+\sqrt{x^2+x+3}\right)}=\dfrac{-7}{6}\) hữu hạn

\(\Rightarrow x=2\) ko phải TCĐ

\(\lim\limits_{x\rightarrow3}\dfrac{2x-1-\sqrt{x^2+x+3}}{x^2-5x+6}=\dfrac{5-\sqrt{15}}{0}=+\infty\)

\(\Rightarrow x=3\) là tiệm cận đứng duy nhất

\(B=\dfrac{6^6+6^3.3^3+3^6}{-73}\)

\(B=\dfrac{46656+5832+729}{-73}\)

\(B=\dfrac{52488+729}{-73}\)

\(B=\dfrac{53217}{-73}=-729\)

\(\dfrac{6^6+6^3\cdot3^3+3^6}{-7^3}\)

\(=\dfrac{46656+46656\cdot27+729}{-343}\)

\(=\dfrac{46656+1259712+729}{-343}\)

\(=\dfrac{1306368+729}{-343}\)

\(=\dfrac{1307097}{-343}\)

\(\dfrac{6^6+6^3.3^3+3^6}{-73}=\dfrac{3^6.2^6+3^3.2^3.3^3+3^6}{-73}=\dfrac{3^6.64+3^6.8+3^6}{-73}=\dfrac{3^6.\left(64+8+1\right)}{-73}=-3^6=-729\)

Giải:

\(\dfrac{6^6+6^3.3^3+3^6}{-73}=\dfrac{2^6.3^6+2^3.3^3.3^3+3^6}{-73}\)

\(=\dfrac{2^6.3^6+2^3.3^6+3^6}{-73}=\dfrac{3^6\left(2^6+2^3+1\right)}{-73}\)

\(=\dfrac{3^6\left(64+8+1\right)}{-73}=\dfrac{3^6.73}{-73}=-3^6=-729.\)

Vậy.....

\(\dfrac{6^6+6^3.3+6^3}{-73}=\dfrac{2^6.3^6+2^3.3^3.3+2^6.3^6}{-73}=\dfrac{2^6.3^6+2^3.3^4+2^6.3^6}{-73}=\dfrac{2^3.3^4.\left(2^{3^{ }}.3^2+1+2^{3^{ }}.3^{2^{ }}\right)}{-73}=\dfrac{93960}{-73}\)

\(=\dfrac{3^6\cdot2^6+3^3\cdot3^3\cdot2^3+3^6}{-73}=\dfrac{3^6\left(2^6+2^3+1\right)}{-73}=-3^6=-729\)

\(\dfrac{6^6+6^3.3^3+3^6}{-73}=\dfrac{46656+\left(6.3\right)^3+729}{-73}=\dfrac{46656+18^3+729}{-73}=\dfrac{46656+5832+729}{-73}=\dfrac{53217}{-73}=-729\)

\(E=\dfrac{98:\left(\dfrac{4}{5}\cdot\dfrac{5}{4}\right)}{\dfrac{16}{25}-\dfrac{1}{25}}+\dfrac{\left(\dfrac{27}{25}-\dfrac{2}{25}\right)\cdot\dfrac{7}{4}}{\left(\dfrac{59}{9}-\dfrac{13}{4}\right)\cdot\dfrac{36}{17}}\\ E=\dfrac{98}{\dfrac{3}{5}}+\dfrac{\dfrac{7}{4}}{\dfrac{119}{36}\cdot\dfrac{36}{17}}\\ E=\dfrac{490}{3}+\dfrac{\dfrac{7}{4}}{7}=\dfrac{490}{3}+\dfrac{1}{4}=\dfrac{1963}{12}\)