MaxWong

Ummmm. I thought of binomial theorem.

$(A+A^b)^b\\ = A^b+\displaystyle C^{b}_{1}A^{2b-1}+C^{b}_{2}A^{3b-2}+C^{b}_{3}A^{4b-3}.......$

$\begin{array}{rll}x+3(x-6)&=&2x-3\\ x+3x-18&=&2x-3\\ 4x-18&=&2x-3\\ 4x-2x&=&18-3\\ 2x&=&15\\ x&=&\dfrac{15}{2}\end{array}$

$2x^2\times 2\\ = 2\times x^2 \times 2\\ = 2\times 2 \times x^2\\ =4\times x^2\\ =4x^2$

Same as Guest #1's answer

$-65+70\\ = +70-65\text{ Yes, you can exchange the two.}\\ = 5$

$\begin{array}{rlllllll}&7&6&8&.&5&0&0\\-&&1&3&.&4&0&2 \\-&-&-&-&-&-&-&-\\ &7&5&5&.&0&9&8\end{array}$

$2 \times 10^0$

$\displaystyle\lim_{x\rightarrow 27}f(x)$

That means f(27) usually.

$\dfrac{x}{x^2-4}+\dfrac{1}{x+2}=3\\ x+1(x-2)=3(x^2-4)\\ x+x-2=3x^2-12\\ 3x^2-2x-10=0\\ x=\dfrac{-(-2)\pm\sqrt{(-2)^2-4(3)(-10)}}{2(3)}\\ \;\;=\dfrac{2\pm\sqrt{124}}{6}\\ \;\;=\dfrac{2\pm2\sqrt{31}}{6}\\ \;\;=\dfrac{1\pm\sqrt{31}}{3}\\ \therefore x=\dfrac{1+\sqrt{31}}{3}\text{ OR }x=\dfrac{1-\sqrt{31}}{3}$

(A+A^b)^b

=A(1+A^(b-1))^b

That could not be further simplified.

3^1 mod 5 = 3

3^2 mod 5 = 4

3^3 mod 5 = 2

3^4 mod 5 = 1

3^5 mod 5 = 3

3^6 mod 5 = 4

You see the sequence right?

Because 200 mod 4 = 0. 3^200 mod 5 = 3^0 mod 5 = 1. Guest #2 is correct.