MaxWong

Name the points. Let the point on the bottom left corner be A, that on the bottom right corner be B, and the one on top be C.

Let H be the point on AB such that CH is perpendicular to AB.

We calculate angle A in terms of n using Law of Cosines.

\(\cos A = \dfrac{(n + 1)^2 + (n + 2)^2 - (n + 3)^2}{2(n+1)(n+2)}\\ \cos A = \dfrac{n-2}{2(n+1)}\)

By this, AH = \(\dfrac{n - 2}2\).

Using Pythagorean theorem on triangle AHC,

\(\left(\dfrac n2 - 1\right)^2 + n^2 = (n + 1)^2\\ \left(\dfrac n2 - 1\right)^2 = 2n+1\\ \dfrac{n^2}4 - 3n = 0\\ n(n - 12) = 0\\ n = 12\)

The perimeter is (12 + 1) + (12 + 2) + (12 + 3) = 42.

It is the sum of geometric series.

It looks like this:

For -1 < r < 1, and an arbitrary a,

\(a+ar+ar^2+ar^3+\cdots+ar^n+\cdots=\dfrac{a}{1 - r}\)\(\)

Notice that \(\triangle RYX \sim \triangle XZQ\) by AA postulate.

Let x be the side length of the square. Then

\(ZQ = 12-x\\ RY=8-x\\ YX=x\\ XZ=x\)

By similar triangles,

\(\dfrac{RY}{YX} = \dfrac{XZ}{ZQ}\\ \dfrac{8-x}{x}=\dfrac{x}{12-x}\)

Solving this equation gives the required answer.

Ans: Side length = 24/5.

As an exercise, you can try to show that when PQ = a, PR = b, the side length of the square is \(\dfrac{ab}{a + b}\).

Np ;)

Noticing the repeating pattern is equal to x.

\(\log_a(x\color{red}\log_a(x\log_a(x\cdots))\color{black}) = x\\ \log_a(x\color{red}x\color{black}) = x\\ \log_a(x^2) = x\\ 2\log_a(x)=x\)

By a simple trial-and-error approach,

(a, x) = (2, 2) is a solution.

Therefore the required positive integer a is 2.

For all x < 0,

\(f(f(x)) = x\text{ (given in the problem)}\\ f\left(-\dfrac1{2x}\right) = x\)

But -1/(2x) > 0, so \(k\left(-\dfrac1{2x}\right) = x\)

Now, let t = -1/(2x). x = -1/(2t).

\(k(t) = -\dfrac1{2t}\\ \boxed{k(x) = -\dfrac1{2x}}\)

No. The 2 shouldn't be distributed to each fraction, and you did the "q" one wrong.

Q2)

By triangle inequality, \(|x| + |y| \geqslant |x + y|\)

By induction, we can generalise this to \(\displaystyle\sum_{i = 1}^n |x_i| \geqslant \left|\sum_{i = 1}^nx_i \right|\).

Therefore 

\(|a- b| + |b-c|+|c-d|+\cdots+|m-n| \geqslant |a-n|\\ |n-o|+\cdots+|y-z|+|z-a| \geqslant |n - a| = |a - n|\)

Adding these inequalities gives

\(|a-b|+|b-c|+|c-d|+\cdots+|m-n|+|n-o|+\cdots+|x-y|+|y-z|+|z-a| \geqslant 2|a-n|\\ 20\geqslant 2|a-n|\\ |a-n|\leqslant 10\)

The maximum value of |a - n| is 10.

Q1)

\(\dfrac{x + 7}{x^2 - x - 2} = \dfrac{A}{x - 2} + \dfrac B{x + 1}\\\dfrac{x + 7}{(x- 2)(x+1)} = \dfrac{A}{x - 2} + \dfrac B{x + 1}\\\)

Multiply by \((x - 2)(x + 1)\) on both sides.

\(x+7=A(x+1)+B(x-2)\\ x+7=(A+B)x+(A-2B)\)

Comparing coefficients, \(\begin{cases}A + B = 1\\A - 2B = 7\end{cases}\)

Solving gives \((A, B) = \left(3,-2\right)\)

You should know that 

\(p^a\cdot p^b = p^{a + b} \qquad (1)\\ \dfrac{p^a}{p^b} = p^{a - b}\qquad (2)\\ \left(p^a\right)^b = p^{ab}\qquad (3)\\ (pq)^a = p^a q^a\qquad (4)\)

Using these three formulae,

\(\quad \dfrac{\left(pq^3\right)^3\cdot \left(4p^2q\right)^2}{\left(2pq^2\right)^3}\\ \stackrel{(4)}{=} \dfrac{p^3 \left(q^3\right)^3\cdot 4^2 \cdot \left(p^2\right)^2\cdot q^2}{2^3 \cdot p^3 \cdot \left(q^2\right)^3}\\ \stackrel{(3)}{=} \dfrac{p^3q^9\cdot16\cdot p^4\cdot q^2}{8p^3q^6}\\ = 2\cdot \dfrac{p^3\cdot p^4}{p^3} \cdot \dfrac{q^9 \cdot q^2}{q^6}\)

Can you do the rest with identities (1) and (2)?