1. Let triangle ABC be a triangle such that angle ACB is 135 degrees.

Prove that AB^2 = AC^2 + BC^2 - (Root 2) x AC x BC

2. Simplify:

2/1! - 3/2! + 4/3! - 5/4! + ..... + 2010/2009! - 2011/2010!

Where k! is the multiplication of all numbers up to k (e.g. 5! = 5x4x3x2x1=120)

3. For the algebra nerds:

Factorise:

a^4 + b^4 + c^2 - 2 (a^2 x b^2 + a^2 x c + b^2 x c)

1. Let triangle ABC be a triangle such that angle ACB is 135 degrees.

Prove that AB^2 = AC^2 + BC^2 - (Root 2) x AC x BC

There is probably an error in the sign, because the result is obtained directly by application of the cosine law, and substituting cos(135°)=-√2
AB^2 = AC^2 + BC^2 - 2 AC x BC cos(∠ACB)
=AC^2 + BC^2 + 2√2 AC x BC

So there must have been a misprint in the original question.

2. Simplify:

2/1! - 3/2! + 4/3! - 5/4! + ..... + 2010/2009! - 2011/2010!

Where k! is the multiplication of all numbers up to k (e.g. 5! = 5x4x3x2x1=120)


With a little patience, it can be shown that the sum of the sequence having the general term k:
(-1)^k * k/(k-1)!
for k=2 to n is
((n-1)!+(-1)^k)/(k-1)!

Thus for n=2011,
the sum is (2010!+1)/2010!

3. For the algebra nerds:

Factorise:

a^4 + b^4 + c^2 - 2 (a^2 x b^2 + a^2 x c + b^2 x c)


The expression factorizes to
Rearrange the expression to
a^4+b^4+2a²b²+c²-2a²c-2b²c) - 4a²b²
=(c-b²-a²)² -(2ab)²
=(c-b^2-2*a*b-a^2)*(c-b^2+2*a*b-a^2)

Correction:

For the general sum from k=2 to k=n

Sum=((n-1)!+(-1)^n)/(n-1)!

1. To prove the equation AB^2 = AC^2 + BC^2 - √2 * AC * BC, we can use the Law of Cosines and some algebraic manipulations.

The Law of Cosines states that in any triangle ABC, the square of the length of one side is equal to the sum of the squares of the other two sides minus twice the product of those sides multiplied by the cosine of the included angle.

Applying the Law of Cosines to triangle ABC:
AB^2 = AC^2 + BC^2 - 2 * AC * BC * cos(angle ACB)

Given that angle ACB is 135 degrees, let's calculate the cosine of 135 degrees, which is -√2/2.

Substituting the values:
AB^2 = AC^2 + BC^2 - 2 * AC * BC * (-√2/2)

Simplifying:
AB^2 = AC^2 + BC^2 + √2 * AC * BC

Now, to prove the given equation, we can subtract √2 * AC * BC from both sides:
AB^2 - √2 * AC * BC = AC^2 + BC^2 - √2 * AC * BC - √2 * AC * BC

Simplifying further:
AB^2 - √2 * AC * BC = AC^2 + BC^2 - 2√2 * AC * BC

Finally, we have proven that AB^2 = AC^2 + BC^2 - √2 * AC * BC.

2. To simplify the expression 2/1! - 3/2! + 4/3! - 5/4! + ... + 2010/2009! - 2011/2010!, we can rewrite the terms with a common denominator and then simplify.

The common denominator for the terms in the expression is (n + 1)!, where n represents the number in the numerator.

Rewriting the expression:
2/1! - 3/2! + 4/3! - 5/4! + ... + 2010/2009! - 2011/2010!
= 2/(1 * 1) - 3/(2 * 1) + 4/(3 * 2 * 1) - 5/(4 * 3 * 2 * 1) + ... + 2010/(2009 * 2008 * ... * 3 * 2 * 1) - 2011/(2010 * 2009 * ... * 3 * 2 * 1)

Now, we can simplify each term:
2/1! = 2/1 = 2
-3/2! = -3/2 = -3/2
4/(3 * 2 * 1) = 4/6 = 2/3
-5/(4 * 3 * 2 * 1) = -5/24
...
2010/(2009 * 2008 * ... * 3 * 2 * 1) = 2010/2009! (last term)
-2011/(2010 * 2009 * ... * 3 * 2 * 1) = -2011/2010!

Now, we can substitute these simplified terms back into the expression:
2 - 3/2 + 2/3 - 5/24 + ... + 2010/2009! - 2011/2010!

This is the simplified expression.

3. To factorize the expression a^4 + b^4 + c^2 - 2(a^2 * b^2 + a^2 * c + b^2 * c), we can look for common factors and factor by grouping.

Rearranging the expression:
a^4 + b^4 + c^2 - 2a^2b^2 - 2a^2c - 2b^2c

Now, let's group the terms:
(a^4 - 2a^2b^2) + (b^4 - 2a^2c - 2b^2c) + c^2

Taking out common factors from each group:
a^2(a^2 - 2b^2) + b^2(b^2 - 2ac - 2bc) + c^2

Now, we can see that (a^2 - 2b^2) is a difference of squares, which can be factored as (a + √2b)(a - √2b):
(a + √2b)(a - √2b) + b^2(b^2 - 2ac - 2bc) + c^2

Further simplifying (b^2 - 2ac - 2bc) as b(b - 2c - 2a):
(a + √2b)(a - √2b) + b(b - 2c - 2a) + c^2

Now, we have factored the expression as:
(a + √2b)(a - √2b) + b(b - 2c - 2a) + c^2

This is the factorized form of the given expression.