I Have just started an engineering course at uni, please help with these maths questions!!!? 10points?

1. Factor first: (2x-1)(x-1)(x+6)/[(2x-1)(2x+1)(x-1)] the factors (2x+1)(x-1) divide into top and bottom to give (x+6)/(2x+1) 2 assuming this is (x-3)/x ÷( x²-9)/x then =(x-3)/x X(x/(x²-9) =(x-3)/x X x/[(x-3)(x+3) = 1/(x+3) 3. √3 - 3-.√(3(9)) + 9=.√3+6-3.√3=6-2√3 4.(a+b)(c+d)=ac+ad+bc+bd so (3+ √3)(1 - 1/√3)=3X1 +3X-1/√3 +√3X1- √3X1/√3 =3- √3+√3-1=2 5. Assuming 1/ (√2 - 1) =(√2+1)/[(√2-1)(√2+1)]=(√2+1)/(2-1)=√2+1 6.Multiply top and bottom by √3, =5√3/(2X3)=5√3/6 7. Multiply top and bottom be (2√3+1), 11(2√3+1)/[(2√3-1)(2√3+1)] =11(2√3+1)/(12 - 1)=11(2√3+1)/11=(2√3+1) I am fed up now! Note that (a+√b)(a-√b)=a^2-b

I'm a bit tired but here's three for starters ! 1) becomes (2x-1)(x-3)(x-2)/((2x-1)((2x+1)(x-1) = (x-3)/(2x+1)(x-1) 2) 3) becomes r3 - 3 - 9r3 + 9 = 6 -8r3 4) becomes 3(1 + 1/r3)(1 - 1/r3) = 3 [1^2 - (1/r3)^2] = 3 x (1 - 1/3) = 2 In 1) they are seeing if you can factorise the quadratic on the top line and the "difference of 2 squares" on the bottom line.. both make an (2x-1) which can then cancel out. In 4) they are seeing if you recognise that by taking a common factor of three out, you end up with a "difference of two squares" ie 1^2 - (1/3)^2. Very hard questions for beginning engineering...I wonder what the point is ? Unfortunately much of top end maths exams have become a battle against tricks rather than testing and recognising core knowledge.

First, you can factorize (x² + 5x - 6) Δ = 5² - 4(1 * - 6) = 25 + 24 = 49 = 7² x1 = (- 5 - 7) / (2 * 1) = - 12/2 = - 6 x2 = (- 5 + 7) / (2 * 1) = 2/2 = 1 → (x² + 5x - 6) = 1 * [x - (- 6)] * [x - (1)] = (x + 6)(x - 1) Second: (4x² - 1) = (2x)² - 1² → it looks like: a² - b² = (a + b)(a - b) → (4x² - 1) = (2x)² - 1² = (2x + 1)(2x - 1) Recall: [(2x - 1)(x² + 5x - 6)] / [(4x² - 1)(x - 1)] = [(2x - 1)(x + 6)(x - 1)] / [(2x + 1)(2x - 1)(x - 1)] = [(2x - 1)(x + 6)(x - 1)] / [(2x + 1)(2x - 1)(x - 1)] = (x + 6)/(2x + 1) = [(x - 3)/x] / [(x² - 9)/2x] → you know that: (x² - 9) = (x² - 3²) = (x + 3)(x - 3) = [(x - 3)/x] / [(x + 3)(x - 3)/2x] = [1/x] / [(x + 3)/2x] → to divide by (a/b) it's equal to multiply the reverse (b/a) = [1/x] * [2x/(x + 3)] = 2x/[x * (x + 3)] = 2/(x + 3) = 3.√3 - √9 - √27 + √81 = 3.√3 - 3 - √27 + 9 = 3.√3 - [√27] + 6 = 3.√3 - [√(9 * 3)] + 6 = 3.√3 - [(√9) * (√3)] + 6 = 3.√3 - [3 * √3] + 6 = 6 = (3 + √3)[1 - (1/√3)] = (3 + √3)[1 - (1 * √3)/(√3 * √3)] = (3 + √3)[1 - (√3)/3] = 3 - √3 + √3 - (√3)²/3 = 3 - √3 + √3 - (3/3) = 3 - 1 = 2 = 1/(√2 - 1) = 1(√2 + 1) / [(√2 - 1)(√2 + 1)] = (√2 + 1) / [(√2 - 1)(√2 + 1)] → you recognize: (a + b)(a - b) = a² - b² = (√2 + 1) / [(√2)² - 1²] = (√2 + 1) / [2 - 1] = √2 + 1 = 5/(2√3) = (5/2) * (1/√3) = (5/2) * [(1 * √3)/(√3 * √3)] = (5/2) * [(√3)/(√3)²] = (5/2) * [(√3)/3] = [5 * (√3)]/(2 * 3) = (5√3)/6 = (5/6).√3 = 11/(2√3 - 1) = [11 * (2√3 + 1)]/[(2√3 - 1)(2√3 + 1)] = [11(2√3 + 1)]/[(2√3)² - 1²] = [11(2√3 + 1)]/[(4 * 3) - 1] = [11(2√3 + 1)]/11 = 2√3 + 1 = (1 - √3/3) / (1 + √3/3) → you recognize: (a + b)(a - b) = a² - b² = (1)² - (√3/3)² = 1 - (3/9) = 1 - (1/3) = (3/3) - (1/3) = 2/3 = [(2√3 + 1) / (2√3 - 1)] + [(2√3 - 1) / (2√3 + 1)] To be more convinient: → Let: a = 2√3 = [(a + 1) / (a - 1)] + [(a - 1) / (a + 1)] → the common denominator is: (a - 1)(a + 1) = a² - 1 = [(a + 1)(a + 1) / (a² - 1)] + [(a - 1)(a - 1) / (a² - 1)] = [(a + 1)(a + 1) + (a - 1)(a - 1)] / (a² - 1) = (a² + 2a + 1 + a² - 2a + 1) / (a² - 1) = 2(a² + 1) / (a² - 1) Recall: a = 2√3 a² + 1 = (2√3)² + 1 = (4 * 3) + 1 = 13 a² - 1 = (2√3)² - 1 = (4 * 3) - 1 = 11 Recall: = 2(a² + 1) / (a² - 1) = 2(13) / (11) = 26/11

I'm a bit tired but here's three for starters ! 1) becomes (2x-1)(x-3)(x-2)/((2x-1)((2x+1)(x-1) = (x-3)/(2x+1)(x-1) 2) 3) becomes r3 - 3 - 9r3 + 9 = 6 -8r3 4) becomes 3(1 + 1/r3)(1 - 1/r3) = 3 [1^2 - (1/r3)^2] = 3 x (1 - 1/3) = 2 In 1) they are seeing if you can factorise the quadratic on the top line and the "difference of 2 squares" on the bottom line.. both make an (2x-1) which can then cancel out. In 4) they are seeing if you recognise that by taking a common factor of three out, you end up with a "difference of two squares" ie 1^2 - (1/3)^2. Very hard questions for beginning engineering...I wonder what the point is ? Unfortunately much of top end maths exams have become a battle against tricks rather than testing and recognising core knowledge.

1. Factor first: (2x-1)(x-1)(x+6)/[(2x-1)(2x+1)(x-1)] the factors (2x+1)(x-1) divide into top and bottom to give (x+6)/(2x+1) 2 assuming this is (x-3)/x ÷( x²-9)/x then =(x-3)/x X(x/(x²-9) =(x-3)/x X x/[(x-3)(x+3) = 1/(x+3) 3. √3 - 3-.√(3(9)) + 9=.√3+6-3.√3=6-2√3 4.(a+b)(c+d)=ac+ad+bc+bd so (3+ √3)(1 - 1/√3)=3X1 +3X-1/√3 +√3X1- √3X1/√3 =3- √3+√3-1=2 5. Assuming 1/ (√2 - 1) =(√2+1)/[(√2-1)(√2+1)]=(√2+1)/(2-1)=√2+1 6.Multiply top and bottom by √3, =5√3/(2X3)=5√3/6 7. Multiply top and bottom be (2√3+1), 11(2√3+1)/[(2√3-1)(2√3+1)] =11(2√3+1)/(12 - 1)=11(2√3+1)/11=(2√3+1) I am fed up now! Note that (a+√b)(a-√b)=a^2-b