MaxWong here it is....

9-a 

look the polynomial 2x²+2x-4 has as roots 1 and -2 (if we replace x by 1 or -2 in the polynomial we'll get 0)

For that, we can be factorized, in an other word, the polynomial can be divided by (x-1) or (x+2)

2x²+2x-4=(x-1)(2x+4)

2x²+2x-4=(x+2)(2x-2)

\(\frac{1}{2}\frac{1}{(x+2)} + \frac{1}{x-1} - \frac{x+5}{2x²+2x-4} = \frac{1}{2}\frac{2x-2}{(x+2)(2x-2)} + \frac{2x+4}{(x-1)(2x+4)} - \frac{x+5}{2x²+2x-4} \)

\( = \frac{1}{2}\frac{2(x-1)}{(x+2)(2x-2)} + \frac{2x+4}{(x-1)(2x+4)} - \frac{x+5}{2x²+2x-4} \)

\( = \frac{2}{2}\frac{x-1}{(x+2)(2x-2)} + \frac{2x+4}{(x-1)(2x+4)} - \frac{x+5}{2x²+2x-4} \)

\( = \frac{x-1}{2x²+2x-4} + \frac{2x+4}{2x²+2x-4} - \frac{x+5}{2x²+2x-4} \)

\( = \frac{2x-2}{2x²+2x-4} = \frac{2(x-1)}{2x²+2x-4} = \frac{2(x-1)}{(x-1)(2x+4) }= \frac{2}{2(x+2) }\)

\(=\frac{1}{(x+2) }\)

首先是多個問題。我真的很感激，如果你分開的問題。第二，我們的論壇不能為你的所有家庭作業一個問題是一件事，但是整個工作表是另一個。最後翻譯您的工作表下一次。

Please don't do all of them Max.

He wants you to do all his homework!

I got question 13...9,10 and 12 are the ones don't know...

Number 9a)

I'm going to sleep now so these questions are solved. There's no need to answer anymore...

\(\displaystyle 2x^{2}+2x-2\neq2(x+2)(x-1)\).

i think that it is just showing you how to do the problem if it isnt im sorry i have no idea how to read japanese

Number 9b)

9(c)    .....in pieces...... evaluating the continued fraction from from bottom to top ......

First part  :    2  −  [x −1]  / x     =   [ 2x  −  (x  − 1)  ] / x   =  [ x + 1 ]  / x

1 /  ( [ x + 1]  / x )   =   x / [x + 1]

Second part :   1 +   x / [x + 1]   =   [ x + 1   + x ]  / [ x + 1]  =   [2x + 1]  / [x + 1]

Last  part  :      1  / ( [2x + 1]  / [x + 1] )   =

[x + 1] / [2x + 1]          [final answer ]

9(d) is very tricky......

x - 2 + 1/x is exactly in the form of a² - 2ab + b² , Therefore we can factor it as (a-b)² .

x - 2 + 1/x = \((\sqrt x - \dfrac{1}{\sqrt x})^2\)

x - 1/x is exactly in the form of a² - b²  . Therefore we can factor it as (a+b)(a-b)

x - 1/x = \((\sqrt x - \dfrac{1}{\sqrt x})(\sqrt x + \dfrac{1}{\sqrt x})\)

The original fraction became:

\(\dfrac{(\sqrt x - \frac{1}{\sqrt x})(\sqrt x - \frac{1}{\sqrt x})}{(\sqrt x + \frac{1}{\sqrt x})(\sqrt x - \frac{1}{\sqrt x})}\)

Did you see any same parts on the denominator and the numerator?

The hardest part is already solved by me.

And don't forget to multiply both the denominator and the numerator by \(\sqrt x\) because we don't want \(\dfrac{1}{\sqrt x}\) to appear in a fraction.

Final answer: \(\dfrac{x-1}{x+1}\)

Try to figure out why.

这个论坛终于有懂中文的人了 :D

我一直不知道马来西亚用简体。 香港是用繁体的， 不过不要紧， 我能看得懂简体中文。 

9 (e) is an easy one though but I will do it too.

\(\dfrac{x}{x-3}-\dfrac{7}{x+2}=\dfrac{x^2+17}{x^2-x-6}\)

x^2 + 17 and x^2 -x - 6 does not have common factors. So we start with making the left hand side have common denominator:

\(\dfrac{x(x+2)-7(x-3)}{(x-3)(x+2)}=\dfrac{x^2+17}{x^2 -x - 6}\)

And (x-3)(x+2) is magically x^2 - x - 6!!

\(x^2 + 2x - 7x +21 = x^2 + 17\)

And the 2 sides both magically only have same number of x^2 so that you don't need to do quadratic equation!!

\(-5x = 17 -21\)

All up to here it became very easy...... You will solve it from here. ;)

I am the smartest cookie in the world :D

Note something else on 9(d), Max.....multipling top/ bottom by x, we have

[x^2 − 2x + 1 ]  / [ x^2  − 1]  =

[ ( x − 1 ) (x  −  1)  ]  / [ (x + 1) ( x −  1) ] =

[x − 1 ] / [ x + 1]

10(a) needs to deal with ranges.......

\(\dfrac{x+3}{3}>\dfrac{x-1}{5}+\dfrac{2}{3}\\ 15 \cdot \dfrac{x+3}{3}>15\cdot\dfrac{x-1}{5}+15\cdot\dfrac{2}{3}\\ 5(x+3)>3(x-1)+10\\ 5x + 15 > 3x - 3 + 10\\ 5x + 15> 3x + 7\\ 5x +8 > 3x\\ x> - 4\)

\(\dfrac{x+1}{3}-\dfrac{x+2}{2}> - \dfrac{1}{6}\\ 6\cdot\dfrac{x+1}{3}-6\cdot\dfrac{x+2}{2}> - 6\cdot\dfrac{1}{6}\\ 2(x+1)-3(x+2)> -1\\ -x -4 > -1\\ x + 4 < 1\\ x < -3\)

x < -3 means -3 > x.

-3 > x and x > -4 means?

x is between what numbers?

something > x > something is the complete answer for this.