MaxWong

Answered in your previous post.

OK Finally back :)

\(\quad d^{-3} \text{ when } d= -2\\ =(-2)^{-3}\\ =\dfrac{1}{(-2)^3}\\ =\dfrac{1}{-8}\\ =-\dfrac{1}{8}\)

You are correct :)

\(\dfrac{\left(70^{262}0.3^{595}7^02^{60}\right)}{\left(21^{599\div \frac{1}{0.5^{634}}}0.2^{692}\right)}\\ =\dfrac{7^{262}\cdot 10^{262}\cdot \frac{3^{595}}{10^{595}}\cdot 1\cdot 2^{60}}{7^{\frac{599}{2^{634}}}\cdot 3^{\frac{599}{2^{634}}}\cdot\frac{2^{692}}{10^{692}}}\\ =\dfrac{7^{262}}{7^{\frac{599}{2^{634}}}}\cdot \dfrac{10^{262}\cdot10^{692}}{10^{595}}\cdot\dfrac{3^{595}}{3^{\frac{599}{2^{634}}}}\cdot\dfrac{2^{60}}{2^{692}}\\ =7^{262-\frac{599}{2^{634}}}\cdot10^{359}\cdot 3^{595-\frac{599}{2^{634}}}\div 2^{632}\)

It cannot be simplified further...... You made it up right?

Me can help!! Me is the smartest cookie in the world!!

\(a^5 b^{-6}\\ =\dfrac{a^5}{b^6}\)

\(\boxed{\color{olive}{a^{-x} \\=\dfrac{1}{a^x}}}\)

Oops.

Didn't realize that......

I factorized it without processing......

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.

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

#1 means

'First it is multiple questions. I will be very thankful if you seperate your questions. Secondly, our forum cannot do ALL your homework, but whole worksheet is a different one(I don't know what he is talking about). Lastly translate your worksheet next time.'

'首先是多个问题。我真的很感激，如果你分开的问题。第二，我们的论坛不能为你的所有家庭作业一个问题是一件事，但是整个工作表是另一个。最后翻译您的工作表下一次。'

#9 means

'You do not deceive someone who knows that you want us to do your homework so that you can be a lazy fool.'

#I Understand Japanese!!

And Sebast1ani5dumb don't be rude.......

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。 

me!! :)