MaxWong

this equals x/(1-x) for all -1

oh lol I am careless... I did it like I am doing a tangent line question :P

Well, even textbooks have mistakes... :P

We can just factorize it :)

4a^2 - 10a + 6 = 0

2a^2 - 5a + 3 = 0

(2a - something)(a - something)=0

We don't know what it is yet, but think of 2 numbers that satisfies:

$-2x-y=-5\\ xy=3$

We can easily see that x = 1 and y = 3.

(2a - 3)(a - 1) = 0

a = 3/2 or a = 1 :)

It's not the simplest form ._.

$-6t \cdot 4m \cdot (-m)\\ =(-6\cdot 4)\cdot(t\cdot m\cdot(-m))\\ =-24(-tm^2)\\ =24tm^2$

Simplest form should be 24tm^2...

$\displaystyle\lim_{x\rightarrow 5}$

LaTeX code: \displaystyle\lim_{x\rightarrow 5}

$\sin(3t+\pi)+2\cos(3t)\\ =\sin(3t)\cos(\pi)+\cos(3t)\sin(\pi)+2\cos(3t)\\ =\sin(3t)\cdot(-1)+0+2\cos(3t)\\ =2\cos(3t)-\sin(3t)$

$x=\dfrac{\pi}{4}\\ \therefore y = \dfrac{1}{\sin\left(2\cdot\dfrac{\pi}{4}\right)}=1\\ \dfrac{dy}{dx}=(\csc(2x))'=-2\csc(2x)\cot(2x)\\ \dfrac{dy}{dx}|_{x=\frac{\pi}{4}}=-2(1)(0)=0\\ y-1=0(x-\dfrac{\pi}{4})\\ y=1\\ \therefore\text{The equation of the normal to }y=\dfrac{1}{\sin(2x)}\text{ when }x=\pi/4\text{ is }\boxed{y=1}$

I think this is the first time I do an application question :P

Gonna solve that all :P

1)

$x^2=x-1\\ x^2-x+1=0\\ \Delta = b^2 - 4ac = (-1)^2-4(1)(1)=-3\\ \therefore\text{No real solutions.}$

2)

$\sqrt{x+5}-2=6\\ \sqrt{x+5}=8\\ x+5=64\\ x=59$

3)

$\text{Note that:}\\\boxed{a^3+b^3=(a+b)(a^2-ab+b^2)}\\ (\sqrt[3]{3}+\sqrt[3]{2})(\sqrt[3]{4}-\sqrt[3]{6}+\sqrt[3]{9})\\ =(\sqrt[3]{3}+\sqrt[3]{2})\left((\sqrt[3]{2})^2-(\sqrt[3]{2})(\sqrt[3]{3})+(\sqrt[3]{3})^2\right)\\ =(\sqrt[3]{3})^3+(\sqrt[3]{2})^3\\ =3+2\\ =5$

4)

$x^4+3x^2+3=1\; |u=x^2\\ u^2+3u+2=0\\ (u+1)(u+2)=0\\ x^2+1=0\text{ or }x^2+2=0\\ x=i,x=-i,x=\sqrt{2}i,x=-\sqrt{2}i$

5) 

$\sqrt{10000}-\sqrt{81}+\sqrt{81}\\ =\sqrt{10000}\\ =100$

6)

$\dfrac{8}{x}=x+\dfrac{1}{x}\\ 8=x^2+1\\ x^2-7=0\\ x=\sqrt7\text{ or }x=-\sqrt7$

7) Not enough information...

Your answer is nice :O

Seems like I still need to learn how to show my points... :D