Correct! Divide both sides by a^2 b^2 to make it look nicer =)?
Correct! Although it's best it you simplify and factorise your final line to make it more clear, like^2 + 4 (2k+1) - 1\\= 4k^2 + 4k + 1 + 8k + 4 - 1\\= 4k^2 + 4k + 8k + 4\\$Since everything are multiples of 4$\\\therefore n^2 + 4n - 1 $ is divisible by 4 if n is an odd integer$)
Another question from my Yr 10 yearly
(hint: start by putting everything into LHS and prove that expression to be greater than or equal to 0)
Yep! Good work - b^4 (a-b) \geq 0 \\ (a^4-b^4)(a-b)\geq 0\\ (a^2+b^2)(a^2-b^2)(a-b)\geq 0 (a^2+b^2)(a+b)(a-b)(a-b) \geq 0 \\ (a^2+b^2)(a+b)(a-b)^2 \geq 0 \\$ Since a and b are positive, and all squares are positive (or equal to 0)\\ so $a^2 + b^2$, $a+b$, $(a-b)^2$ are positive (or equal to 0) $\\ \therefore a^5-a^4b - ab^4+b^5\geq 0 $ and $ a^5-a^4b\geq ab^4-b^5 $is true.$)
thanks for the adviceYep! Good work
It might be a good idea to mention that
I know it's pretty obvious, but its good to acknowledge the question's criteria for a and b somewhere
Also, don't forget about this questionsimplify
(a-c)} + \frac{1}{(b-a)(b-c)} + \frac{1}{(c-a)(c-b)})
Here's an interesting intimidating (yet very easy) question on algebraic factorisation.
^2 (x-y\sqrt{d})^2 $can be written as the difference of two squares.$ )
does anyone have any good "simplifying expressions" questions? they're my favourite
lol the actual answer to this isn't within the year 10 course.Find da value of this
New to latex.... I tried XDAnother maths yearly question!
Next time, just copy the code from the editor and paste it here with {tex} and {/tex} around it, replacing the {}s with []s
