Question

fomd the lcm x^2+10x,x^2-10x

Answer 1

Okay, let's factor both!
the first is x(x+10), and the second is x(x-10)
Then we need to have both the x+10 and the x-10 each once; also we need one factor of x, because that is the highest power that appears in either factorization.
Hence it is x(x-10)(x+10) = x(x^2 - 100) = x^3 - 100x

Answer 2

yeah

Answer 3

i need ti solve 2/1-x=3/1+x

Answer 4

Okay, so we will multiply by (1-x) and (1+x) to clear the fractions. Then we have
\[{2 \over 1-x} (1+x)(1-x) = {3 \over 1+x} (1+x)(1-x)\]Cancelling and then simplifying gives
\[2(1+x) = 3(1-x)\]\[2 + 2x = 3 - 3x\]\[5x = 1\]\[x = {1 \over 5}\]And we can quickly verify that this is valid, by seeing that it does not make either denominator 0 (if it did we would have a problem)