Pls solve cus im not smartimage given

Mathematics

1 answer:

otez555 [7]2 years ago

3 0

From the factoring, it is possible to find the following answers:

Letter A: $\frac{x\left(x+2\right)}{x-3}$

Letter B: $-\frac{2x^5}{\left(x^2+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)}$

Letter C: $x^2+y^2$

<h3>Factoring</h3>

In math, factoring or factorization is used to write an algebraic expression in factors. There are some rules for factorization. One of them is a factor out a common term for example: x²-x= x(x-1), where x is a common term.

Letter a

Here you should factor the given expression.

$\frac{x^3-4x}{x^2-5x+6} \\ \\ \frac{x\left(x+2\right)\left(x-2\right)}{x^2-5x+6}\\ \\ \frac{x\left(x+2\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}\\ \\ \frac{x\left(x+2\right)}{x-3}\\ \\ Then,\\ \\ \frac{x^3-4x}{x^2-5x+6} =\frac{x\left(x+2\right)}{x-3}$

Letter b

Firstly, you should replace the variables A, B and C for the given expressions.

$\frac{1}{1-x+x^2}-\frac{1}{1+x+x^2}-\frac{2x}{1+x^2}$

After that, you should find the least common multiple.

$\frac{\left(x^2+1\right)\left(x^2+x+1\right)}{\left(x^2+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)}-\frac{\left(x^2+1\right)\left(x^2-x+1\right)}{\left(x^2+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)}-\frac{2x\left(x^2-x+1\right)\left(x^2+x+1\right)}{\left(x^2+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)}$

Finally, you can simplify the expression

$\frac{\left(x^2+1\right)\left(x^2+x+1\right)-\left(x^2+1\right)\left(x^2-x+1\right)-2x\left(x^2-x+1\right)\\ \\ \left(x^2+x+1\right)}{\left(x^2+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)}\\ \\ \frac{-2x^5}{\left(x^2+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)}= -\frac{2x^5}{\left(x^2+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)}$

Letter c

Firstly, you should replace the variables P, Q and R for the given expressions.

$\frac{x^4-y^4}{x^2+y^2-2xy}\cdot \frac{\left(x+y\right)^2-4xy}{x^3-y^3}\div \frac{x+y}{x^2+y^2+xy}$

Rewriting

$\frac{\frac{x^4-y^4}{x^2+y^2-2xy}\cdot \frac{\left(x+y\right)^2-4xy}{x^3-y^3}}{\frac{x+y}{x^2+y^2+xy}}=\frac{\frac{x^4-y^4}{x^2+y^2-2xy}\cdot \frac{\left(x+y\right)^2-4xy}{x^3-y^3}\left(x^2+y^2+xy\right)}{x+y}\\ \\$