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# Simplify the expression $\frac{5\left(\frac{x^2+3x+5}{2x-1}\right)^4\left(2x^2-2x-13\right)}{4x^2-4x+1}$

## Step-by-step Solution

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###  Solution

$\frac{5\left(x^2+3x+5\right)^4\left(2x^2-2x-13\right)}{\left(2x-1\right)^4\left(4x^2-4x+1\right)}$
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##  Step-by-step Solution 

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The power of a quotient is equal to the quotient of the power of the numerator and denominator: $\displaystyle\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}$

$\frac{5\left(\frac{\left(x^2+3x+5\right)^4}{\left(2x-1\right)^4}\right)\left(2x^2-2x-13\right)}{4x^2-4x+1}$

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$\frac{5\left(\frac{\left(x^2+3x+5\right)^4}{\left(2x-1\right)^4}\right)\left(2x^2-2x-13\right)}{4x^2-4x+1}$

Learn how to solve problems step by step online. Simplify the expression (5((x^2+3x+5)/(2x-1))^4(2x^2-2x+-13))/(4x^2-4x+1). The power of a quotient is equal to the quotient of the power of the numerator and denominator: \displaystyle\left(\frac{a}{b}\right)^n=\frac{a^n}{b^n}. Multiplying the fraction by 5\left(2x^2-2x-13\right). Divide fractions \frac{\frac{5\left(x^2+3x+5\right)^4\left(2x^2-2x-13\right)}{\left(2x-1\right)^4}}{4x^2-4x+1} with Keep, Change, Flip: \frac{a}{b}\div c=\frac{a}{b}\div\frac{c}{1}=\frac{a}{b}\times\frac{1}{c}=\frac{a}{b\cdot c}.

$\frac{5\left(x^2+3x+5\right)^4\left(2x^2-2x-13\right)}{\left(2x-1\right)^4\left(4x^2-4x+1\right)}$

##  Explore different ways to solve this problem

Solving a math problem using different methods is important because it enhances understanding, encourages critical thinking, allows for multiple solutions, and develops problem-solving strategies. Read more

SimplifyWrite in simplest formFactorFactor by completing the squareFind the integralFind the derivativeFind 5((x^2+3x)/(2x-1))^4/(4x^2+-4x) using the definitionFind the rootsFind break even pointsFind the discriminant

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◻/◻
/
÷
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e
π
ln
log
log
lim
d/dx
Dx
|◻|
θ
=
>
<
>=
<=
sin
cos
tan
cot
sec
csc

asin
acos
atan
acot
asec
acsc

sinh
cosh
tanh
coth
sech
csch

asinh
acosh
atanh
acoth
asech
acsch

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