Simplify the expression of the following rational functions.
g\left(x\right)=\dfrac{x-3}{x^2-9}
For real numbers a,b we have the following algebraic identity:
a^2-b^2=\left(a-b\right)\left(a+b\right)
The denominator of the fraction equals:
x^2-9=\left(x-3\right)\left(x+3\right)
Therefore:
g\left(x\right) = \dfrac{\left(x-3\right)}{\left(x-3\right)\left(x+3\right)} =\dfrac{1}{x+3}
g\left(x\right)=\dfrac{1}{x+3}
g\left(x\right)=\dfrac{x^2-1}{\left(x+1\right)^2}
For real numbers a,b we have the following algebraic identity:
a^2-b^2=\left(a-b\right)\left(a+b\right)
The numerator of the fraction equals:
x^2-1=\left(x-1\right)\left(x+1\right)
Therefore:
g\left(x\right) = \dfrac{\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x+1\right)} =\dfrac{x-1}{x+1}
g\left(x\right)=\dfrac{x-1}{x+1}
g\left(x\right)=\dfrac{x^2-4x+4}{x^2-4}
For real numbers a,b we have the following algebraic identities:
a^2-2ab+b^2 = \left(a-b\right)^2
a^2-b^2=\left(a-b\right)\left(a+b\right)
The numerator of the fraction equals:
x^2-4x+4=\left(x-2\right)^2
The denominator of the fraction equals:
x^2-4=\left(x-2\right)\left(x+2\right)
Hence we have:
g\left(x\right) = \dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}= \dfrac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}= \dfrac{\left(x-2\right)}{\left(x+2\right)}
g\left(x\right)=\dfrac{x-2}{x+2}
g\left(x\right)=\dfrac{x-1}{x^3-x}
For real numbers a,b we have the following algebraic identity:
a^2-b^2=\left(a-b\right)\left(a+b\right)
The denominator of the fraction equals:
x^3-x= x\left(x^2-1\right)=x\left(x-1\right)\left(x+1\right)
Therefore:
g\left(x\right) = \dfrac{x-1}{x\left(x-1\right)\left(x+1\right)}= \dfrac{1}{x\left(x+1\right)}
g\left(x\right) = \dfrac{1}{x\left(x+1\right)}
g\left(x\right)=\dfrac{x^4-x^3}{x^5-x^4}
The numerator of the fraction equals:
x^4-x^3=x^3\left(x-1\right)
The denominator of the fraction equals:
x^5-x^4 = x^4\left(x-1\right)
Therefore:
g\left(x\right)=\dfrac{x^4-x^3}{x^5-x^4}=\dfrac{x^3\left(x-1\right)}{x^4\left(x-1\right)}= \dfrac{1}{x}
g\left(x\right)=\ \dfrac{1}{x}
g\left(x\right)=\dfrac{x^5-x^3}{x^3\left(x-1\right)^2}
The numerator of the fraction equals:
x^5-x^3= x^3\left(x^2-1\right)
For real numbers a,b we have the following algebraic identity:
a^2-b^2=\left(a-b\right)\left(a+b\right)
The numerator of the fraction equals:
x^3\left(x^2-1\right) =x^3\left(x-1\right)\left(x+1\right)
Therefore:
g\left(x\right)=\dfrac{x^5-x^3}{x^3\left(x-1\right)^2}=\dfrac{x^3\left(x-1\right)\left(x+1\right)}{x^3\left(x-1\right)\left(x-1\right)}=\dfrac{x+1}{x-1}
g\left(x\right)=\dfrac{x+1}{x-1}
g\left(x\right)=\dfrac{x^3-1}{x^2-1}
For real numbers a,b we have the following algebraic identities:
a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)
a^2-b^2=\left(a-b\right)\left(a+b\right)
The numerator of the fraction equals:
x^3-1=\left(x-1\right)\left(x^2+x+1\right)
The denominator of the fraction equals:
x^2-1=\left(x-1\right)\left(x+1\right)
Therefore:
g\left(x\right) = \dfrac{\left(x-1\right)\left(x^2+x+1\right)}{\left(x-1\right)\left(x+1\right)}= \dfrac{x^2+x+1}{x+1}
g\left(x\right)=\dfrac{x^2+x+1}{x+1}