Simplify the following expressions in which a, x and y are real numbers greater than 0.
\dfrac{a^3.a^{-2}.a^{1/4}}{a^2.a^{-1}}
For any real numbers a, b and c with a\gt0 we have:
a^b\times a^c=a^{b+c}
Therefore:
\dfrac{a^3.a^{-2}.a^{1/4}}{a^2.a^{-1}} = \dfrac{a^{3+\left(-2\right)+1/4}}{a^{2+\left(-1\right)}} = \dfrac{a^{5/4}}{a}
Furthermore, for any real numbers a, b and c with a\gt0 we have:
\dfrac{a^b}{a^c}=a^{b-c}
Therefore:
\dfrac{a^3.a^{-2}.a^{1/4}}{a^2.a^{-1}} =a^{5/4-1}
\dfrac{a^3.a^{-2}.a^{1/4}}{a^2.a^{-1}}= a^{1/4}
\dfrac{2\times2^{-2}\times2^{4}}{2^{3}\times2^{1/2}}
For any real numbers a, b and c with a\gt0 we have:
a^b\times a^c=a^{b+c}
Therefore:
\dfrac{2\times2^{-2}\times2^{4}}{2^{3}\times2^{1/2}} = \dfrac{2^{1+\left(-2\right)+4}}{2^{3+1/2}} = \dfrac{2^3}{2^{7/2}}
Furthermore, for any real numbers a, b and c with a\gt0 we have:
\dfrac{a^b}{a^c}=a^{b-c}
Therefore:
\dfrac{2\times2^{-2}\times2^{4}}{2^{3}\times2^{1/2}} =2^{3-7/2} =2^{-1/2} =\dfrac{1}{\sqrt{2}} = \dfrac{\sqrt{2}}{2}
\dfrac{2\times2^{-2}\times2^{4}}{2^{3}\times2^{1/2}} = \dfrac{\sqrt{2}}{2}
\dfrac{a^3.a^{-1/2}}{a^2.a^{1/2}.a}
For any real numbers a, b and c with a\gt0 we have:
a^b\times a^c=a^{b+c}
Therefore:
\dfrac{a^3.a^{-1/2}}{a^2.a^{1/2}.a} = \dfrac{a^{3-1/2}}{a^{2+1/2+1}} = \dfrac{a^{5/2}}{a^{7/2}}
Furthermore, for any real numbers a, b and c with a\gt0 we have:
\dfrac{a^b}{a^c}=a^{b-c}
Therefore:
\dfrac{a^3.a^{-1/2}}{a^2.a^{1/2}.a} = a^{5/2 - 7/2}= a^{-1}= \dfrac{1}{a}
\dfrac{a^3.a^{-1/2}}{a^2.a^{1/2}.a} = \dfrac{1}{a}
\dfrac{2^{3/2}.x^3.y^{-2}}{2^{5/2}.x^{1/3}.y^{1/2}}
To avoid confusion, we write this expression as a product of three expressions of each type:
\dfrac{2^{3/2}x^3.y^{-2}}{2^{5/2}x^{1/3}.y^{1/2}} = \dfrac{2^{3/2}}{2^{5/2}} . \dfrac{x^{3}}{x^{1/3}} .\dfrac{y^{-2}}{y^{1/2}}
For any real numbers a, b and c with a\gt0 we have:
\dfrac{a^b}{a^c}=a^{b-c}
Therefore:
\dfrac{2^{3/2}x^3.y^{-2}}{2^{5/2}x^{1/3}.y^{1/2}} \\=2^{{3/2}-{5/2}} . {x^{3}-{1/3}} .y^{-2-1/2} \\= 2^{-1} . x^{8/3} .y^{-5/2}\\=\dfrac{1}{2} . x^{8/3} .y^{-5/2}
\dfrac{2^{3/2}x^3.y^{-2}}{2^{5/2}x^{1/3}.y^{1/2}} =\dfrac{1}{2} . x^{8/3} .y^{-5/2}
\dfrac{\left(a^3.a^{-1/4}\right)^4}{\left(a^{1/3}.a^{-1/2}\right)^3}\\\\\\
For any real numbers a, b and c with a\gt0 we have:
a^b\times a^c=a^{b+c}
Therefore:
\dfrac{\left(a^3.a^{-1/4}\right)^4}{\left(a^{1/3}.a^{-1/2}\right)^3} = \dfrac{\left(a^{3-1/4}\right)^4}{\left(a^{1/3-1/2}\right)^3} = \dfrac{\left(a^{11/4}\right)^4}{\left(a^{-1/6}\right)^3}
Furthermore, for any real numbers a, b and c with a\gt0 we have:
\left(a^b\right)^c = a^{bc}
Thus:
\dfrac{\left(a^{11/4}\right)^4}{\left(a^{-1/6}\right)^3}= \dfrac{a^{11/4 \times 4}}{a^{-1/6 \times 3}}= \dfrac{a^{11}}{a^{-1/2}}
On the other hand, for any real numbers a, b and c we have:
\dfrac{a^b}{a^c}=a^{b-c}
Therefore:
\dfrac{a^{1}}{a^{-1/2}} =a^{11+1/2}= a^{23/2}
Thus:
\dfrac{\left(a^3.a^{-1/4}\right)^4}{\left(a^{1/3}.a^{-1/2}\right)^3} = a^{23/2}
\dfrac{4\times 5^{-1/2}\times 5^{3}}{5^{4}\times 4^{1/2}}
We write this as a product of two expressions:
\dfrac{4\times 5^{-1/2}\times 5^{3}}{5^{4}\times 4^{1/2}} = \dfrac{4}{4^{1/2}}\times \dfrac{ 5^{-1/2} 5^{3}}{5^4}
For any real numbers a, b and c with a\gt0 we have:
a^b\times a^c=a^{b+c}
Thus:
5^{-1/2} 5^{3} = 5^{5/2}
Also, observe that:
4^{1/2} =2
Therefore:
\dfrac{4\times 5^{-1/2}\times 5^{3}}{5^{4}\times 4^{1/2}} = \dfrac{4}{4^{1/2}}\times \dfrac{ 5^{-1/2} 5^{3}}{5^4} = 2 \times \dfrac{5^{5/2}}{5^4}
Furthermore, for any real numbers a, b and c with a\gt0 we have:
\dfrac{a^b}{a^c}=a^{b-c}
Therefore:
\dfrac{5^{5/2}}{5^4} = 5^{5/2-4} = 5^{-3/2}
Thus we have:
\dfrac{4\times 5^{-1/2}\times 5^{3}}{5^{4}\times 4^{1/2}} = 2 \times 5^{-3/2}
\dfrac{\left(a^3.b^{-1/6}\right)^3}{\left(a^{-2}.b^{-1/2}\right)^2}\\\\\\
For any real numbers a, b and c with a\gt0 we have:
\left(a.b\right)^c = a^c .b^c
So, we have:
\dfrac{\left(a^3.b^{-1/6}\right)^3}{\left(a^{-2}.b^{-1/2}\right)^2} = \dfrac{\left(a^3\right)^3. \left(b^{-1/6}\right)^3}{\left(a^{-2}\right)^2. \left(b^{-1/2}\right)^2}
On the other hand, for any real numbers a, b and c with a\gt0 we have:
\left(a^b\right)^c = a^{bc}
Thus:
\dfrac{\left(a^3.b^{-1/6}\right)^3}{\left(a^{-2}.b^{-1/2}\right)^2} = \dfrac{a^9.b^{-1/2}}{a^{-4}b^{-1}} = \dfrac{a^9}{a^{-4}}\times \dfrac{b^{-1/2}}{b^{-1}}
Furthermore, for any real numbers a, b and c with a\gt0 we have:
\dfrac{a^b}{a^c}=a^{b-c}
Therefore:
\dfrac{a^9}{a^{-4}}\times \dfrac{b^{-1/2}}{b^{-1}} = a^{9-\left(-4\right)} . b^{-1/2 -\left(-1\right)} = a^{13}b^{1/2}
\dfrac{\left(a^3.b^{-1/6}\right)^3}{\left(a^{-2}.b^{-1/2}\right)^2} = a^{13} b^{1/2}