For x ne 0,{left( {{{{x^l}} over {{x^m}}}} ri | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(a) ${\left( {{{{x^l}} \over {{x^m}}}} \right)^{{l^2} + lm + {m^2}}}{\left( {{{{x^m}} \over {{x^n}}}} \right)^{{m^2} + nm + {n^2}}}{\left( {{{{x^n}} \

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(a) ${\left( {{{{x^l}} \over {{x^m}}}} \right)^{{l^2} + lm + {m^2}}}{\left( {{{{x^m}} \over {{x^n}}}} \right)^{{m^2} + nm + {n^2}}}{\left( {{{{x^n}} \over {{x^l}}}} \right)^{{n^2} + nl + {l^2}}}$

$={({x^{l - m}})^{({l^2} + lm + {m^2})}}{({x^{m - n}})^{{m^2} + nm + {n^2}}}$${({x^{n - l}})^{{n^2} + nl + {l^2}}}$

$={x^{{l^3} - {m^3}}}.{x^{{m^3} - {n^3}}}.{x^{{n^3} - {l^3}}}$=${x^{{l^3} - {m^3} + {m^3} - {n^3} + {n^3} - {l^3}}} = {x^0}=1$

For $x \ne 0,{\left( {{{{x^l}} \over {{x^m}}}} \right)^{({l^2} + lm + {m^2})}}$${\left( {{{{x^m}} \over {{x^n}}}} \right)^{({m^2} + nm + {n^2})}}{\left( {{{{x^n}} \over {{x^l}}}} \right)^{({n^2} + nl + {l^2})}}=$

Similar Questions

${({x^5})^{1/3}}{(16{x^3})^{2/3}}$${\left( {{1 \over 4}{x^{4/9}}} \right)^{ - 3/2}} = $

${a^{m{{\log }_a}n}} = $

Solution of the equation $\sqrt {(x + 10)} + \sqrt {(x - 2)} = 6$ are

If ${2^x} = {4^y} = {8^z}$ and $xyz = 288,$ then ${1 \over {2x}} + {1 \over {4y}} + {1 \over {8z}} = $

The number of integers $q , 1 \leq q \leq 2021$, such that $\sqrt{ q }$ is rational, and $\frac{1}{ q }$ has a terminating decimal expansion, is