Question

solve
${5x}^{0} - 1 \div {7x}^{0} - 3$
​

Answer 1

Answer:

$\dfrac{5x^0-1}{7x^0-3}=1$

Step-by-step explanation:

$\dfrac{5x^0-1}{7x^0-3}$

$\mathrm{Apply\:rule}\:a^0=1,\:a\ne \:0$

$x^0=1$

$=\dfrac{5\cdot \:1-1}{7\cdot \:1-3}$

$5\cdot \:1-1=4$

$=\dfrac{4}{7\cdot \:1-3}$

$7\cdot \:1-3=4$

$=\dfrac{4}{4}$

$\mathrm{Apply\:rule}\:\dfrac{a}{a}=1$

$=1$