Question

show that (x+y) ^1*(x^-1+y^-1)=(xy) ^1​

Answer 1

SOLUTION:

We have to prove that:

$\rm \hookrightarrow {(x + y)}^{ - 1} \times ( {x}^{ - 1} + {y}^{ - 1}) = {(xy)}^{ - 1}$

Taking Left Hand Side, we get:

$\rm = {(x + y)}^{ - 1} \times ( {x}^{ - 1} + {y}^{ - 1})$

$\rm = \dfrac{1}{x + y} \times \bigg ( \dfrac{1}{x} + \dfrac{1}{y} \bigg)$

$\rm = \dfrac{1}{x + y} \times \dfrac{y + x}{xy}$

$\rm = \dfrac{1}{x + y} \times \dfrac{x +y }{xy}$

$\rm = \dfrac{1}{xy}$

$\rm=(xy)^{-1}$

Now, taking Right Hand Side, we get:

$\rm=(xy)^{-1}$

Therefore:

→ LHS = RHS

Hence Proved..!!

LEARN MORE:

Laws of exponents.

$\rm 1. \: \: {a}^{m} \times {a}^{n} = {a}^{m + n}$

$\rm 2. \: \: ({a}^{m})^{n} = {a}^{mn}$

$\rm 3. \: \: \dfrac{ {a}^{m} }{ {a}^{n} } = {a}^{m - n}$

$\rm4. \: \: {a}^{m} \times {b}^{m} = {(ab)}^{m}$

$\rm5. \: \: \bigg(\dfrac{a}{b} \bigg)^{m} = \dfrac{ {a}^{m} }{ {b}^{m} }$

$\rm6. \: \: {a}^{ - n} = \dfrac{1}{ {a}^{n} }$

$\rm7. \: \: {a}^{n} = {b}^{n} \rightarrow a = b, n \neq0$

$\rm8. \: \: {a}^{m} = {a}^{n} \rightarrow m = n, a \neq 1$