Question

When simplified (x-1 + y-1)-1 is equal to
xy
а) ху
b) x + y
c)
d)*+y
x + y
xy​

Answer 1

If we simply $(x-1 + y-1)-1$ then it would be equal to $x + y - 3$. Let's see how.

$\implies (x-1 + y-1)-1 \\ \\ \implies x-1 + y-1-1 \\ \\ \implies x+(-1-1-1)+y \\ \\ \implies x+(-3)+y \\ \\ \implies x - 3 + y \\ \\ \implies \boxed{x + y - 3}$

Hence, if we simply $(x-1 + y-1)-1$ we will get $x + y - 3$.

Answer 2

Answer:

Appropriate Question :-

$\leadsto \sf When\: simplified\: (x^{- 1} + y^{- 1})^{- 1}\: is\: equal\: to\: ?$

Options :

$\bullet \: \: \sf a)\: xy$

$\bullet \: \: \sf b) x + y$

$\bullet \: \: \sf c) \dfrac{xy}{x + y}$

$\bullet \: \: \sf d) \dfrac{1}{xy}$

Given :

$\leadsto \sf (x^{- 1} + y^{- 1})^{- 1}$

To Find :-

$\leadsto$ Simplified.

Solution :-

$\dashrightarrow \bf (x^{- 1} + y^{- 1})^{- 1}$

$\dashrightarrow \sf \bigg\{\dfrac{1}{x^{- 1} + y^{- 1}}\bigg\}$

$\dashrightarrow \sf \Bigg\{\dfrac{1}{\dfrac{1}{x} + \dfrac{1}{y}}\Bigg\}$

$\dashrightarrow \sf \Bigg\{\dfrac{1}{\dfrac{x + y}{xy}}\Bigg\}$

$\dashrightarrow \sf \dfrac{xy}{x + y} \times \dfrac{1}{1}$

$\dashrightarrow \sf\boxed{\bold{\red{\dfrac{xy}{x + y}}}}$

$\therefore$ $\sf\bold{\purple{ The\: value\: of\: (x^{- 1} + y^{- 1})^{- 1}\: is\: \dfrac{xy}{x + y}\: .}}$

Hence, the correct options is option no c) xy/x + y .

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EXTRA INFORMATION ON INDICE LAW :-

$\bigstar \: \: \bf x^0 =\: 1$ [where x ≠ 0]

$\bigstar \: \: \bf x^{- n} =\: \dfrac{1}{x^n}$

$\bigstar \: \: \bf x^n .\: x^m =\: x^{n + m}$

$\bigstar \: \: \bf \dfrac{x^n}{x^m} =\: x^{n - m}$

$\bigstar \: \: \bf (x^n)^m =\: x^{m . n}$

$\bigstar \: \: \bf x^{\frac{n}{m}} =\: \sqrt[m]{x^n}$