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$\left[2-3\left(2-3\right)^{-1}\right]^{-1} =$ কত?

Options :

  • (A) $5$
  • (B) $-5$
  • (C) $\frac15$
  • (D) $-\frac15$

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$\left[2-3\left(2-3\right)^{-1}\right]^{-1}$

= $\left[2-3\frac1{\left(2-3\right)}\right]^{-1}$

= $\left[2-3\frac1{-1}\right]^{-1}$

= $\left[2-3\left(-1\right)\right]^{-1}$

= $\left[2+3\right]^{-1}$

= $\left[5\right]^{-1}$

= $\frac15$

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