Evaluate (2/3 + 2/6) to the first power, plus [(3/5 - 1/10) squared divided by (1 - 1/3) squared], minus (1/2 + 2/3) to the zero power, minus (5/16) to the fifth divided by (5/16) to the fourth.

Step-by-step solution

1. 1

   Simplify the first term. $\tfrac26 = \tfrac13$, so $\tfrac23+\tfrac13 = 1$, and raising to the power $1$ leaves it as $1.$

2. 2

   Simplify the third term with the zero-exponent rule. $\tfrac12+\tfrac23 = \tfrac76 \neq 0$, so $\left(\tfrac76\right)^0 = 1$. The base never needs to be computed beyond checking it is nonzero.

3. 3

   Simplify the last term with the quotient rule. Same base, so $\left(\frac5{16}\right)^5:\left(\frac5{16}\right)^4 = \left(\frac5{16}\right)^{5-4} = \frac5{16}.$ Expanding the fifth power here would be a wasted page of arithmetic.

4. 4

   Work inside the middle bracket. $\frac35-\frac1{10} = \frac6{10}-\frac1{10} = \frac12, \qquad 1-\frac13 = \frac23.$

5. 5

   Divide the two squares. $\left(\frac12\right)^2:\left(\frac23\right)^2 = \frac14:\frac49 = \frac14\cdot\frac94 = \frac9{16}.$

6. 6

   Combine all four terms. $1+\frac9{16}-1-\frac5{16} = \frac9{16}-\frac5{16} = \frac4{16} = \frac14.$ The two $1$s cancel, leaving a single subtraction over the denominator $16$.