Answer:
[tex]\large\boxed{\dfrac{13}{36}}[/tex]
Step-by-step explanation:
Use PEMDAS:
P Parentheses first
E Exponents (ie Powers and Square Roots, etc.)
MD Multiplication and Division (left-to-right)
AS Addition and Subtraction (left-to-right)
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[tex]\dfrac{1}{4}\left(1-\dfrac{2}{3}\right)^2+\dfrac{1}{3}=\dfrac{1}{4}\left(\dfrac{3}{3}-\dfrac{2}{3}\right)^2+\dfrac{1}{3}=\dfrac{1}{4}\left(\dfrac{3-2}{3}\right)^2+\dfrac{1}{3}\\\\=\dfrac{1}{4}\left(\dfrac{1}{3}\right)^2+\dfrac{1}{3}=\dfrac{1}{4}\cdot\dfrac{1^2}{3^2}+\dfrac{1}{3}=\dfrac{1}{4}\cdot\dfrac{1}{9}+\dfrac{1}{3}=\dfrac{1}{36}+\dfrac{1}{3}\qquad(*)\\\\\text{the common denominator is}\ 36.\\\\36=3\cdot12\to\dfrac{1}{3}=\dfrac{1\cdot12}{3\cdot12}=\dfrac{12}{36}\\\\(*)\qquad=\dfrac{1}{36}+\dfrac{12}{36}=\dfrac{1+12}{36}=\dfrac{13}{36}[/tex]
