Вправа 349 алгебра Бевз гдз 7 клас
$$\begin{aligned}
&\text { № } 349\\
&\begin{aligned}
& \text { a) }\left(2-\frac{1}{2}\right)^2-3^2-\left(1-3^3\right)-1,5^2= \\
& =2,25-9+26-2,25=15 ; \\
& \text { б) } \frac{2}{3}-\frac{1}{2}+\frac{1}{6}-\frac{1}{2} \cdot\left(\frac{2}{3}-\frac{1}{2}\right)+\left(\frac{1}{3}\right)^2= \\
& =\frac{4}{6}-\frac{3}{6}+\frac{1}{6}-\frac{1}{2} \cdot\left(\frac{4}{6}-\frac{3}{6}\right)+\frac{1}{9}=\frac{2}{6}-\frac{1}{2} \cdot \frac{1}{6}+\frac{1}{9}= \\
& =\frac{1}{3}-\frac{1}{12}+\frac{1}{9}=\frac{12}{36}-\frac{3}{36}+\frac{4}{36}=\frac{13}{36} .
\end{aligned}
\end{aligned}
$$
вправи поруч
