Re: Help needed with fractions/brackets/parentheses and powers.
Jackie In Italy said:
$\left(-\frac{1}{2}-\frac{1}{3}\right)^2 \cdot \left(\frac{2}{5}-2\right)^2 +\frac{\frac{7}{9}-\frac{2}{3}}{\frac{1}{2}-\frac{9}{22}}-\frac{5}{6} (2+\frac{2}{5} )$
In this case, what you've to exercise cautious with is all about the order of operations. It's a standard that defines the order in which you should simplify the given expression with a combination of different operations.
First, we should always simplify the inside of parentheses before dealing with the exponent of the set of parentheses.
Second, we simplify the exponent of a set of parentheses before we multiply, divide, add, or subtract it.
Next, we simplify multiplication and division in the order that they appear from left to right.
Last, we simplify addition and subtraction in the order that they appear from left to right.
$=\color{red}\left(-\dfrac{1}{2}-\dfrac{1}{3}\right)^2 \cdot \left(\dfrac{2}{5}-2\right)^2 \color{black}+\dfrac{\dfrac{7}{9}-\dfrac{2}{3}}{\dfrac{1}{2}-\dfrac{9}{22}}-\dfrac{5}{6} \color{red}(2+\dfrac{2}{5} )$
$=\color{red}\left(-\dfrac{5}{6}\right)^2 \cdot \left(-\dfrac{8}{5}\right)^2 \color{black}+\dfrac{\dfrac{7}{9}-\dfrac{2}{3}}{\dfrac{1}{2}-\dfrac{9}{22}}-\dfrac{5}{6} \color{red}(\dfrac{12}{5} )$
$=\color{red}\left(\dfrac{5}{6}\right)\left(\dfrac{5}{6}\right) \cdot \left(\dfrac{8}{5}\right)\left(\dfrac{8}{5}\right) \color{black}+\dfrac{\dfrac{7}{9}-\dfrac{2}{3}}{\dfrac{1}{2}-\dfrac{9}{22}}-\dfrac{\cancel{5}^1}{\cancel{6}} \color{red}(\dfrac{\cancel{12}^2}{\cancel{5}} )$
$=\color{red}\left(\dfrac{\cancel{5}}{\cancel{6}^3}\right)\left(\dfrac{\cancel{5}}{\cancel{6}^3}\right) \cdot \left(\dfrac{\cancel{8}^4}{\cancel{5}}\right)\left(\dfrac{\cancel{8}^4}{\cancel{5}}\right) \color{black}+\dfrac{\dfrac{7}{9}-\dfrac{2}{3}}{\dfrac{1}{2}-\dfrac{9}{22}}-\color{red}\dfrac{2}{1}$
$=\color{red}\left(\dfrac{4\cdot 4}{3\cdot 3}\right) \color{black}+\dfrac{\dfrac{7}{9}-\dfrac{2}{3}}{\dfrac{1}{2}-\dfrac{9}{22}}-\color{red}2$
$=\color{red}\dfrac{16}{9} \color{black}+\color{blue}\dfrac{\dfrac{1}{9}}{\dfrac{1}{11}}-\color{red}2$
$=\color{red}\dfrac{16}{9} \color{black}+\color{blue}\dfrac{1}{9}\div\dfrac{1}{11}-\color{red}2$
$=\color{red}\dfrac{16}{9} \color{black}+\color{blue}\dfrac{1}{9}\times\dfrac{11}{1}-\color{red}2$
$=\color{red}\dfrac{16}{9} \color{black}+\color{blue}\dfrac{11}{9}-\color{red}2$
$=\color{red}\dfrac{16+11}{9} \color{black}-\color{red}2$
$=\color{red}\dfrac{27}{9} \color{black}-\color{red}2$
$=\color{red}3 \color{black}-\color{red}2$
$=1$