Câu 1:
\(\begin{array}{l}a)\,\,\,A = 4\frac{1}{3} - {\left( { - \frac{1}{2}} \right)^2} - \frac{{\sqrt {100} }}{{15}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{13}}{3} - \frac{1}{4} - \frac{{10}}{{15}} = \frac{{41}}{{12}}.\\b)\,\,\,B = {\left( {\frac{3}{7}} \right)^0}{.1^{20}} + \frac{7}{9}:{\left( {\frac{2}{3}} \right)^2} - \sqrt {\frac{{16}}{{25}}} \\\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 + \frac{7}{9}:\frac{4}{9} - \frac{4}{5} = 1 + \frac{7}{9}.\frac{9}{4} - \frac{4}{5}\\\,\,\,\,\,\,\,\,\,\,\,\,\, = 1 + \frac{7}{4} - \frac{4}{5} = \frac{{39}}{{20}}.\end{array}\)
Câu 2:
a) \(y = f\left( x \right) = 4x + b\)
Ta có: \(f\left( {\frac{1}{2}} \right) = 1 \Rightarrow 4.\frac{1}{2} + b = 1 \Leftrightarrow 2 + b = 1 \Leftrightarrow b = - 1.\)
Vậy \(b = - 1.\)
b) Tìm \(x:\)
\(\begin{array}{l}\,\,\,\,\,\,\,\left| {0,25 - x} \right| - 0,75 = 3,5\\ \Leftrightarrow \left| {0,25 - x} \right| = 3,5 + 0,75\\ \Leftrightarrow \left| {0,25 - x} \right| = 4,25\\ \Leftrightarrow \left[ \begin{array}{l}0,25 - x = 4,25\\0,25 - x = - 4,25\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = - 4\\x = 4,5\end{array} \right.\end{array}\)
Vậy \(x = - 4\) hoặc \(x = 4,5.\)
c) Ta có \(x,\,y\) là hai đại lượng tỉ lệ thuận nên \(y = ax\,\,\,\left( {a \ne 0} \right).\)
+) Với \({y_2} = - 15,\,\,{x_2} = - 8\)
\(\begin{array}{l} \Rightarrow - 15 = - 8a \Leftrightarrow a = \frac{{15}}{8}\,\,\,\,\left( {tm} \right)\\ \Rightarrow y = \frac{{15}}{8}x.\end{array}\)
+) Với \({y_1} = 10 \Rightarrow \frac{{15}}{8}{x_1} = 10 \Leftrightarrow {x_1} = 10:\frac{{15}}{8} = \frac{{16}}{3}.\)
