C

\(\dfrac{-15}{4}-\dfrac{6}{7}+\dfrac{15}{4}\\ =\left(\dfrac{-15}{4}+\dfrac{15}{4}\right)-\dfrac{6}{7}\\ =0-\dfrac{6}{7}\\ =\dfrac{-6}{7}\)

\(295>29a\Rightarrow295>294;293;292;291;290\\ \Rightarrow a=4;3;2;1;0\)

\(\dfrac{2}{5}\times\dfrac{3}{4}+\dfrac{1}{4}\times\dfrac{2}{5}\\ =\dfrac{2}{5}\times\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\\ =\dfrac{2}{5}\times1\\ =\dfrac{2}{5}\)

\(\dfrac{3}{4}+\dfrac{1}{4}\times\dfrac{2}{5}-\dfrac{1}{2}\\ =\dfrac{3}{4}+\dfrac{1}{10}-\dfrac{1}{2}\\ =\dfrac{15}{20}+\dfrac{2}{20}-\dfrac{10}{20}\\ =\dfrac{15-10+2}{2}\\ =\dfrac{7}{2}\)

\(2:x-\dfrac{1}{3}=\dfrac{2}{5}\\ 2:x=\dfrac{2}{5}+\dfrac{1}{3}\\ 2:x=\dfrac{11}{15}\\ x=2:\dfrac{11}{15}\\ x=\dfrac{30}{11}\)

Số đường còn lại là:

\(50-10=40\left(kg\right)\)

Buổi chiều bán được số kg đường là:

\(40\times\dfrac{3}{8}=15\left(kg\right)\)

Cả 2 buổi bán được số kg đường là:

\(10+15=25\left(kg\right)\)

\(\left(3,5\times11,25+11,25\times6,5\right)\times\left(65\times11-65:0,1:65\right)\\ =\left[11,25\times\left(3,5+6,5\right)\right]\times\left(65\times11-10\right)\\ =\left[11,25\times10\right]\times\left[715-10\right]\\ =112,5\times705\\ =79312,5\)

C