\(\frac{a}{b}=\frac{c}{d}\Rightarrow\left(\frac{a}{b}\right)^3=\left(\frac{c}{d}\right)^3=\frac{a^3}{b^3}=\frac{c^3}{d^3}=\frac{a^3+c^3}{b^3+d^3}\left(1\right)\)

\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\Rightarrow\frac{a^3}{b^3}=\frac{c^3}{d^3}=\frac{\left(a+c\right)^3}{\left(b+d\right)^3}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{a^3+c^3}{b^3+d^3}=\frac{\left(a+c\right)^3}{\left(b+d\right)^3}\left(dpcm\right)\)

a) \(\left(-2\right)^3+\left(-1\frac{1}{2}+2\right)^2=-8+\left(-\frac{3}{2}+2\right)^2=-8+\left(\frac{1}{2}\right)^2=-8+\frac{1}{4}=-\frac{31}{4}\)

b) \(\frac{2}{5}.\frac{1}{3}-\frac{4}{3}.\frac{2}{5}=\frac{2}{5}.\left(\frac{1}{3}-\frac{4}{3}\right)=\frac{2}{5}.\left(-1\right)=-\frac{2}{5}\)

c) \(\frac{1}{2}.\sqrt{\frac{49}{9}}-12^0=\frac{1}{2}.\frac{7}{3}-1=\frac{7}{6}-1=\frac{1}{6}\)

Gọi số kg giấy lớp 7A, 7B, 7C nhặt được lần lượt là \(x,y,z\)

Ta có: \(x,y,z=3:5:7\Leftrightarrow\frac{x}{3}=\frac{y}{5}=\frac{z}{7};x+y+z=105\)

Áp dụng tính chất dãy tỉ số bằng nhau là:

\(\frac{x}{3}=\frac{y}{5}=\frac{z}{7}=\frac{x+y+z}{3+5+7}=\frac{105}{15}=7\)

\(\Leftrightarrow\hept{\begin{cases}\frac{x}{3}=7\Leftrightarrow x=21\\\frac{y}{5}=7\Leftrightarrow y=35\\\frac{z}{7}=7\Leftrightarrow z=49\end{cases}}\)

Vậy: ...