a) thể tích của hình ABCD,A'B'C'D' là:

V=Sđáy . h = 5.3.3=15.3=45(cm³)
b) thể tích của hinhd ABCD,EHGF là:
V=Sđáy .h=[(3+4).3:2].5=[7.3:2].5=[21:2].5=10,5 .5= 52,5(cm³)
diện tích xung quanh hình ABCD,EHGF là:
Sxung quanh =Cđáy .h=(3+3+4+4).5=14.5=70(cm²)

\(x+\dfrac{1}{2}=\dfrac{6}{4} x=\dfrac{3}{2}-\dfrac{1}{2} x=1\)

x⁴.3⁵=27³
x⁴ .3⁵=(3³)3
x⁴.3⁵=3⁹
x⁴=3⁹:3⁵
x⁴=3^{4
x=3
\(\dfrac{8}{3}.\left(\dfrac{5}{24}-x\right)=\dfrac{-1}{3}\)
      \(\left(\dfrac{5}{24}-x\right)=\dfrac{-1}{3}:\dfrac{8}{3}\)
      \(\left(\dfrac{5}{24}-x\right)=\dfrac{-1}{3}.\dfrac{3}{8}\)
     \(\left(\dfrac{5}{24}-x\right)=\dfrac{-1}{8}\)
       \(\dfrac{5}{24}-x=\dfrac{-1}{8}\)
               \(x=\dfrac{5}{24}-\dfrac{-1}{8}\)
              \(x=\dfrac{5}{24}-\dfrac{-3}{24}\)
             \(x=\dfrac{2}{24}=\dfrac{1}{12}\)}
 

\(\dfrac{5}{4}-\dfrac{1}{4}=\dfrac{4}{4}=1\)

\(\dfrac{2}{3}.\dfrac{-3}{2}+\dfrac{-7}{2}.\dfrac{2}{3}=\dfrac{2}{3}.\left(\dfrac{-3}{2}+\dfrac{-7}{2}\right)=\dfrac{2}{3}.\dfrac{-5}{1}=\dfrac{-10}{3}\)

\(\left(\dfrac{1}{5}+\dfrac{4}{13}\right)+\left(\dfrac{-2}{5}+\dfrac{7}{13}\right)-\left(\dfrac{4}{5}-\dfrac{2}{13}\right)=\dfrac{1}{5}+\dfrac{4}{13}-\dfrac{2}{5}+\dfrac{7}{13}-\dfrac{4}{5}+\dfrac{2}{13}=\left(\dfrac{1}{5}-\dfrac{2}{5}-\dfrac{4}{5}\right)+\left(\dfrac{4}{13}+\dfrac{7}{13}+\dfrac{2}{13}\right)=-1+1=0\)