\(A=\frac{35^3+13^3}{48}-35.13\)

\(=\frac{\left(35+13\right)\left(35^2-35.13+13^2\right)}{48}-35.13\)

\(=\frac{48.\left(35^2-35.13+13^2\right)}{48}-35.13\)

\(=35^2-35.13+13^2-35.13\)

\(=\left(35-13\right)^2\)

\(=22^2=484\)

\(P=\frac{\left(35-13\right)\left(35^2+35.13+13^2\right)}{48}-35.13\)

\(=\frac{11\left(35^2+13^2+35.13\right)}{24}-\frac{24.35.13}{24}\)

\(=\frac{11.35^2+11.13^2+11.35.13-24.35.13}{24}\)

\(=\frac{11.35^2+11.13^2-13^2.24}{24}=\frac{11.35^2-13^3}{24}\)

Đến đay tính là ra

a) Ta có: A= \(\frac{35^3+13^3}{48}-35\cdot13\)=\(\frac{\left(35+13\right)\left(35^2-35\cdot13+13^2\right)}{48}-35\cdot13\)

                 =\(35^2-35\cdot13+13^2+35\cdot13\)=\(35^2+13^2=1394\)

b) Ta có: B=\(\frac{68^3-52^3}{16}+68\cdot52\)=\(\frac{\left(68-52\right)\left(68^2+68\cdot52+52^2\right)}{16}+68\cdot52\)

                  =\(68^2+2\cdot68\cdot52+52^2\)= \(\left(68+52\right)^2=120^2=14400\)

\(B=\frac{\left(68-52\right)\left(68^2+68.52+52^2\right)}{16}+68.52=\frac{16\left(68^2+68.52+52^2\right)}{16}+68.52\)

\(B=68^2+2.68.52+52^2=\left(68+52\right)^2=120^2\)

Câu tiếp theo làm tương tự

\(A=\dfrac{\left(35+13\right)\left(35^2-35\cdot13+13^2\right)}{48}-35\cdot13\)

\(=35^2-2\cdot35\cdot13+13^2\)

\(=22^2=484\)

a: Sửa đề: \(A=\dfrac{35^3+13^3}{48}-35\cdot13\)

\(=\dfrac{\left(35+13\right)\left(35^2-35\cdot13+13^2\right)}{48}-35\cdot13\)

\(=35^2-2\cdot35\cdot13+13^2=22^2\)

b: \(B=\dfrac{\left(68-52\right)\left(68^2+68\cdot52+52^2\right)}{16}+68\cdot52\)

\(=68^2+2\cdot68\cdot52+52^2\)

\(=110^2\)

a: Sửa đề: \(A=\dfrac{35^3+13^3}{48}-35\cdot13\)

\(=\dfrac{\left(35+13\right)\left(35^2-35\cdot13+13^2\right)}{48}-35\cdot13\)

\(=35^2-2\cdot35\cdot13+13^2=22^2\)

b: \(B=\dfrac{\left(68-52\right)\left(68^2+68\cdot52+52^2\right)}{16}+68\cdot52\)

\(=68^2+2\cdot68\cdot52+52^2\)

\(=110^2\)