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

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

Câu tiếp theo làm tương tự

a) 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=\(\frac{68^3-52^3}{16}+68\cdot52\)=\(\frac{\left(68-52\right)\left(68^2+68\cdot52+52^2\right)}{16}+68\cdot52\)

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

cho mk hỏi tại sao ở dòng thứ nhất là trừ 35.13 xong dưới lại là cộng 

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

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

\(=110^2\)

A= \(\dfrac{42875+2197}{48}-35.13\)

A=\(\dfrac{42875+2197}{48}-455\)

A=\(\dfrac{45072}{48}-455\)

A= 939-455

Vậy A= 484

B=\(\dfrac{68^3-52^3}{16}+\dfrac{6852}{100}\)

B= \(\dfrac{68^3-52^3}{16}+\dfrac{1713}{25}\)

Vậy B= 10932,52

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: Ta có: \(A=\dfrac{35^3+13^3}{48}-35\cdot13\)

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

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

\(=22^2=484\)

b: Ta có: \(B=\dfrac{68^3-52^3}{16}+68\cdot52\)

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

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