A=\(\frac{35^3+13^3}{48}-35.13\)
A=\(\frac{35^3+13^3}{48}\)-35.13
Giúp mk vs.
= \(\frac{35^3+13^3}{48}-35.13\)
=\(\frac{\left(35+13\right)\left(35^2-35.13+13^2\right)}{48}-35.13\)
=\(\frac{48.\left(35^2+13^2\right)-48.35.13-48.35.13}{48}\)
=352-2.35.13+132=(35-13)2=222=484
Vậy A = 484
\(A=\frac{35^3+13^3}{48}-35.13\)
\(A=\frac{42875+2197}{48}-455\)
\(A=939-455\)
\(A=484\)
THEO BÀI TA CÓ :
=> A=\(\frac{\left(35+13\right)^3}{48}\)-35.13
=> A = \(\frac{48^3}{48}\) -35.13
=> A = 483:48-35.13
=> A = 482-35.13
=> A = 2304-455
=> A = 1579
VẬY A = 1579
Tính
P=\(\frac{35^3-13^3}{48}-35.13\)
\(P=\frac{\left(35-13\right)\left(35^2+35.13+13^2\right)}{48}-35.13\)
\(=\frac{11\left(35^2+13^2+35.13\right)}{24}-\frac{24.35.13}{24}\)
\(=\frac{11.35^2+11.13^2+11.35.13-24.35.13}{24}\)
\(=\frac{11.35^2+11.13^2-13^2.24}{24}=\frac{11.35^2-13^3}{24}\)
Đến đay tính là ra
a)A=\(\frac{35^3+13^3}{48}-35.13\)
b)B=\(\frac{68^3-52^3}{16}+68.52\)
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=\(\frac{68^3-52^3}{16}+68\cdot52\)=\(\frac{\left(68-52\right)\left(68^2+68\cdot52+52^2\right)}{16}+68\cdot52\)
=\(68^2+2\cdot68\cdot52+52^2\)= \(\left(68+52\right)^2=120^2=14400\)
Tính
B=\(\frac{68^3-52^3}{16}+68.52\)
A=\(\frac{35^3+13^3}{48}-35.13\)
\(B=\frac{\left(68-52\right)\left(68^2+68.52+52^2\right)}{16}+68.52=\frac{16\left(68^2+68.52+52^2\right)}{16}+68.52\)
\(B=68^2+2.68.52+52^2=\left(68+52\right)^2=120^2\)
Câu tiếp theo làm tương tự
tính nhanh
\(A=\frac{35^3+13^3}{48}-35.13\)
\(B=\frac{68^3-52^3}{16}+68.52\)
A=\(\dfrac{35^3+13^3}{48}-35.13\)
\(A=\dfrac{\left(35+13\right)\left(35^2-35\cdot13+13^2\right)}{48}-35\cdot13\)
\(=35^2-2\cdot35\cdot13+13^2\)
\(=22^2=484\)
A= 355+135/48-35.13 B= 683-523/16+68.52
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= 355+135/48-35.13 B= 683-523/16+68.52
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: \(B=\dfrac{\left(68-52\right)\left(68^2+68\cdot52+52^2\right)}{16}+68\cdot52\)
\(=68^2+2\cdot68\cdot52+52^2\)
\(=110^2\)
Tính nhanh
a) A = \(\frac{35^3+15^3}{48}-35.13\)
b) B = \(\frac{68^3-52^3}{16}+68.52\)