a) ( 100   +   1 ) 3   =   100 3   +   3 . 100 2 + 3.100 + 1 = 1030301.

b) ( 47   +   3 ) 3   =   50 3  = 125000.

c) ( 300   –   1 ) 3  = 26730899.

d) ( 1008   –   23 ) 3   =   1000 3   =   10 9 .

a)510 

b)........

Ta có; \(D=\frac{5}{6\cdot37}+\frac{1}{6\cdot43}+\frac{6}{7\cdot43}+\frac{10}{7\cdot59}\)

\(=7\left(\frac{5}{37\cdot42}+\frac{1}{42\cdot43}+\frac{6}{43\cdot49}+\frac{10}{49\cdot59}\right)\)

\(=7\left(\frac{1}{37}-\frac{1}{42}+\frac{1}{42}-\frac{1}{43}+\frac{1}{43}-\frac{1}{49}+\frac{1}{49}-\frac{1}{59}\right)\)

\(=7\left(\frac{1}{37}-\frac{1}{59}\right)\)

Ta có: \(E=\frac{8}{9\cdot37}+\frac{2}{9\cdot47}+\frac{3}{10\cdot47}+\frac{9}{10\cdot59}\)

\(=5\left(\frac{8}{37\cdot45}+\frac{2}{45\cdot47}+\frac{3}{47\cdot50}+\frac{9}{50\cdot59}\right)\)

\(=5\left(\frac{1}{37}-\frac{1}{45}+\frac{1}{45}-\frac{1}{47}+\frac{1}{47}-\frac{1}{50}+\frac{1}{50}-\frac{1}{59}\right)\)

\(=5\left(\frac{1}{37}-\frac{1}{59}\right)\)

Do đó: \(\frac{D}{E}=\frac75\)

(145 x 99 + 145 ) - ( 143 x 102 – 143 × 2 ) + 54 x 47 - 47 x 53 - 20 - 27

= 145 × (99 + 1) – 143 × (102 – 2) + 47 × (54 – 53) – (20 + 27)

= 145 × 100 – 143 × 100 + 47 × 1 – 47

= 100 × (145 – 143) + (47 – 47)

= 100 × 2 + 0

= 200

(145 x 99 + 145 ) - ( 143 x 102 – 143 × 2 ) + 54 x 47 - 47 x 53 - 20 - 27

= 145 × (99 + 1) – 143 × (102 – 2) + 47 × (54 – 53) – (20 + 27)

= 145 × 100 – 143 × 100 + 47 × 1 – 47

= 100 × (145 – 143) + (47 – 47)

= 100 × 2 + 0

= 200

giỏi

a/ \(=47^3+3.47^2.3+3.47.3^2+3^3\)

\(=\left(47+3\right)^3=50^3=125000\)

b/ \(=102^3-3.102^2.2+3.102.2^2-2^3\)

\(=\left(102-2\right)^3=100^3=1000000\)

(145 x 99 + 145 ) - ( 143 x 102 – 143 × 2 ) + 54 x 47 - 47 x 53 - 20 - 27

= 145 × (99 + 1) – 143 × (102 – 2) + 47 × (54 – 53) – (20 + 27)

= 145 × 100 – 143 × 100 + 47 × 1 – 47

= 100 × (145 – 143) + (47 – 47)

= 100 × 2 + 0

= 200

= 145 x (99+1) - 143 x ( 102-2) + 47 x ( 54 - 53) -20-27

= 145 x 100 - 143 x 100 + 47 x 1 -20-27

= 100 x ( 145 - 143 ) + 47 - 20 - 27

= 200 + 47 - 20 - 27

= 200