\(\dfrac{2^2}{1\times3}\times\dfrac{3^2}{2.4}\times\dfrac{4^2}{3.5}\times\dfrac{5^2}{4.6}=\dfrac{2^2.3^2.4^2.5^2}{1.3.2.4.3.5.4.6}=\dfrac{2^2.3^2.4^2.5^2}{1.2.3.3.4.4.5.2.3}=\dfrac{2^2.3^2.4^2.5^2}{3^3.2^2.4^2.5.1}=\dfrac{5}{3.1}=\dfrac{5}{3}\)

\(\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot\dfrac{5^2}{4.6}\\ =\dfrac{2^2\cdot3^2\cdot4^2\cdot5^2}{1\cdot3\cdot2\cdot4\cdot3\cdot5\cdot4\cdot6}\\ =\dfrac{2^2\cdot3^2\cdot4^2\cdot5^2}{1\cdot2\cdot4^2\cdot4^2\cdot5\cdot6}\\ =\dfrac{2\cdot5}{6}=\dfrac{5}{3}\)

b: Sửa đề: \(B=1\cdot3+2\cdot4+3\cdot5+4\cdot6+\cdots+2018\cdot2020\)

\(=1\left(1+2\right)+2\left(2+2\right)+\cdots+2018\left(2018+2\right)\)

\(=\left(1^2+2^2+\cdots+2018^2\right)+2\left(1+2+\cdots+2018\right)\)

\(=\frac{2018\left(2018+1\right)\left(2\cdot2018+1\right)}{6}+2\cdot\frac{2018\cdot2019}{2}\)

\(=\frac{2018\cdot2019\cdot4037}{6}+2018\cdot2019=1009\cdot673\cdot4037+2018\cdot2019\)

\(=1009\cdot673\left(4037+2\cdot3\right)=1009\cdot673\cdot4043\)

=2745427451

c: Ta có: \(C=\left(1+2+3+\cdots+2020\right)\left(20192019\cdot2020-2019\cdot20202020\right)+\left(15-12\right)^2\)

\(=\frac{2020\left(2020+1\right)}{2}\cdot2019\cdot2020\cdot\left(10001-10001\right)+3^2\)

=0+9

=9

b: Sửa đề: \(B=1\cdot3+2\cdot4+3\cdot5+4\cdot6+\cdots+2018\cdot2020\)

\(=1\left(1+2\right)+2\left(2+2\right)+\cdots+2018\left(2018+2\right)\)

\(=\left(1^2+2^2+\cdots+2018^2\right)+2\left(1+2+\cdots+2018\right)\)

\(=\frac{2018\left(2018+1\right)\left(2\cdot2018+1\right)}{6}+2\cdot\frac{2018\cdot2019}{2}\)

\(=\frac{2018\cdot2019\cdot4037}{6}+2018\cdot2019=1009\cdot673\cdot4037+2018\cdot2019\)

\(=1009\cdot673\left(4037+2\cdot3\right)=1009\cdot673\cdot4043\)

=2745427451

c: Ta có: \(C=\left(1+2+3+\cdots+2020\right)\left(20192019\cdot2020-2019\cdot20202020\right)+\left(15-12\right)^2\)

\(=\frac{2020\left(2020+1\right)}{2}\cdot2019\cdot2020\cdot\left(10001-10001\right)+3^2\)

=0+9

=9

\(A=\frac{2^2}{1.3}\cdot\frac{3^2}{2.4}....\frac{999^2}{998.1000}\)

\(A=\frac{2^2.3^2....999^2}{1.3.2.4.998.100}=\frac{\left(2.3.....999\right)\left(2.3....999\right)}{\left(1.2....998\right)\left(3.4....1000\right)}\)

\(A=999\cdot\frac{1}{500}=\frac{999}{500}\)( khúc này mk làm tắt, bn bỏ dấu ở trên rồi bỏ từng tử)

=?????????????????????????????????????????????????????????????????????????????????????????????????????????????????

a: \(274+351+126-151\)

\(=274+126+351-151\)

=400+200

=600

b: \(5^7:5^5\cdot3\cdot7\cdot4\)

\(=5^2\cdot21\cdot4\)

\(=21\cdot100=2100\)

c: \(164\cdot27+164\cdot73-400\)

\(=164\left(27+73\right)-400\)

=16400-400

=16000

d: \(300-2\cdot\left[\left(6^2-32\right):2^2+99\right]\)

\(=300-2\left[\dfrac{\left(36-32\right)}{4}+99\right]\)

=300-200

=100

= 1.( 2 + 1 ) + 2 ( 3 + 1 ) + 3( 4 + 1 ) + ..... + 98( 99 + 1 )

 = 1 . 2 + 1 + 2 . 3 + 2 + 3 . 4 + 3 + ...... 98 . 99 + 98

= ( 1 . 2 + 2 . 3 + 3 . 4 + ..... + 98 . 99 ) + ( 1 + 2 + 3 + ..... + 98 )

= 323400 + 4851 

= 328351

bn ủng hộ vũ khánh linh nha