\(a,=\left[\left(-65\right)+\left(-39\right)\right]+\left(92+42\right)=-104+134=30\\ b,=\left(-57+17\right)+\left(40+21\right)=-40+61=21\\ c,=\left(137-157\right)+\left(2040-40\right)=-20+2000=1980\)

B = 2²⁰²¹  - 2²⁰²⁰ - 2²⁰¹⁹ -...- 2 -1

B = 2²⁰²¹ - (2²⁰²⁰ + 2²⁰¹⁹ +...+2 +1)

Đặt         C =              2²⁰²⁰ + 2²⁰¹⁹ +...+ 2 + 1

             2C = 2²⁰²¹ + 2²⁰²⁰ + 2²⁰¹⁹+....+ 2 + 1

       2C - C = 2²⁰²¹ - 1

               C = 2²⁰²¹ - 1

B = 2²⁰²¹ - (2²⁰²¹ -1)

B = 2²⁰²¹ - 2²⁰²¹ + 1

B  = 1

(-157). (127 – 316) – 127. (316 – 157)

= (- 157). 127 + (-157). (-316) + (–127). 316 + (-127). (-157)

= -157. 127 + 157. 316 – 127. 316 + 127. 157

= [- (127. 157) + 127. 157] + (157. 316 – 127. 316)

= 0 + 316. (157 – 127)

= 316. 30

= 9 480.

\(\left(-157\right)\left(127-316\right)-127\left(316-157\right)=-157.127+157.316-127.316+157.127\)

\(=157.316-127.316=316\left(157-127\right)=316.30=9480\)

a) = 3807 - 3800 = 7

b) = 1999

c) = -900

a, (-16).75.125.(-2)

= (-8). 2 . 125. 75. (-2)

= (-8).125 . (-2). 2.75

= -1000. (-300)

= 300 000

b,(-137).73 - 137.27

= (-137). 73 + (-137). 27

= (-137). (73+27)

= -137. 100

= - 13700

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

gợi ý 

a, phá ngoặc và nhóm những hạng tử sao cho phù hợp

b,nhân phân phối như thế này nè :=156x(-137)+156x(-56)-56x(-156)-(-56)x(-137) rồi nhóm nhưng hạng tử sao cho phủ hợp

Nếu bạn ko hiểu thì cứ nói nha mình sẽ giải chi tiết cho(thấy đúng thì t.i.c.k minh nha)

\(\frac{1}{3}+\frac{2}{3^2}\frac{3}{3^3}+...+\frac{2019}{3^{2019}}\)