Ta có: \(\dfrac{101+100+99+...+3+2+1}{101-100+99-98+...+3-2+1}\)

\(=\dfrac{101+\left(100+1\right)\cdot50}{101-\left[100-99+98-97+...+2-1\right]}\)

\(=\dfrac{101\cdot51}{101-1\cdot50}\)

\(=\dfrac{101\cdot51}{101-50}=101\)

c.mơn bn nhá. ~THANK YOU~

c, 4C= (1.2.3+2.3.4+3.4.5+...+8.9.10) .4

==> 4C= [1.2.3.(4-0) + 2.3.4-(5-1) + 8.9.10.(11-7)

==>4C= 1.2.3.4 - 1.2.3.4+ 2.3.4.5-2.3.4.5 + 7.8.9.10- 7.8.9.10 + 8.9.10.11

==> 4C= 8.9.10.11=7920

==> C= 7920 :4=1980

a, Ta có: 3A= 1.2.3+2.3.3+3.4.3+...+99.100.3

               3A=1.2.(3-0) + 2.3.(4-1)+ 3.4.(5-2)+ ... + 99.100.( 101-98)

               3A=(1.2.3 + 2.3.4+ 3.4.5+ 99.100.101) - (0.1.2 +1.2.3+ 2.3.4 + ... + 98.99.100)

               3A= 99.100.101 - 0.1.2

               3A= 999900 - 0

               3A= 999900

    ==> A= 999900 : 3

   ==> A= 333300

2A=1-1/2+1/2^2-...+1/2^98-1/2^99

=>3A=1-1/2^100

=>\(A=\dfrac{2^{100}-1}{3\cdot2^{100}}\)

Ta có: \(1\cdot2\cdot3\cdot\ldots\cdot2015-1\cdot2\cdot3\cdot\ldots\cdot2014-1\cdot2\cdot3\cdot\ldots\cdot2014^2\)

\(=1\cdot2\cdot\ldots\cdot2014\left(2015-1-2014\right)\)

\(=1\cdot2\cdot\ldots\cdot2014\cdot0=0\)

 2.-6.12²:4.-6.12

=-1728:-288

=6

Vậy giá trị của biểu thức là 6

(2ab²):abc=[2.4.(-6)²]: [4.(-6).12]

=(8.36):(-288)

=288:(-288)

=-1

Học tốt!!!

Ta có : \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)

\(\Leftrightarrow2B=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{18.19.20}\)

\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+....+\frac{1}{18.19}-\frac{1}{19.20}\)

\(=\frac{1}{2}-\frac{1}{19.20}=\frac{189}{380}\)

\(\Rightarrow B=\frac{189}{760}\)

\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{18.19.20}\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}\right)+\frac{1}{2}\left(\frac{1}{2.3}-\frac{1}{3.4}\right)+...+\frac{1}{2}\left(\frac{1}{18.19}-\frac{1}{19.20}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\right)\)

\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{19.20}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{2}-\frac{1}{380}\right)\)

\(=\frac{1}{2}.\frac{189}{380}=\frac{189}{760}\)

A=2¹⁰⁰-2⁹⁹-2⁹⁸-...-2²-2-1

\(\Rightarrow\)2A=2¹⁰¹-2¹⁰⁰-2⁹⁹-...-2³-2²-2

\(\Rightarrow\)2A+A=(2¹⁰¹-2¹⁰⁰-2⁹⁹-...-2³-2²-2)+(2¹⁰⁰-2⁹⁹-2⁹⁸-...-2²-2-1)

\(\Rightarrow\)3A=2¹⁰¹+1

\(\Rightarrow\)A=\(\frac{2^{101}+1}{3}\)

Vậy A=\(\frac{2^{101}+1}{3}\).

Ta có : A = 2¹⁰⁰ - 2⁹⁹ - 2⁹⁸ - ... - 2² - 2 - 1 

=> 2A = 2¹⁰¹ - 2¹⁰⁰ - 2⁹⁹ - ... - 2³ - 2² - 2

Lấy A - 2A = (2¹⁰⁰ - 2⁹⁹ - 2⁹⁸ - ... - 2² - 2 - 1) - (2¹⁰¹ - 2¹⁰⁰ - 2⁹⁹ - ... - 2³ - 2² - 2)

   => - A     = 2¹⁰⁰ + 2¹⁰⁰ - 2¹⁰¹ - 1

   => - A     = 2.2¹⁰⁰ - 2¹⁰¹ - 1

   => - A     = 2¹⁰¹ - 2¹⁰¹ - 1

   => - A     = - 1

  => A = 1