\(B=1-\left(\dfrac{1}{2.6}+\dfrac{1}{4.9}+\dfrac{1}{6.12}+...+\dfrac{1}{35.67}+\dfrac{1}{38.60}\right)\left(1\right)\)

Đặt \(S=\dfrac{1}{2.6}+\dfrac{1}{4.9}+\dfrac{1}{6.12}+...+\dfrac{1}{35.67}+\dfrac{1}{38.60}\)

\(S=\dfrac{1}{2.3.\left(1.2\right)}+\dfrac{1}{2.3.\left(2.3\right)}+\dfrac{1}{2.3.\left(3.4\right)}+...+\dfrac{1}{2.3.\left(18.19\right)}+\dfrac{1}{2.3.\left(19.20\right)}\)

\(S=\dfrac{1}{6}.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{18.19}+\dfrac{1}{19.20}\right)\)

\(S=\dfrac{1}{6}.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{18}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{20}\right)\)

\(S=\dfrac{1}{6}.\left(1-\dfrac{1}{20}\right)=\dfrac{1}{6}.\dfrac{19}{20}=\dfrac{19}{120}\)

\(\left(1\right)\Rightarrow B=1-\dfrac{19}{120}=\dfrac{101}{120}\)

Đạ biểu thức trong dấu ngoặc đơn là A

\(A=\dfrac{1}{2.1.3.2}+\dfrac{1}{2.2.3.3}+\dfrac{1}{2.3.3.4}+\dfrac{1}{2.4.3.5}+...+\dfrac{1}{2.18.3.19}+\dfrac{1}{2.19.3.20}=\)

\(=\dfrac{1}{2.3}\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{18.19}+\dfrac{1}{19.20}\right)=\)

Đặt biểu thức trong dấu ngoặc đơn là C

\(C=\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+\dfrac{4-3}{3.4}+...+\dfrac{20-19}{19.20}=\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{19}-\dfrac{1}{20}=\)

\(=1-\dfrac{1}{20}=\dfrac{19}{20}\)

\(\Rightarrow B=1-\dfrac{1}{6}.C=1-\dfrac{1}{6}.\dfrac{19}{20}=\dfrac{101}{120}\)

Đặt BT là A

\(\Rightarrow A=2016-\left(\frac{1}{1.2.6}+\frac{1}{2.3.6}+\frac{1}{3.4.6}+....+\frac{1}{19.20.6}\right)\)

\(\Rightarrow A=2016-\frac{1}{6}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{19}-\frac{1}{20}\right)\)

\(\Rightarrow A=2016-\frac{1}{6}\left(1-\frac{1}{20}\right)\)

\(A=2016-\frac{1}{6}.\frac{19}{20}=2016-\frac{19}{120}=\frac{241901}{120}\)

đúng ko bn

2a/ Ta có: \(\left|x+1\right|\ge0\Rightarrow A=\left|x+1\right|+5\ge5\) 

Đẳng thức xảy ra khi: |x + 1| = 0  => x = -1

Vậy giá trị nhỏ nhất của A là 5 khi x = -1

ta có \(\frac{1+5y}{5x}\)=\(\frac{1+7y}{4x}\)

=>      4x(1+5y)=5x(1+7y)

=>      4x+20xy=5x+35xy

=>      4x-5x    =35xy-20xy

=>      -x          =15xy

=>      -1          =15y

=>      y           =\(\frac{-1}{15}\)

có y roi thi có thể dễ dàng tìm được x=-2

A= 3x³ - (3x -2)x²  - 2x(x+1)

A= 3x³ - 3x³ + 2x² - 2x² -2x

A= -2x

Thay x =-20 vào A ta được:

A = -2.(-20) = 40

Vậy A= 40 khi x = -20 

b) C= x(2x+1) - x²(x+2) + x³ -x + 3

C= 2x² + x - x³ - 2x² + x³ -x +3

C= (2x² - 2x²) + (x-x) - (x³ -x³) +3 

C = 3

Vậy C= 3

a. ta có : \(\frac{5}{-3}=\frac{15}{-9}=-\frac{15}{9}\)

b.\(-\frac{1}{5}< 0< \frac{1}{100}\Rightarrow-\frac{1}{5}< \frac{1}{100}\)

c.\(\hept{\begin{cases}2^3=8\\3^2=9\end{cases}\Rightarrow2^3< 3^2}\)