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

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

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

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

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

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

Có sai đề ko bạn

\(S=\frac{101}{102}+\frac{1}{1.2.2.3}+\frac{1}{2.3.2.3}+\frac{1}{3.4.2.3}+...+\frac{1}{17.18.2.3}=\frac{101}{102}+\frac{1}{6}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{17.18}\right)\)

Đặt BT trong ngoặc đơn là A

\(A=\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{18-17}{17.18}=1-\frac{1}{18}=\frac{17}{18}\)

\(S=\frac{101}{120}+\frac{1}{6}.\frac{17}{18}\)

Nhầm

\(\frac{1}{36.57}=\frac{1}{18.19.2.3}\) Rồi làm lại như trên

Đặt BT là A

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

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

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

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

9/120 nha bạn

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