\(S=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{1001\cdot1003}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{1001}-\dfrac{1}{1003}\)

\(=1-\dfrac{1}{1003}< 1\)

a: Những tia trên hình vẽ là Ex,Ey,Em,En,Ct,CK,Cn

Đoạn thẳng: EK,EC,CK

b: Các cặp tia đối nhau là:

Ex;Ey

Kx;Ky

Cn;CE

CK,Ct

\(A=\dfrac{1}{2}+\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2+...+\left(\dfrac{3}{2}\right)^{2023}\)

Đặt: \(C=\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2+...+\left(\dfrac{3}{2}\right)^{2023}\)

\(\dfrac{3}{2}C=\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+...+\left(\dfrac{3}{2}\right)^{2024}\)

\(\dfrac{3}{2}C-C=\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+...+\left(\dfrac{3}{2}\right)^{2024}-\dfrac{3}{2}-\left(\dfrac{3}{2}\right)^2-...-\left(\dfrac{3}{2}\right)^{2023}\)

\(\dfrac{1}{2}C=\left(\dfrac{3}{2}\right)^{2024}-\dfrac{3}{2}\) 

\(C=2\left(\dfrac{3}{2}\right)^{2024}-3\)

\(\Rightarrow A=\dfrac{1}{2}+2\left(\dfrac{3}{2}\right)^{2024}-3\)

\(=2\left(\dfrac{3}{2}\right)^{2024}-\dfrac{5}{2}\)

\(\Rightarrow A-B=2\left(\dfrac{3}{2}\right)^{2024}-\dfrac{5}{2}-2\left(\dfrac{3}{2}\right)^{2024}=-\dfrac{5}{2}\)

Xác suất thực nghiệm không phải mặt 4 chấm là:

\(\dfrac{40-13}{40}=\dfrac{27}{40}\)

a) O nằm giữa A và B nên:

\(AB=OA+OB\)

\(\Rightarrow OA=1,5+3=4,5\left(cm\right)\)

b) C nằm giữa O và B 

\(OB=OC+BC\)

\(\Rightarrow BC=OB-OC\)

\(\Rightarrow BC=3-1,5=1,5\left(cm\right)\)

\(OC=BC=1,5\left(cm\right)\)

TH1: với \(a>b\) 

\(\dfrac{a}{b}-1=\dfrac{a-b}{b}\)

\(\dfrac{a+n}{b+n}-1=\dfrac{a+n-\left(b+n\right)}{b+n}=\dfrac{a-b}{b+n}\)

Mà: \(b+n>b\)

\(\Rightarrow\dfrac{a-b}{b+n}< \dfrac{a-b}{b}\)

\(\Rightarrow\dfrac{a+n}{b+n}-1< \dfrac{a}{b}-1\)

\(\Rightarrow\dfrac{a+n}{b+n}< \dfrac{a}{b}\)

TH2: với \(a< b\)

\(1-\dfrac{a}{b}=\dfrac{b-a}{b}\)

\(1-\dfrac{a+n}{b+n}=\dfrac{\left(b+n\right)-\left(a+n\right)}{b+n}=\dfrac{b-a}{b+n}\)

Mà: \(b+n>b\)

\(\Rightarrow\dfrac{b-a}{b+n}< \dfrac{b-a}{b}\)

\(\Rightarrow1-\dfrac{a+n}{b+n}< 1-\dfrac{a}{b}\)

\(\Rightarrow\dfrac{a+n}{b+n}>\dfrac{a}{b}\) 

TH3: \(a=b\)

\(\dfrac{a+n}{b+n}=\dfrac{a+n}{a+n}=1\)

\(\dfrac{a}{b}=\dfrac{a}{a}=1\)

\(\Rightarrow\dfrac{a}{b}=\dfrac{a+n}{b+n}=1\)

\(S=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{512}\\ 2S=1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{256}\\ 2S-S=\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+...+\dfrac{1}{256}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{512}\right)\\ S=1-\dfrac{1}{512}=\dfrac{511}{512}\)

Lời giải:

$S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}$

$2S=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^8}$

$\Rightarrow 2S-S=1-\frac{1}{2^9}$

$\Rightarrow S=1-\frac{1}{2^9}$

\(S=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)\left(1-\dfrac{1}{5^2}\right)\left(1-\dfrac{1}{6^2}\right)...\left(1-\dfrac{1}{99^2}\right)\\ =\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)...\left(1-\dfrac{1}{9801}\right)\\ =\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}...\dfrac{9800}{9801}\\ =\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}...\dfrac{98.100}{99.99}\\ =\dfrac{1.2.3...98}{2.3.4...99}.\dfrac{3.4.5...100}{2.3.4...99}\\ =\dfrac{1}{99}.\dfrac{100}{2}=\dfrac{50}{99}\)

Lời giải:

\(S=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{98.100}{99^2}\\ =\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}....\frac{98.100}{99^2}\\ =\frac{1.3.2.4.3.5....98.100}{2^2.3^2.4^2...99^2}\\ =\frac{(1.2.3...98)(3.4.5..100)}{(2.3.4...99)(2.3.4...99)}\)

\(=\frac{1.2.3...98}{2.3.4...99}.\frac{3.4.5..100}{2.3.4...99}=\frac{1}{99}.\frac{100}{2}=\frac{100}{198}\)