Question

[Ngọc Minh]

Bài 1 : Tính giá trị chưa biết :

\(1^1_{30}:(24^1_6-24^1_5)-1^1_2-\dfrac{3}{4}:4x-\dfrac{1}{2}=(-1^1_{15}):\left(8^1_5-8^1_3\right)\)

Bài 2 : Chứng Minh Rằng

A = \(\dfrac{36}{1.3.5}+\dfrac{36}{3.5.7}+...+\dfrac{36}{25.27.29}< 3\)

B= \(\dfrac{1}{12}+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}< 2\)

Bài 3 : Cho P = \(\dfrac{2}{60.63}+\dfrac{2}{63.66}+...+\dfrac{2}{117+120}+\dfrac{2}{2011}\)

Q =\(\dfrac{5}{40.44}+\dfrac{5}{44.48}+...+\dfrac{5}{76.80}+\dfrac{5}{2011}\)

So sánh P và Q

Answer 1

[Mới vô]

2.

\(A=\dfrac{36}{1\cdot3\cdot5}+\dfrac{36}{3\cdot5\cdot7}+...+\dfrac{36}{25\cdot27\cdot29}\\ =9\cdot\left(\dfrac{4}{1\cdot3\cdot5}+\dfrac{4}{3\cdot5\cdot7}+...+\dfrac{4}{25\cdot27\cdot29}\right)\\ =9\cdot\left(\dfrac{1}{1\cdot3}-\dfrac{1}{3\cdot5}+\dfrac{1}{3\cdot5}-\dfrac{1}{5\cdot7}+...+\dfrac{1}{25\cdot27}-\dfrac{1}{27\cdot29}\right)\\ =9\cdot\left(\dfrac{1}{1\cdot3}-\dfrac{1}{27\cdot29}\right)\\ =9\cdot\left(\dfrac{1}{3}-\dfrac{1}{783}\right)\\ =9\cdot\dfrac{1}{3}-9\cdot\dfrac{1}{783}\\ =3-\dfrac{1}{87}< 3\)

Vậy \(A< 3\)

b,

\(B=\dfrac{1}{1^2}+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}\\ B=1+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}\\ B< 1+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\\ B< 1+\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ B< 1+\dfrac{1}{1}-\dfrac{1}{50}\\ B< 2-\dfrac{1}{50}< 2\)

Vậy \(B< 2\)

Answer 2

[Mới vô]

\(P=\dfrac{2}{60\cdot63}+\dfrac{2}{63\cdot66}+...+\dfrac{2}{117\cdot120}+\dfrac{2}{2011}\\ =\dfrac{2}{3}\cdot\left(\dfrac{3}{60\cdot63}+\dfrac{3}{63\cdot66}+...+\dfrac{3}{117\cdot120}+\dfrac{3}{2011}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{1}{60}-\dfrac{1}{63}+\dfrac{1}{63}-\dfrac{1}{66}+...+\dfrac{1}{117}-\dfrac{1}{120}+\dfrac{3}{2011}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{1}{60}-\dfrac{1}{120}+\dfrac{3}{2011}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{1}{2}+\dfrac{3}{2011}\right)\)

\(Q=\dfrac{5}{40\cdot44}+\dfrac{5}{44\cdot48}+...+\dfrac{5}{76\cdot80}+\dfrac{5}{2011}\\ =\dfrac{5}{4}\cdot\left(\dfrac{4}{40\cdot44}+\dfrac{4}{44\cdot48}+...+\dfrac{4}{76\cdot80}+\dfrac{4}{2011}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{40}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{48}+...+\dfrac{1}{76}-\dfrac{1}{80}+\dfrac{4}{2011}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{40}-\dfrac{1}{80}+\dfrac{4}{2011}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{2}+\dfrac{4}{2011}\right)\)

\(\dfrac{3}{2011}< \dfrac{4}{2011}\Rightarrow\dfrac{1}{2}+\dfrac{3}{2011}< \dfrac{1}{2}+\dfrac{4}{2011}\left(1\right)\)

\(\dfrac{2}{3}< \dfrac{5}{4}\left(2\right)\)

Từ (1) và (2) ta có: \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{2011}\right)< \dfrac{5}{4}\left(\dfrac{1}{2}+\dfrac{4}{2011}\right)\Leftrightarrow P< Q\)

Vậy P < Q