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a, \(M=\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{97.99}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{97}-\dfrac{1}{99}\)
\(=\dfrac{1}{3}-\dfrac{1}{99}\)
\(=\dfrac{32}{99}\)
Vậy \(M=\dfrac{32}{99}\)
b, Ta có: \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2012^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2011.2012}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2011}-\dfrac{1}{2012}\)
\(=1-\dfrac{1}{2012}< 1\) (1)
Do mỗi phân số đều lớn hơn 0 nên \(A>0\) (2)
Từ (1), (2) \(\Rightarrow0< A< 1\)
\(\Rightarrow A\notin N\left(đpcm\right)\)
Vậy...
a, \(M=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{97.99}\\ =\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{2}{97}-\dfrac{2}{99}\\ =\dfrac{1}{3}-\dfrac{2}{99}=\dfrac{31}{99}\)
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2012^2}+\dfrac{1}{2013^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{2011.2012}+\dfrac{1}{2012.2013}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2012}-\dfrac{1}{2013}\)
\(=1-\dfrac{1}{2013}\)
\(\Rightarrow A< 1-\dfrac{1}{2013}\)
\(\Rightarrow A< 1\) ( đpcm )
mình gợi ý nè :
Chứng minh A <\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
Bài 1:
a) Ta có:
\(3^{200}=\left(3^2\right)^{100}=9^{100}\)
\(2^{300}=\left(2^3\right)^{100}=8^{100}\)
Vì \(9^{100}>8^{100}\Rightarrow3^{200}>2^{300}\)
b) Ta có:
\(71^{50}=\left(71^2\right)^{25}=5041^{25}\)
\(37^{75}=\left(37^3\right)^{25}=50653^{25}\)
Vì \(5041^{25}< 50653^{25}\Rightarrow71^{50}< 37^{75}\)
c) Ta có:
\(\frac{201201}{202202}=\frac{201.1001}{202.1001}=\frac{201}{202}\)
\(\frac{201201201}{202202202}=\frac{201.1001001}{202.1001001}=\frac{201}{202}\)
\(\Rightarrow\frac{201201}{202202}=\frac{201201201}{202202202}\)
Bài 2:
a) \(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{50^2}\)
Ta có: \(\frac{1}{1^2}=1;\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};....;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(\Rightarrow A< 1+1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow A< 1+1-\frac{1}{50}\)
\(\Rightarrow A< 2-\frac{1}{50}< 2\)
b) \(B=2^1+2^2+2^3+...+2^{30}\) (Có 30 số hạng)
\(\Rightarrow B=\left(2^1+2^2+...+2^5+2^6\right)+\left(2^7+2^8+2^9+...+2^{12}\right)+...+\left(2^{25}+2^{26}+...+2^{29}+2^{30}\right)\)
(có \(30:6=5\) nhóm)
\(\Rightarrow B=1\left(2^1+2^2+...+2^6\right)+2^6\left(2^1+2^2+...+2^6\right)+.....+2^{24}\left(2^1+2^2+...+2^6\right)\)
\(\Rightarrow B=1.126+2^6.126+2^{12}.126+...+2^{24}.126\)
\(\Rightarrow B=126.\left(1+2^6+2^{12}+...+2^{24}\right)\)
\(\Rightarrow B=21.6.\left(1+2^6+2^{12}+...+2^{24}\right)⋮21\)
\(\Rightarrow B⋮21\)
3/ Chu vi hình chữ nhật:
\(\left(\dfrac{1}{4}+\dfrac{3}{10}\right)\cdot2=\dfrac{11}{10}\) (chưa biết đơn vị)
Diện tích hình chữ nhật:
\(\dfrac{1}{4}\cdot\dfrac{3}{10}=\dfrac{11}{20}\) (chưa biết đơn vị)
\(A=2.\left(\dfrac{1}{2.5}+\dfrac{1}{5.8}+...+\dfrac{1}{95.98}\right)\)
\(A=\dfrac{2}{3}\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+....+\dfrac{3}{95.98}\right)\)
\(A=\dfrac{2}{3}\left(\dfrac{1}{2}-\dfrac{1}{98}\right)\)
\(A=\dfrac{2}{3}\dfrac{24}{49}=\dfrac{16}{49}\)
Ta có: A=\(\dfrac{2}{2.5}+\dfrac{2}{5.8}+\dfrac{2}{8.11}+...+\dfrac{2}{95.98}\)
\(\Rightarrow A=\dfrac{3}{2}.\left(\dfrac{3}{2.5}+\dfrac{3}{5.8}+\dfrac{3}{8.11}+...+\dfrac{3}{95.98}\right)\)\(\Rightarrow A=\dfrac{3}{2}.\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+...+\dfrac{1}{95}-\dfrac{1}{98}\right)\)\(\Rightarrow A=\dfrac{3}{2}.\left(\dfrac{1}{2}-\dfrac{1}{98}\right)\)
\(\Rightarrow A=\dfrac{3}{2}.\left(\dfrac{49}{98}-\dfrac{1}{98}\right)\)
\(\Rightarrow A=\dfrac{3}{2}.\dfrac{48}{98}\)
\(\Rightarrow A=\dfrac{3.2.2.12}{2.2.49}\)
\(\Rightarrow A=\dfrac{36}{49}\)
a) Nếu:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in Z\right)\)
\(\Rightarrow B=\dfrac{5^{12}+2}{5^{13}+2}< 1\)
\(B< \dfrac{5^{12}+2+48}{5^{13}+2+48}\Rightarrow B< \dfrac{5^{12}+50}{5^{13}+50}\Rightarrow B< \dfrac{5^2\left(5^{10}+2\right)}{5^2\left(5^{11}+2\right)}\Rightarrow B< \dfrac{5^{10}+2}{5^{11}+2}=A\)\(B< A\)
bạn ơi thế còn phần b thì sao? Mong bạn có câu trả lời sớm tớ cảm ơn bạn nhiều lắm