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\(\dfrac{1}{38}>\dfrac{1}{40}>\dfrac{1}{42}>...>\dfrac{1}{50}\)
=>\(\dfrac{1}{38}+\dfrac{1}{40}+\dfrac{1}{42}+\dfrac{1}{44}+\dfrac{1}{46}+\dfrac{1}{48}+\dfrac{1}{50}< 7\cdot\dfrac{1}{38}=\dfrac{7}{38}< 1\)
Vậy tổng trên bé hơn 1
A=-1-3-5-...-2017
=-(1+3+5+...+2017)
Xét tổng B=1+3+5+...+2017
Tổng B có:(2017-1):2+1=1009(số hạng)
Tổng B=\(\dfrac{\left(2017+1\right)\cdot1009}{2}=1009\cdot1009=1018081\)
=>A=-B=-1018081
Bài 2:
Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1.2};\dfrac{1}{3^2}< \dfrac{1}{2.3};....;\dfrac{1}{100^2}< \dfrac{1}{99.100}\)
\(\Rightarrow A< 1+\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}=1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=2-\dfrac{1}{100}< 2\)
Vậy A < 2
Bài 3:
D = \(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right)....\left(1-\dfrac{1}{2015}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}......\dfrac{2014}{2015}\)
\(=\dfrac{1.2......2014}{2.3......2015}=\dfrac{1}{2015}\)
Bài 4:
A = \(\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}......\dfrac{899}{900}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}.\dfrac{3.5}{4.4}........\dfrac{29.31}{30.30}\)
\(=\dfrac{1.2.3......29}{2.3.4.......30}.\dfrac{3.4.5......31}{2.3.4.....30}\)
\(=\dfrac{1}{30}.\dfrac{31}{2}=\dfrac{31}{60}\)
Câu hỏi của Dung Van - Toán lớp 6 | Học trực tuyến tìm kĩ trước khi hỏi nhé.
2) Tinh nhanh:
a) \(\dfrac{5}{23}\) . \(\dfrac{17}{26}\) + \(\dfrac{5}{23}\) . \(\dfrac{10}{26}\) - \(\dfrac{5}{23}\)
= \(\dfrac{5}{23}\) . \(\left(\dfrac{17}{26}+\dfrac{10}{26}-1\right)\)
= \(\dfrac{5}{23}\) . \(\left(\dfrac{27}{26}-1\right)\) = \(\dfrac{5}{23}\) . \(\dfrac{1}{26}\)
= \(\dfrac{5}{598}\)
b) \(\dfrac{1}{7}.\dfrac{5}{9}+\dfrac{5}{9}.\dfrac{2}{7}+\dfrac{5}{9}.\dfrac{1}{7}+\dfrac{5}{9}.\dfrac{3}{7}\)
= \(\dfrac{5}{9}.\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
= \(\dfrac{5}{9}\) . 1= \(\dfrac{5}{9}\)
Áp dụng tính chất : \(\dfrac{a}{b}< 1\Leftrightarrow\dfrac{a}{b}< \dfrac{a+m}{b+m}\) (\(a;b,m\in N\)*)
Ta có :
\(A=\dfrac{100^{2007}+1}{100^{2008}+1}< \dfrac{100^{2007}+1+99}{100^{2008}+1+99}=\dfrac{100^{2007}+100}{100^{2008}+100}=\dfrac{100\left(100^{2006}+1\right)}{100\left(100^{2007}+1\right)}=\dfrac{100^{2006}+1}{100^{2007}+1}=B\)
\(\Rightarrow A< B\)
1) \(19\dfrac{5}{8}:\dfrac{7}{12}-15\dfrac{1}{4}:\dfrac{7}{12}\)
\(=\dfrac{157}{8}\cdot\dfrac{12}{7}-\dfrac{61}{4}\cdot\dfrac{12}{7}\\ =\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{61}{4}\right)\\ =\dfrac{12}{7}\cdot\dfrac{35}{8}\\ =\dfrac{15}{2}\)
2) \(\dfrac{2}{5}\cdot\dfrac{1}{3}-\dfrac{2}{15}:\dfrac{1}{5}+\dfrac{3}{5}\cdot\dfrac{1}{3}\)
\(=\dfrac{1}{3}\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{2}{15}\cdot5\\ =\dfrac{1}{3}\cdot1-\dfrac{2}{3}\\ =\dfrac{1}{3}-\dfrac{2}{3}\\ =-\dfrac{1}{3}\)
3) \(\dfrac{4}{9}\cdot19\dfrac{1}{3}-\dfrac{4}{9}\cdot39\dfrac{1}{3}\)
\(=\dfrac{4}{9}\left(19\dfrac{1}{3}-39\dfrac{1}{3}\right)\\ =\dfrac{4}{9}\cdot\left(\dfrac{58}{3}-\dfrac{118}{3}\right)\\ =\dfrac{4}{9}\cdot\left(-20\right)\\ =-\dfrac{80}{9}\)
\(\dfrac{1}{9}+\dfrac{1}{10}< \dfrac{1}{8}+\dfrac{1}{8}=\dfrac{1}{4}\) và \(\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{40}+\dfrac{1}{40}=\dfrac{1}{20}\)
Suy ra:
\(S=\dfrac{1}{5}+\dfrac{1}{9}+\dfrac{1}{10}+\dfrac{1}{41}+\dfrac{1}{42}< \dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}=\dfrac{1}{2}\)
Các câu dễ bạn tự làm nha:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(A=\dfrac{2017^{2017}+1}{2017^{2018}+1}< 1\)
\(A< \dfrac{2017^{2017}+1+2016}{2017^{2018}+1+2016}\Rightarrow A< \dfrac{2017^{2017}+2017}{2017^{2018}+2017}\Rightarrow A< \dfrac{2017\left(2017^{2016}+1\right)}{2017\left(2017^{2017}+1\right)}\Rightarrow A< \dfrac{2017^{2016}+1}{2017^{2017}+1}=B\)\(A< B\)
Ta có : \(\dfrac{1}{9}=\dfrac{1}{9}\)
\(\dfrac{1}{10}< \dfrac{1}{9}\)
.....
\(\dfrac{1}{19}< \dfrac{1}{9}\)
\(\Rightarrow\dfrac{1}{9}+\dfrac{1}{10}+...+\dfrac{1}{19}< \dfrac{1}{9}+\dfrac{1}{9}+...+\dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{1}{9}+\dfrac{1}{10}+..+\dfrac{1}{19}< \dfrac{11}{9}\)
Hay \(\dfrac{1}{9}+\dfrac{1}{10}+..+\dfrac{1}{19}< \dfrac{9}{9}=1\)
Đặt biểu thức trên là A.
Ta có A có 11 số hạng, chia A thành 2 nhóm, mỗi nhóm có 5 số hạng còn thừa 1 số hạng như sau:
\(A=\dfrac{1}{9}+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{14}\right)+\left(\dfrac{1}{15}+\dfrac{1}{16}+...+\dfrac{1}{19}\right)\)
Lại có: \(\dfrac{1}{10}=\dfrac{1}{10};\dfrac{1}{11}< \dfrac{1}{10};...;\dfrac{1}{14}< \dfrac{1}{10}\) \(\Rightarrow\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{14}< \dfrac{1}{10}+\dfrac{1}{10}+...+\dfrac{1}{10}\) (5 số hạng)
\(\Rightarrow\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{14}< \dfrac{1}{10}.5=\dfrac{1}{2}\) (1)
\(\dfrac{1}{15}=\dfrac{1}{15};\dfrac{1}{16}< \dfrac{1}{15};...;\dfrac{1}{19}< \dfrac{1}{15}\)
\(\Rightarrow\dfrac{1}{15}+\dfrac{1}{16}+...+\dfrac{1}{19}< \dfrac{1}{15}+\dfrac{1}{15}+...+\dfrac{1}{15}\) (5 số hạng)
\(\Rightarrow\dfrac{1}{15}+\dfrac{1}{16}+...+\dfrac{1}{19}< \dfrac{1}{15}.5=\dfrac{1}{3}\)(2)
\(\dfrac{1}{9}=\dfrac{1}{9}\left(3\right)\)
Từ (1) và (2) ta suy ra:
\(\dfrac{1}{9}+\left(\dfrac{1}{10}+\dfrac{1}{11}+...+\dfrac{1}{14}\right)+\left(\dfrac{1}{15}+\dfrac{1}{16}+...+\dfrac{1}{19}\right)< \dfrac{1}{9}+\dfrac{1}{2}+\dfrac{1}{3}\) \(\Rightarrow A< \dfrac{1}{9}+\dfrac{1}{2}+\dfrac{1}{3}\)
\(\Rightarrow A< \dfrac{2}{18}+\dfrac{9}{18}+\dfrac{6}{18}\)
\(\Rightarrow A< \dfrac{2+9+6}{18}\)
\(\Rightarrow A< \dfrac{17}{18}< \dfrac{18}{18}=1\)
\(\Rightarrow A< 1\left(đpcm\right)\)