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\(\dfrac{1}{1\cdot2}>\dfrac{1}{2^2}>\dfrac{1}{2\cdot3},\dfrac{1}{2\cdot3}>\dfrac{1}{3^2}>\dfrac{1}{3\cdot4},...,\dfrac{1}{8\cdot9}>\dfrac{1}{9^2}>\dfrac{1}{9\cdot10}\)
\(\Rightarrow\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{8\cdot9}>\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}>\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\) \(\Rightarrow1-\dfrac{1}{9}>A>\dfrac{1}{2}-\dfrac{1}{10}\) \(\Rightarrow\dfrac{8}{9}>A>\dfrac{2}{5}\)
Câu 1.8: Giải
*Ta có: \(\dfrac{1}{2^2}=\dfrac{1}{2.2}>\dfrac{1}{2.3}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}>\dfrac{1}{3.4}\)
...
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}< \dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)
\(A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(A>\dfrac{1}{2}-\dfrac{1}{10}\)
\(A>\dfrac{2}{5}\) (1)
*Ta có: \(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
...
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}< \dfrac{1}{8.9}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{8.9}\)
\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
\(A< 1-\dfrac{1}{9}\)
\(A< \dfrac{8}{9}\) (2)
Từ (1) và (2) \(\Rightarrow\dfrac{2}{5}< A< \dfrac{8}{9}\)
`A=(8 2/7-4 2/7)-3 4/9`
`=8+2/7-4-2/7-3-4/9`
`=4-3-4/9`
`=1-4/9=5/9`
`B=(10 2/9-6 2/9)+2 3/5`
`=10+2/9-6-2/9+2+3/5`
`=4+2+3/5`
`=6+3/5=33/5`
Bài 2:
`a)5 1/2*3 1/4`
`=11/2*13/4`
`=143/8`
`b)6 1/3:4 2/9`
`=19/3:38/9`
`=19/3*9/38=3/2`
`c)4 3/7*2`
`=31/7*2`
`=62/7`
Bài 1:
\(A=\left(8\dfrac{2}{7}-4\dfrac{2}{7}\right)-3\dfrac{4}{9}\)
\(A=\left(\dfrac{58}{7}-\dfrac{30}{7}\right)-\dfrac{31}{9}\)
\(A=4-\dfrac{31}{9}\)
\(A=\dfrac{5}{9}\)
\(B=\left(10\dfrac{2}{9}-6\dfrac{2}{9}\right)+2\dfrac{3}{5}\)
\(B=\left(\dfrac{92}{9}-\dfrac{56}{9}\right)+\dfrac{13}{5}\)
\(B=4+\dfrac{13}{5}\)
\(B=\dfrac{33}{5}\)
\(a,\dfrac{3}{5}+\dfrac{-5}{9}=\dfrac{27-25}{45}=\dfrac{2}{49}.\)
\(c,\dfrac{-27}{23}+\dfrac{5}{21}+\dfrac{4}{23}+\dfrac{16}{21}+\dfrac{1}{2}=\dfrac{-23}{23}+\dfrac{21}{21}+\dfrac{1}{2}=-1+1+\dfrac{1}{2}=\dfrac{1}{2}.\)
\(d,\dfrac{-8}{9}+\dfrac{1}{9}.\dfrac{2}{9}+\dfrac{1}{9}.\dfrac{7}{9}=\dfrac{-8}{9}+\dfrac{1}{9}.\left(\dfrac{2}{9}+\dfrac{7}{9}\right)=\dfrac{-8}{9}+\dfrac{1}{9}.1=\dfrac{-8+1}{9}=\dfrac{-7}{9}.\)
a) Ta có: \(-3\dfrac{1}{4}\cdot x-75\%+\dfrac{3x}{2}=-1.2:\dfrac{-9}{10}-1\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{-13x}{4}-\dfrac{3}{4}+\dfrac{3x}{2}=\dfrac{-6}{5}\cdot\dfrac{10}{-9}-\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{-13x-3+6x}{4}=\dfrac{4}{3}-\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{-7x-3}{4}=\dfrac{1}{12}\)
\(\Leftrightarrow-7x-3=\dfrac{1}{3}\)
\(\Leftrightarrow-7x=\dfrac{10}{3}\)
hay \(x=-\dfrac{10}{21}\)
b) Ta có: \(\dfrac{5}{3}+\dfrac{5}{15}+\dfrac{5}{35}+...+\dfrac{5}{x\left(x+2\right)}=2\dfrac{8}{17}\)
\(\Leftrightarrow\dfrac{5}{2}\left(\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{x\left(x+2\right)}\right)=2\dfrac{8}{17}\)
\(\Leftrightarrow\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=2+\dfrac{8}{17}\)
\(\Leftrightarrow\left(1-\dfrac{1}{x+2}\right)=\dfrac{42}{17}:\dfrac{5}{2}\)
\(\Leftrightarrow\dfrac{x+1}{x+2}=\dfrac{42}{17}\cdot\dfrac{2}{5}=\dfrac{84}{85}\)
\(\Leftrightarrow85x+85=84x+168\)
\(\Leftrightarrow x=83\)
\(P=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}\)
\(P=\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{9.9}\)
\(P< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{8.9}\)
\(P=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
\(P=1-\dfrac{1}{9}=\dfrac{8}{9}\)
\(\Rightarrow P< \dfrac{8}{9}\)
\(P=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{9^2}\)
\(P=\dfrac{1}{2.2}+\dfrac{1}{3.3}+...+\dfrac{1}{9.9}\)
\(P>\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)
\(P=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...\dfrac{1}{9}-\dfrac{1}{10}\)
\(P=\dfrac{1}{2}-\dfrac{1}{10}=\dfrac{2}{5}\)
\(\Rightarrow P>\dfrac{2}{5}\)
a)\(A=\dfrac{1}{2^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2^2-1}+\dfrac{1}{4^2-1}+...+\dfrac{1}{100^2-1}\)
\(A< \dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{99\cdot101}\)
\(A< \dfrac{1}{2}\cdot\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(A< \dfrac{1}{2}\cdot\left(1-\dfrac{1}{101}\right)=\dfrac{1}{2}\cdot\dfrac{100}{101}=\dfrac{50}{101}< \dfrac{50}{100}=\dfrac{1}{2}\)
Vậy \(A< \dfrac{1}{2}\)
b)B=\(\dfrac{3}{4}+\dfrac{8}{9}+...+\dfrac{2499}{2500}\)
49-B=\(\dfrac{1}{4}+\dfrac{1}{9}+...+\dfrac{1}{2500}=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{50^2}\)
\(49-B< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{49.50}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(49-B< 1-\dfrac{1}{50}< 1\Leftrightarrow49< 1+B\Leftrightarrow B>48\)(ĐPCM)
b) Đặt :
\(A=\dfrac{3}{4}+\dfrac{8}{9}+\dfrac{15}{16}+............+\dfrac{2499}{2500}\)
\(\Rightarrow A=\dfrac{4}{4}-\dfrac{1}{4}+\dfrac{9}{9}-\dfrac{1}{9}+.........+\dfrac{2500}{2500}-\dfrac{1}{2500}\)
\(A=1-\dfrac{1}{2^2}+1-\dfrac{1}{3^2}+...........+1-\dfrac{1}{50^2}\)
\(A=\left(1+1+....+1\right)-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+......+\dfrac{1}{50^2}\right)\)(\(49\) chữ số \(1\))
\(A=49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+........+\dfrac{1}{50^2}\right)\)
Lại có :
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+.....+\dfrac{1}{50^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{49.50}\)
Mà :
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.....+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=1-\dfrac{1}{50}< 1\)
\(\Rightarrow-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+......+\dfrac{1}{50^2}\right)>-1\)
\(\Rightarrow49-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+............+\dfrac{1}{50^2}\right)>49-1\)\(=48\)
\(\Rightarrow A>48\) \(\rightarrowđpcm\)
Ta có: \(\dfrac{1}{2^2}>\dfrac{1}{2\cdot3}\)
\(\dfrac{1}{3^2}>\dfrac{1}{3\cdot4}\)
\(\dfrac{1}{4^2}>\dfrac{1}{4\cdot5}\)
..................
\(\dfrac{1}{9^2}>\dfrac{1}{9\cdot10}\)
\(\Rightarrow\) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}>\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+...+\dfrac{1}{9\cdot10}\)
\(\Rightarrow\) \(A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(\Rightarrow\) \(A>\dfrac{1}{2}-\dfrac{1}{10}\)
\(\Rightarrow\) \(A>\dfrac{5}{10}-\dfrac{1}{10}\)
\(\Rightarrow\) \(A>\dfrac{2}{5}\) (1)
Ta có: \(\dfrac{1}{2^2}< \dfrac{1}{1\cdot2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2\cdot3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3\cdot4}\)
...................
\(\dfrac{1}{9^2}< \dfrac{1}{8\cdot9}\)
\(\Rightarrow\) \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}< \dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{8\cdot9}\)
\(\Rightarrow\) \(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
\(\Rightarrow\) \(A< 1-\dfrac{1}{9}\)
\(\Rightarrow\) \(A< \dfrac{9}{9}-\dfrac{1}{9}\)
\(\Rightarrow\) \(A< \dfrac{8}{9}\) (2)
Từ (1) và (2) ta được: \(\dfrac{2}{5}< A< \dfrac{8}{9}\)
Vậy \(\dfrac{2}{5}< A< \dfrac{8}{9}\).
Mà đề phần kết luận sai nhé, nếu \(\dfrac{1}{n^2}\) thì A đâu lớn hơn \(\dfrac{2}{5}\), phải thay \(\dfrac{1}{n^2}\) thành \(\dfrac{1}{9^2}\) nha