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Câu hỏi của Trần Minh Hưng - Toán lớp | Học trực tuyến
(1 - 1/7) x (1 - 2/7) x ............(1 - 10/7)
= 6/7 x 5/7 x ..... x -3/7
=6/7 x 5/7x 4/7 x 3/7x 2/7 x 1/7 x 0 x -1/7x -2/7 x -3/7
=0
nhớ tk cho mình nha
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}.\)
Ta có :
\(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3}\)
..................................
\(\frac{1}{8^2}=\frac{1}{8.8}< \frac{1}{7.8}\)
Vậy:
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{8^2}< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{7.8}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{8^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{7}-\frac{1}{8}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{8^2}< 1-\frac{1}{8}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{8^2}< \frac{7}{8}\)
Ta có : \(\frac{7}{8}< 1\)mà \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}< \frac{7}{8}\)
\(\Rightarrow\)\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}< 1\)
Ta có : \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+\frac{1}{8^2}\)
\(=\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6}+\frac{1}{7.7}+\frac{1}{8.8}\)
\(< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=1-\frac{1}{8}< 1\)(đpcm)
A) \(\frac{1}{2}\cdot\left(\frac{2}{9}+\frac{3}{7}-\frac{5}{27}\right)\)
\(=\frac{1}{2}\cdot\frac{1}{2}\)
\(=\frac{1}{4}\)
B) \(\left(\frac{-5}{28}+1.75+\frac{8}{35}\right):\left(-3\frac{9}{20}\right)\)
\(=\left(\frac{-5}{28}+\frac{7}{4}+\frac{8}{35}\right):\frac{-69}{20}\)
\(=\frac{14}{5}:\frac{-69}{20}\)
\(=\frac{-56}{69}\)
Bài làm:
Xét: \(\frac{1}{5^2}>\frac{1}{5.6}\) ; \(\frac{1}{6^2}>\frac{1}{6.7}\) ; ... ; \(\frac{1}{100^2}>\frac{1}{100.101}\)
=> \(A>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)
\(=\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{1}{6}\) (1)
Lại có: \(\frac{1}{5^2}< \frac{1}{4.5}\) ; \(\frac{1}{6^2}< \frac{1}{5.6}\) ; ... ; \(\frac{1}{100^2}< \frac{1}{99.100}\)
=> \(A< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}\) (2)
Từ (1) và (2) => \(\frac{1}{6}< A< \frac{1}{4}\)
1,\(\left(\frac{7}{2}-2x\right).\frac{4}{3}=\frac{22}{3}\)
\(x.\left(\frac{7}{2}-2\right)=\frac{22}{3}:\frac{4}{3}=\frac{22}{3}.\frac{3}{4}=\frac{11}{2}\)
\(x.\frac{3}{2}=\frac{11}{2}\)
\(x=\frac{11}{2}:\frac{3}{2}=\frac{11}{2}.\frac{2}{3}=\frac{11}{3}\)
a) =-5/7 +7/8-2/7+1/8- -1/12+ -13/12
=(-5/7-2/7)+(7/8+1/8)-(-1/12--13/12)
=-7/7+8/8 - 12/12
= -1+1+1
=1
b)= ( -3/8+11/8)-(12/11+ -1/11)+(-3/5- 2/5)
= 1- 1 + (-1)
=-1
\(A=\frac{1}{7^2}-\frac{1}{7^4}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}\)
\(\Rightarrow7^2.A=\frac{1}{1}-\frac{1}{7^2}+...+\frac{1}{7^{96}}-\frac{1}{7^{98}}\)
\(\Rightarrow49A+A=1-\frac{1}{7^{100}}\)
\(50A=1-\frac{1}{7^{100}}
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