Chứng minh rằng :
A=\(9\left(\frac{1}{10!}+\frac{1}{11!}+\frac{1}{12!}+...+\frac{1}{2020!}\right)< \frac{1}{9!}\)
Chứng minh rằng :
B=\(9\left(\frac{1}{10!}+\frac{1}{11!}+\frac{1}{12!}+...+\frac{1}{2020!}\right)< \frac{1}{9!}\)
nè mọi người cho mình biết bài này dải thế nào: Tính diện tích tứ giác MBND. Biết hình chữ nhật ABCD có chiều dài DC=36cm chiều rộng d=20cm, AM=13 MB, BM=NC
Phần trả lời của người ta mà ghi câu hỏi vào là sao hả ? Bùi Duy Hưng
\(B=9\left(\frac{1}{10!}+\frac{1}{11!}+\frac{1}{12!}+...+\frac{1}{2020!}\right)\)
\(=\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...+\frac{9}{2020!}\)
Ta có : \(\frac{9}{10!}=\frac{9}{10!}\)
\(\frac{9}{11!}< \frac{10}{11!}\)
\(\frac{9}{12!}< \frac{11}{12!}\)
...
\(\frac{9}{2020!}< \frac{2019}{2020!}\)
\(\Rightarrow B< \frac{9}{10!}+\frac{10}{11!}+\frac{11}{12!}+...+\frac{2019}{2020!}\)
\(B< \frac{10-1}{9!}+\frac{11-1}{10!}+\frac{12-1}{12!}+...+\frac{2020-1}{2020!}\)
\(B< \frac{1}{9!}-\frac{1}{10!}+\frac{1}{10!}-\frac{1}{11!}+\frac{1}{11!}-\frac{1}{12!}+...+\frac{1}{2019!}-\frac{1}{2020!}\)
\(B< \frac{1}{9!}-\frac{1}{2020!}< \frac{1}{9!}\)
\(\Rightarrow B< \frac{1}{9!}\)(đpcm)
Tính bằng cách hợp lí nhất :
1/ \(\left(\frac{4}{9}-\frac{5}{11}\right):\frac{3}{10}+\left(\frac{3}{9}-\frac{9}{11}\right):\frac{3}{10}-\left(\frac{2}{9}-\frac{2}{8}\right).\frac{-10}{3}\)
2/ \(\frac{1}{2}.\frac{1}{-3}+\frac{1}{-3}.\frac{1}{4}+\frac{1}{4}.\frac{1}{-5}+\frac{1}{-5}.\frac{1}{6}\)
3/ \(-\frac{7}{4}\left(\frac{33}{12}+\frac{3333}{2020}+\frac{333333}{303030}+\frac{33333333}{42424242}\right)\)
App giải toán không cần nhập đề chỉ cần chụp ảnh cho cả nhà đây: https://www.facebook.com/watch/?v=485078328966618
Chứng minh rằng \(\left(\frac{9}{11}-0,81\right)^{2008}=\left(\frac{9}{11}\right)^{2008}\times\frac{1}{10^{4016}}\)
Chứng minh rằng: \(\left(\frac{9}{11}-0,81\right)^{2008}=\left(\frac{9}{11}\right)^{2008}\times\frac{1}{10^{4016}}\)
Có: \(\left(\frac{9}{11}-0,81\right)^{2008}=\left(\frac{9}{1100}\right)^{2008}\)
\(\left(\frac{9}{11}\right)^{2008}\times\frac{1}{10^{4016}}=\frac{9^{2008}}{11^{2008}\times\left(10^2\right)^{2008}}=\frac{9^{2008}}{11^{2008}\times100^{2008}}=\frac{9^{2008}}{\left(11\times100\right)^{2008}}=\frac{9^{2008}}{1100^{2008}}=\left(\frac{9}{1100}\right)^{2008}\)
Vì: \(\left(\frac{9}{1100}\right)^{2008}=\left(\frac{9}{1100}\right)^{2008}\Rightarrow\left(\frac{9}{11}-0,81\right)^{2008}=\left(\frac{9}{11}\right)^{2008}\times\frac{1}{10^{4016}}\)
1,Chứng minh rằng
\(\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...+\frac{9}{1000!}< \frac{1}{9!}\)
a.\(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+_{\frac{x-11}{12}}\)
b.\(\frac{x-1}{2020}+\frac{x-2}{2019}-\frac{x-3}{2018}=\frac{x-4}{2017}\)
c.\(\left(\frac{3}{4}x+3\right)-\left(\frac{2}{3}x-4\right)-\left(\frac{1}{6}x+1\right)=\left(\frac{1}{3}x+4\right)-\left(\frac{1}{3}x-3\right)\)
giúp mk vs mk đg cần gấp
Thank you
a) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
=> \(\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)
=> \(\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}+\frac{x+1}{12}=0\)
=> \(\left(x+1\right)\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)=0\)
=> x + 1 = 0
=> x = -1
b) \(\frac{x-1}{2020}+\frac{x-2}{2019}-\frac{x-3}{2018}=\frac{x-4}{2017}\)
=> \(\left(\frac{x-1}{2020}-1\right)+\left(\frac{x-2}{2019}-1\right)-\left(\frac{x-3}{2018}-1\right)=\left(\frac{x-4}{2017}-1\right)\)
=> \(\frac{x-2021}{2020}+\frac{x-2021}{2019}-\frac{x-2021}{2018}=\frac{x-2021}{2017}\)
=> \(\left(x-2021\right)\left(\frac{1}{2020}+\frac{1}{2019}-\frac{1}{2018}-\frac{1}{2017}\right)=0\)
=> x - 2021 = 0
=> x = 2021
c) \(\left(\frac{3}{4}x+3\right)-\left(\frac{2}{3}x-4\right)-\left(\frac{1}{6}x+1\right)=\left(\frac{1}{3}x+4\right)-\left(\frac{1}{3}x-3\right)\)
=> \(\frac{3}{4}x+3-\frac{2}{3}x+4-\frac{1}{6}x-1=\frac{1}{3}x+4-\frac{1}{3}x+3\)
=> \(-\frac{1}{12}x+6=7\)
=> \(-\frac{1}{12}x=1\)
=> x = -12
Chứng minh rằng:
a) \(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}<1\)
b) \(\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...+\frac{9}{1000!}<\frac{1}{9!}\)
Bạn tham khảo nhé
\(a)\)Đặt \(A=\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}\)
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A< 1-\frac{1}{100}=\frac{100-1}{100}=\frac{99}{100}< 1\) ( đpcm )
Vậy \(A< 1\)
Chứng minh rằng:
a) \(\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}<1\)
b) \(\frac{9}{10!}+\frac{9}{11!}+\frac{9}{12!}+...+\frac{9}{1000!}<\frac{1}{9!}\)
Chứng minh: \(\frac{2}{x^2-1}+\frac{4}{x^2-4}+...+\frac{20}{x^2-100}=\frac{11}{\left(x-10\right)\left(x+1\right)}+\frac{11}{\left(x-9\right)\left(x+2\right)}+...+\frac{11}{\left(x-1\right)\left(x+10\right)}\)
Vê trái:
\(=\frac{2}{\left(x-1\right)\left(x+1\right)}+\frac{4}{\left(x-2\right)\left(x+2\right)}+...+\frac{20}{\left(x-10\right)\left(x+10\right)}\)
\(=\frac{\left(x+1\right)-\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{\left(x+2\right)-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+...+\frac{\left(x+10\right)-\left(x-10\right)}{\left(x+10\right)\left(x-10\right)}\)
\(=\frac{1}{x-1}-\frac{1}{x+1}+\frac{1}{x-2}-\frac{1}{x+2}+...+\frac{1}{x-10}-\frac{1}{x+10}\)
\(=\left(\frac{1}{x-1}+\frac{1}{x-2}+...+\frac{1}{x-10}\right)-\left(\frac{1}{x+1}+\frac{1}{x+2}+...+\frac{1}{x+10}\right)\)
Vế phải:
\(=\frac{\left(x+1\right)-\left(x-10\right)}{\left(x-10\right)\left(x+1\right)}+\frac{\left(x+2\right)-\left(x-9\right)}{\left(x-9\right)\left(x+2\right)}+...+\frac{\left(x+10\right)-\left(x-1\right)}{\left(x-1\right)\left(x+10\right)}\)
\(=\frac{1}{x-10}-\frac{1}{x+1}+\frac{1}{x-9}-\frac{1}{x+2}+...+\frac{1}{x-1}-\frac{1}{x+10}\)
\(=\left(\frac{1}{x-1}+\frac{1}{x-2}+...+\frac{1}{x-10}\right)-\left(\frac{1}{x+1}+\frac{1}{x+2}+...+\frac{1}{x+10}\right)\) = vế phải
=> đpcm
Bài 1 thực hiện phép tính
a)\(\frac{45}{19}-\left(\frac{1}{2}+\left(\frac{1}{3}+\left(\frac{1}{4}\right)^{-1}\right)^{-1}\right)^{-1}.\)
b) \(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^{10}.6^{19}-7.2^{29}.27^6}.\)
Bài 2. tìm x, biết:
a) 2(x-1) - 3(2x+2) - 4(2x+3) =16
b) \(3\frac{1}{2}:\left|2x-1\right|=\frac{21}{22}\)
c) |x2+|x-1|| = x2+2
Bài 3. Chứng minh rằng số có dạng abcabc luôn chia hết cho 11
Bài 4.tính:
a) A = \(\left(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\right):\frac{2012}{2013}\)
b) B =\(4.\left(-\frac{1}{2}\right)^2-2.\left(-\frac{1}{2}\right)^2+3.\left(-\frac{1}{2}\right)+1\)
c) C =\(\frac{1}{2}:\left(-1\frac{1}{2}\right):1\frac{1}{3}:\left(-1\frac{1}{4}\right):1\frac{1}{5}:\left(-1\frac{1}{6}\right):...:\left(-1\frac{1}{100}\right)\)
d) D =\(\frac{4^6.9^5+6^9.120}{-8^4.3^{12}+6^{11}}\)