Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
.a, \(\frac{x+1}{999}+\frac{x+2}{998}=\frac{x+3}{997}+\frac{x+4}{996}\)
.\(< =>\frac{x+1}{999}+1+\frac{x+2}{998}+1=\frac{x+3}{997}+1+\frac{x+4}{996}+1\)
.\(< =>\frac{x+1}{999}+\frac{999}{999}+\frac{x+2}{998}+\frac{998}{998}=\frac{x+3}{997}+\frac{997}{997}+\frac{x+4}{996}+\frac{996}{996}\)
.\(< =>\frac{x+1+999}{999}+\frac{x+2+998}{998}=\frac{x+3+997}{997}+\frac{x+4+996}{996}\)
.\(< =>\frac{x+1000}{999}+\frac{x+1000}{998}-\frac{x+1000}{997}-\frac{x+1000}{996}=0\)
.\(< =>\left(x+1000\right)\left(\frac{1}{999}+\frac{1}{998}-\frac{1}{997}-\frac{1}{996}\right)=0\)
.Do \(\frac{1}{999}+\frac{1}{998}-\frac{1}{997}-\frac{1}{996}\ne0\)
.Suy ra \(x+1000=0\Leftrightarrow x=-1000\)
.b, \(\frac{x+1}{1001}+\frac{x+2}{1002}=\frac{x+3}{1003}+\frac{x+4}{1004}\)
.\(< =>\frac{x+1}{1001}-1+\frac{x+2}{1002}-1=\frac{x+3}{1003}-1+\frac{x+4}{1004}-1\)
.\(< =>\frac{x+1}{1001}-\frac{1001}{1001}+\frac{x+2}{1002}-\frac{1002}{1002}=\frac{x+3}{1003}-\frac{1003}{1003}+\frac{x+4}{1004}-\frac{1004}{1004}\)
.\(< =>\frac{x+1-1001}{1001}+\frac{x+2-1002}{1002}=\frac{x+3-1003}{1003}+\frac{x+4-1004}{1004}\)
.\(< =>\frac{x-1000}{1001}+\frac{x+1000}{1002}-\frac{x+1000}{1003}-\frac{x+1000}{1004}=0\)
.\(< =>\left(x-1000\right)\left(\frac{1}{1001}+\frac{1}{1002}-\frac{1}{1003}-\frac{1}{1004}\right)=0\)
.Do \(\frac{1}{1001}+\frac{1}{1002}-\frac{1}{1003}-\frac{1}{1004}\ne0\)
.Suy ra \(x-1000=0\Leftrightarrow x=1000\)
\(A=\frac{9}{8}-\frac{8}{9}+\frac{3}{25}+\frac{1}{4}-\frac{5}{16}+\frac{19}{25}-\frac{1}{9}+\frac{2}{25}-\frac{1}{81}\)
\(A=\left(\frac{9}{8}+\frac{1}{4}-\frac{5}{16}\right)-\left(\frac{8}{9}+\frac{1}{9}-\frac{1}{81}\right)+\left(\frac{3}{25}+\frac{19}{25}+\frac{2}{25}\right)\)
\(A=\frac{17}{16}-\frac{80}{81}+\frac{24}{25}\)
\(A=\frac{33529}{32400}\)
\(C=\frac{3.8.15....80.99}{4.9.16.81.100}\)
\(=\frac{1.3.2.4.3.5...8.10.9.11}{2.2.3.3.4.4...9.9.10.10}\)
\(=\frac{\left(1.2.3....9\right).\left(3.4.5...10.11\right)}{\left(2.3.4.5...10\right).\left(2.3.4...10\right)}\)
\(=\frac{11}{10}\)
trả lời
c=\(\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot....\cdot\frac{99}{100}\)
C=\(\frac{3.8.15....99}{4.9.16.100}\)
C=\(\frac{1.3.2.4.3.5.....9.11}{2.2.3.3.4.4....10.10}\)
C=\(\frac{\left(1.2.....9\right)}{2.3....10}.\left(\frac{3.4....11}{2.3...10}\right)\)
C=\(\frac{1}{10}\cdot\frac{11}{2}=\frac{11}{20}\)
a) \(\left(\frac{1}{3}\right)^n=\frac{1}{81}\)
\(\Rightarrow\left(\frac{1}{3}\right)^n=\frac{1^4}{3^4}\)
\(\Rightarrow\left(\frac{1}{3}\right)^n=\left(\frac{1}{3}\right)^4\)
\(\Rightarrow n=4\)
Vậy n = 4
b) \(\frac{-512}{343}=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow\frac{-8^3}{7^3}=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow\left(\frac{-8}{7}\right)^3=\left(\frac{-8}{7}\right)^n\)
\(\Rightarrow n=3\)
Vậy n = 3
theo bài ra ta có:
\(C=\left(\frac{3}{4}-81\right)\left(\frac{3^2}{5}-81\right)\left(\frac{3^3}{6}-81\right)...\left(\frac{3^{2013}}{2016}-81\right)\)
C = \(\left(\frac{3}{4}-81\right)...\left(\frac{3^6}{9}-81\right)...\left(\frac{3^{2013}}{2016}-81\right)\)
\(C=\left(\frac{3}{4}-81\right)...\left(81-81\right)...\left(\frac{3^{2013}}{2016}-81\right)\)
C = \(\left(\frac{3}{4}-81\right)...0...\left(\frac{3^{2013}}{2016}-81\right)\)
C = 0
vậy C = 0
(3/4 -81 )(3^2/5 -81 )(3^3/6 -81)......(3^2000/2003 -81)
ta viết tiếp dãy số (3/4 -81 )(3^2/5 -81 )(3^3/6 -81)(3^4/7 - 81 ) (3^5/8 -81)(3^6/9 -81).........(3^2000/2003 -81) thì thấy 3^6/9=81 ->3^6/9 -81=0 -> dãy số bằng 0 -> (3/4 -81 )(3^2/5 -81 )(3^3/6 -81)......(3^2000/2003 -81) =0
Minh k hiểu cho lắm. Bạn viết theo công thức toán olm cho sẵn đi cho dễ đọc