![](https://rs.olm.vn/images/avt/0.png?1311)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2.THPT\)
\(A=\frac{9}{1.2}+\frac{9}{2.3}+\frac{9}{3.4}+...+\frac{9}{98.99}+\frac{9}{99.100}\)
\(A=9\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(A=9\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=9\left(1-\frac{1}{100}\right)\)
\(A=9.\frac{99}{100}\)
\(A=\frac{891}{100}\)
\(B=\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{93.95}\)
\(B=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{93}-\frac{1}{95}\)
\(B=\frac{1}{5}-\frac{1}{95}\)
\(B=\frac{18}{95}\)
\(D=\frac{5}{2.7}+\frac{4}{7.11}+\frac{3}{11.14}+\frac{1}{14.15}+\frac{13}{15.28}\)
\(D=\frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+\frac{1}{14}-\frac{1}{15}+\frac{1}{15}-\frac{1}{28}\)
\(D=\frac{1}{2}-\frac{1}{28}\)
\(D=\frac{13}{28}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a) Ta có: \(\frac{-5}{7}+\frac{2}{7}+\frac{4}{-9}+\frac{4}{9}\)
\(=-\frac{3}{7}+\frac{-4}{9}+\frac{4}{9}\)
\(=-\frac{3}{7}\)
b) Ta có: \(\left(\frac{1}{2}:\frac{3}{4}\right)^2\)
\(=\left(\frac{1}{2}\cdot\frac{4}{3}\right)^2\)
\(=\left(\frac{2}{3}\right)^2=\frac{4}{9}\)
c) Ta có: \(\frac{1}{2}+\frac{3}{4}-\left(\frac{4}{5}+\frac{3}{4}\right)\)
\(=\frac{1}{2}+\frac{3}{4}-\frac{4}{5}-\frac{3}{4}\)
\(=\frac{1}{2}-\frac{4}{5}\)
\(=\frac{5}{10}-\frac{8}{10}=\frac{-3}{10}\)
d) Ta có: \(5^6:5^4+2^3\cdot2^2-225:15^2\)
\(=5^2+2^5-\frac{15^2}{15^2}\)
\(=25+32-1\)
\(=56\)
e) Ta có: \(\frac{7}{23}+\frac{4}{17}-\frac{7}{23}+\frac{13}{17}\)
\(=\frac{4}{17}+\frac{13}{17}\)
\(=\frac{17}{17}=1\)
g) Ta có: \(19\frac{1}{4}\cdot\frac{7}{12}-15\frac{1}{4}\cdot\frac{7}{12}\)
\(=\frac{7}{12}\left(19+\frac{1}{4}-15-\frac{1}{4}\right)\)
\(=\frac{7}{12}\cdot4=\frac{7}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=3+3^2+3^3+.....+3^{2006}\)
\(\Rightarrow3B=3^2+3^3+....+3^{2007}\)
\(\Rightarrow2B=3^{2007}-3\)
\(\Rightarrow B=\frac{3^{2007}-3}{2}\)
\(2B+3=3^x\)
\(\Rightarrow2.\frac{3^{2007}-3}{2}+3=3^x\)
\(\Rightarrow3^{2007}-3+3=3^x\Rightarrow3^{2007}=3^x\Rightarrow x=2007\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x+5}{3}=\frac{y-7}{4}\)
áp dụng t\c của dãy tỉ số bằng nhau ta có :
\(\frac{x+5}{3}=\frac{y-7}{4}=\frac{x+5+y-7}{3+4}=\frac{23-2}{7}=\frac{21}{7}=3\)
\(\Rightarrow\hept{\begin{cases}x=3\cdot3-5=4\\y=3\cdot4+7=19\end{cases}}\)
đặt \(k=\frac{x+5}{3}=\frac{y-7}{4}\)
\(\Rightarrow\hept{\begin{cases}x=3k-5\\y=4k+7\end{cases}}\)
\(\Rightarrow x+y=3k-5+4k+7=7k+2=23\)
\(\Rightarrow k=\frac{23-2}{7}=3\)
\(\Rightarrow\hept{\begin{cases}x=4\\y=19\end{cases}}\)
các câu tiếp theo tương tự
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=-2.\frac{2}{3}.\frac{1}{3}:\left(\frac{-1}{6}+0,5\right)-\left(-2009^0\right)-\left(-2\right)^2\)
\(=\frac{4}{3}.\frac{1}{3}:\left(\frac{-1}{6}+\frac{1}{2}\right)-1.4\)
\(=\frac{4}{3}.\frac{1}{3}+4\)
\(=4+4\)
\(=8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Trl :
\(\frac{1}{9}.27^n=3^{n+2}\)
\(3^{-2}.\left(3^3\right)^n=3^{n+2}\)
\(3^{-2}.3^{3n}=3^{n+2}\)
\(\Rightarrow-2+3n=n+2\)
\(\Rightarrow3n=n+4\)
\(\Rightarrow2n=4\)\(\Rightarrow n=2\)
Hok tốt
Trl :
\(\frac{1}{9}3^4.3^n=3^7\)
\(3^{-2}.3^4.3^n=3^7\)
\(\Rightarrow-2+4+n=7\)
\(\Rightarrow2+n=7\)
\(\Rightarrow n=7-2\)
\(\Rightarrow n=5\)
Hok tốt !
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{-9}{4}\).\(19\frac{2}{5}\)+\(\left(\frac{-3}{2}\right)^2\).\(\left(-14\frac{3}{5}\right)\)-\(\left(\frac{99}{100}\right)^0\)
=\(\frac{-9}{4}\).\(\frac{97}{5}\)+\(\frac{9}{4}\).\(\frac{-73}{5}\)-1
=\(\frac{-9}{4}\).\(\frac{97}{5}\)+\(\frac{-9}{4}\).\(\frac{73}{5}\)-1
=\(\frac{-9}{4}\).(\(\frac{97}{5}\)+\(\frac{73}{5}\))
=\(\frac{-9}{4}\).34
=\(\frac{-153}{2}\)
Học tốt
\(4:\left(x+\frac{2}{3}\right)^2=9\)
=> \(\left(x+\frac{2}{3}\right)^2=\frac{9}{4}\)=>\(\orbr{\begin{cases}x=\frac{3}{2}-\frac{2}{3}=\frac{5}{6}\\x=\frac{-3}{2}-\frac{2}{3}=\frac{-13}{6}\end{cases}}\)