\(F= \frac{5}{9} + \frac{4}{9} . \frac{3}{7} + \frac{4}{9}. \frac{4}{7}\)
1.Thực hiện phép tính:(hepl me)
a)\(4\frac{5}{9}:\left(\frac{-5}{7}\right)+\frac{49}{9}:\left(\frac{-5}{7}\right)\)
b)\(\left(\frac{-3}{5}+\frac{4}{9}\right):\frac{7}{11}+\left(\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)
c)\(\left(\frac{3}{4}\right)^4\times\left(\frac{8}{9}\right)^2\)
d)\(\left(\frac{-3}{5}\right)^6\times\left(-\frac{5}{3}\right)^5\)
e)\(\frac{8^{14}}{4^4\times64^5}\)
f)\(\frac{9^{10}\times27^7}{81^7\times3^{15}}\)
a) \(4\frac{5}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)=\frac{41}{9}:\left(-\frac{5}{7}\right)+\frac{49}{9}:\left(-\frac{5}{7}\right)\)
\(=\frac{41}{9}\cdot\left(-\frac{7}{5}\right)+\frac{49}{9}\cdot\left(-\frac{7}{5}\right)=\left(\frac{41}{9}+\frac{49}{9}\right)\cdot\left(-\frac{7}{5}\right)=10\cdot\left(-\frac{7}{5}\right)=-14\)
b) \(\left(\frac{-3}{5}+\frac{4}{9}\right):\frac{7}{11}+\left(\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(\frac{-3}{5}+\frac{4}{9}+\frac{-2}{5}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(\frac{-3}{5}+\frac{-2}{5}+\frac{4}{9}+\frac{5}{9}\right):\frac{7}{11}\)
\(=\left(-1+1\right):\frac{7}{11}=0\cdot\frac{11}{7}=0\)
c) \(\left(\frac{3}{4}\right)^4\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\right)^2\cdot\left(\frac{3}{4}\right)^2\cdot\left(\frac{8}{9}\right)^2=\left(\frac{3}{4}\cdot\frac{3}{4}\cdot\frac{8}{9}\right)^2\)
\(=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)
d) \(\left(-\frac{3}{5}\right)^6\cdot\left(-\frac{5}{3}\right)^5=\left(-\frac{3}{5}\right)^5\cdot\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)^5=\left[\left(-\frac{3}{5}\right)\cdot\left(-\frac{5}{3}\right)\right]^5\cdot\left(-\frac{3}{5}\right)\)
\(=1^5\cdot\left(-\frac{3}{5}\right)=1\cdot\left(-\frac{3}{5}\right)=-\frac{3}{5}\)
e) \(\frac{8^{14}}{4^4\cdot64^5}=\frac{\left(2^3\right)^{14}}{\left(2^2\right)^4\cdot\left(2^6\right)^5}=\frac{2^{42}}{2^8\cdot2^{30}}=\frac{2^{42}}{2^{38}}=2^4=16\)
f) \(\frac{9^{10}\cdot27^7}{81^7\cdot3^{15}}=\frac{\left(3^2\right)^{10}\cdot\left(3^3\right)^7}{\left(3^4\right)^7\cdot3^{15}}=\frac{3^{20}\cdot3^{21}}{3^{28}\cdot3^{15}}=\frac{3^{41}}{3^{43}}=3^{-2}=\frac{1}{3^2}=\frac{1}{9}\)
Bài 1: Tính(hợp lý nếu có thể) e) \(\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{9}{13}-\frac{5}{9}.\frac{3}{13}\) f)\(\frac{12}{19}\cdot\frac{7}{15}\cdot\frac{-13}{17}\cdot\frac{19}{12}\cdot\frac{17}{13}\) g) \(\left(\frac{-4}{5}+\frac{4}{3}\right)+\left(\frac{-5}{4}+\frac{14}{5}\right)-\frac{7}{3}\)
Rút gọn
1. \(\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{9}{13}-\frac{5}{9}.\frac{3}{13}\)
2. \(\left(\frac{1+\frac{1}{5}+\frac{1}{7}+\frac{1}{11}}{2+\frac{2}{5}+\frac{2}{7}+\frac{2}{17}}:\frac{4-\frac{4}{7}+\frac{4}{9}-\frac{4}{13}}{1-\frac{1}{7}+\frac{1}{9}-\frac{1}{13}}\right):\frac{838383}{808080}\)
1. \(\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{9}{13}-\frac{5}{9}.\frac{3}{13}\)
= \(\frac{5}{9}\) .(\(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\) )
= \(\frac{5}{9}\) . 1 = \(\frac{5}{9}\)
\(\left(8-\frac{9}{4}+\frac{2}{7}\right)-\left(-6-\frac{3}{7}+\frac{5}{4}\right)-\left(3+\frac{2}{4}-\frac{9}{7}\right)\)\(\frac{9}{7}\))
\(\frac{1}{3}-\frac{3}{5}+\frac{5}{7}-\frac{7}{9}+\frac{9}{11}-\frac{11}{13}+\frac{13}{15}+\frac{11}{13}-\frac{9}{11}+\frac{7}{9}-\frac{5}{7}+\frac{3}{5}-\frac{1}{3}\)
\(\frac{1}{2014}-\frac{1}{2014.2013}-\frac{1}{2013.2012}-\frac{1}{2012.2011}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
thực hiện phép tính :
a, \(\frac{0,4-\frac{2}{9}+\frac{2}{11}}{1,4-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{\frac{7}{6}-58+5+0,7}\)
b, \(\frac{\frac{1}{9}-\frac{1}{7}-\frac{1}{11}}{\frac{4}{9}-\frac{4}{7}-\frac{4}{11}}+\frac{\frac{3}{5}-\frac{3}{25}-\frac{3}{125}-\frac{3}{625}}{\frac{4}{5}-\frac{4}{25}-\frac{4}{125}-\frac{4}{625}}\)
c, \(\frac{\frac{3}{7}-\frac{3}{17}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)
Mong các bạn giúp đỡ nhé
\(\frac{3}{4}+\frac{5}{7}+\frac{4}{5}.\frac{7}{9}+\frac{5}{7}.\frac{1}{4}+\frac{4}{5}+\frac{2}{9}\)=?
\(a.6\frac{4}{5}-\left(1\frac{2}{3}+3\frac{4}{5}\right)\) \(b.6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)\) \(c.7\frac{5}{9}-\left(2\frac{3}{4}+3\frac{5}{9}\right)\)
\(d.7\frac{5}{11}\left(2\frac{3}{7}+3\frac{5}{11}\right)\) \(e.\frac{-3}{5}.\frac{5}{7}+\frac{-3}{5}.\frac{3}{7}+\frac{-3}{5}.\frac{6}{7}\) \(f.\frac{1}{3}.\frac{4}{5}+\frac{1}{3}.\frac{6}{5}-\frac{4}{3}\)
\(g.\frac{4}{19}.\frac{-3}{7}+\frac{-3}{7}.\frac{15}{19}+\frac{5}{7}\) \(h.\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{9}{13}-\frac{5}{9}.\frac{3}{13}\)
AI GIAI SOM VA RO RANG NHAT MINH SE TICK CHO NHE! CAM ON NHIEU LAM!
\(-\frac{3}{5}.\frac{5}{7}+-\frac{3}{5}.\frac{3}{7}+-\frac{3}{5}.\frac{6}{7}=-\frac{3}{5}\left(\frac{5}{7}+\frac{3}{7}+\frac{6}{7}\right)=-\frac{3}{5}.2=-\frac{6}{5}\)
\(\frac{1}{3}.\frac{4}{5}+\frac{1}{3}.\frac{6}{5}-\frac{4}{3}=\frac{1}{3}\left(\frac{4}{5}+\frac{6}{5}\right)-\frac{4}{3}=\frac{1}{3}.2-\frac{4}{3}=\frac{2}{3}-\frac{4}{3}=-\frac{2}{3}\)
\(\frac{4}{19}.\frac{-3}{7}+-\frac{3}{7}.\frac{15}{19}+\frac{5}{7}=-\frac{3}{7}\left(\frac{4}{19}+\frac{15}{19}\right)+\frac{5}{7}=-\frac{3}{7}+\frac{5}{7}=\frac{2}{7}\)
\(\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{9}{13}-\frac{5}{9}.\frac{3}{13}=\frac{5}{9}\left(\frac{7}{13}+\frac{9}{13}-\frac{3}{13}\right)=\frac{5}{9}\)
\(a,6\frac{4}{5}-\left(1\frac{2}{3}+3\frac{4}{5}\right)=6\frac{4}{5}-1\frac{2}{3}-3\frac{4}{5}=6\frac{4}{5}-3\frac{4}{5}-1\frac{2}{3}=3-1\frac{2}{3}=\frac{4}{3}\)
\(b,6\frac{5}{7}-\left(1\frac{3}{4}+2\frac{5}{7}\right)=6\frac{5}{7}-2\frac{5}{7}-1\frac{3}{4}=\frac{9}{4}\)
\(c,7\frac{5}{9}-\left(2\frac{3}{4}+3\frac{5}{9}\right)=7\frac{5}{9}-3\frac{5}{9}-2\frac{3}{4}=4-2\frac{3}{4}=\frac{5}{4}\)
mk nghĩ là phần d như thế này cơ \(7\frac{5}{11}\left(2\frac{3}{7}+3\frac{5}{11}\right)\)
\(7\frac{5}{11}-\left(2\frac{3}{7}+3\frac{5}{11}\right)=7\frac{5}{11}-3\frac{5}{11}-2\frac{3}{7}=4-2\frac{3}{7}=\frac{11}{7}\)
\(\frac{3}{\frac{5}{\frac{3}{4}.7^2-\frac{3}{4}.5^7+\frac{3}{4}.3^9}}.7^2-3.5^6+\frac{3}{5}.3^9\)