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\(C=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}+\frac{\frac{3}{5}+\frac{3}{13}-0,9}{\frac{7}{91}+0,2-\frac{3}{10}}\)
\(=\frac{5\cdot\left(31-\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}{13\cdot\left(31-\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}+\frac{\frac{3}{5}+\frac{3}{13}-\frac{9}{10}}{\frac{1}{13}+\frac{1}{5}-\frac{3}{10}}\)
\(=\frac{5}{13}+\frac{3\cdot\left(\frac{1}{5}+\frac{1}{13}-\frac{3}{10}\right)}{\frac{1}{5}+\frac{1}{13}-\frac{3}{10}}\)
\(=\frac{5}{13}+3\)
\(=\frac{44}{13}\)
Ta có: M=\(\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}\)+\(\frac{\frac{3}{5}+\frac{3}{13}-0,9}{\frac{7}{91}+0,2-\frac{3}{10}}\)
=\(\frac{5.\left(31-\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}{13.\left(31-\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}\)+\(\frac{\frac{9}{15}+\frac{9}{39}-\frac{9}{10}}{\frac{1}{13}+\frac{1}{5}+\frac{3}{10}}\)
=\(\frac{5}{13}\)+\(\frac{9.\left(\frac{1}{15}+\frac{1}{39}-\frac{1}{10}\right)}{3.\left(\frac{1}{39}+\frac{1}{15}-\frac{1}{10}\right)}\)
=\(\frac{5}{13}\)+\(\frac{9}{3}\)
=\(\frac{5}{13}\)+3
=\(\frac{44}{13}\)
a, 1 - 7x = 3x - 4
=> -7x - 3x = - 4 - 1
=> - 10x = - 5
=> x = 1/2
vậy_
b, đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)
\(3A-A=1-\frac{1}{3^{99}}\)
\(A=\frac{1-\frac{1}{3^{99}}}{2}\)
mk chỉ bt lm mấy phần hui à!
d)\(\frac{5}{17}+\frac{-4}{7}-\frac{20}{31}+\frac{12}{17}-\frac{11}{31}\)\(=\left(\frac{5}{17}+\frac{12}{17}\right)+\left(\frac{-20}{31}-\frac{11}{31}\right)+\frac{-4}{7}\)
\(=\frac{17}{17}+\frac{-31}{31}+\frac{-4}{7}\)\(=1+\left(-1\right)+\frac{-4}{7}\)\(=0+\frac{-4}{7}\)\(=-\frac{4}{7}\)
e)\(\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{20}{7}-\frac{13}{3}+\frac{13}{23}}\)
Trả lời
\(A=\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{2.\left(\frac{1}{6}-\frac{1}{8}-\frac{1}{10}\right)}{\frac{7}{6}-\frac{7}{8}-\frac{7}{10}}\right):\left(1^2+2^2+...+2015^2\right).\)
\(A=\left(\frac{2}{7}-\frac{2}{7}\right):\left(1^2+2^2+3^2+...+2015^2\right)\)
\(A=0:\left(1^2+2^2+3^2+.....+2015^2\right)\)
A=0
Study well
\(A=...\)
\(=\left(\frac{\frac{2}{5}-\frac{2}{9}+\frac{2}{11}}{\frac{7}{5}-\frac{7}{9}+\frac{7}{11}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\right):\left(1^2+2^2+...+2015^2\right)\)
\(=\left[\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\right]:\left(1^2+2^2+...+2015^2\right)\)
\(=\left(\frac{2}{7}-\frac{1}{\frac{7}{2}}\right):\left(1^2+2^2+...+2015^2\right)\)
\(=\left(\frac{2}{7}-\frac{2}{7}\right):\left(1^2+2^2+...+2015^2\right)\)
\(=0:\left(1^2+2^2+...+2015^2\right)=0\)
\(A=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}-\frac{\frac{3}{5}+\frac{3}{13}-0,9}{\frac{7}{91}+0,2-\frac{3}{10}}\)
\(A=\frac{155-5\left(\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}{403-13\left(\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}-\frac{\frac{3}{5}+\frac{3}{13}-\frac{9}{10}}{\frac{7}{91}+\frac{2}{10}-\frac{3}{10}}\)
\(A=\frac{155-5}{403-13}-\frac{3\left(\frac{1}{5}+\frac{1}{13}\right)-\frac{9}{10}}{\frac{7}{91}+\left(-\frac{1}{10}\right)}\)
\(A=\frac{5}{13}-\frac{\left(-\frac{9}{130}\right)}{\left(-\frac{3}{130}\right)}=\frac{5}{13}-\frac{\frac{9}{130}}{\frac{3}{130}}\)
\(A=\frac{5}{13}-\frac{9}{130}\cdot\frac{130}{3}\)
\(A=\frac{5}{13}-3=-\frac{34}{13}\)
\(B=\frac{30\cdot4^7\cdot3^{29}-5\cdot14^5\cdot2^{12}}{54\cdot6^{14}\cdot9^7-12\cdot8^5\cdot7^5}\)
\(B=\frac{30\cdot\left(2^2\right)^7\cdot3^{29}-5\cdot\left(2\cdot7\right)^5\cdot2^{12}}{54\cdot\left(2\cdot3\right)^{14}\cdot\left(3^2\right)^7-12\cdot\left(2^3\right)^5\cdot7^5}\)
\(B=\frac{30\cdot2^{14}\cdot3^{29}-5\cdot2^5\cdot7^5\cdot2^{12}}{54\cdot2^{14}\cdot3^{14}\cdot3^{14}-12\cdot2^{15}\cdot7^5}\)
\(B=\frac{30\cdot3^{29}-5\cdot2^{17}\cdot7^5}{54\cdot3^{28}-12\cdot2^{15}\cdot7^5}=\frac{30\cdot3-5\cdot2^2}{54-12}=\frac{5}{3}\)
bạn kieu cao dương nghich quá các bạn đè bạn ấy xuống đi