a) Số số hạng của tổng là:

         (407-1):2+1=204(số hạng )

Từ đó ,ta có:\(\Rightarrow\)S=1+3-5-7+9+11-...-405-407

                    \(\Rightarrow\)(1+3-5-5)+(9+11-13-15)+...+(401+403-405-407)                                                                                                                                  \(\Rightarrow\)(-8)+(-8)+...+(-8) (51 số -8)

                    \(\Rightarrow\)-8.51=-408

a) Số số hạng của tổng là:

         ( 407 - 1 ) : 2 + 1 = 204 ( số hạng )

Từ đó ,ta có:\(⇒\)S = 1 + 3 - 5 - 7 + 9 + 11 - ... - 405 - 407

                    ⇒( 1 + 3 - 5 - 5 ) + ( 9 + 11 - 13 - 15 ) + ... + ( 401 + 403 - 405 - 407 )                                                                                                                                  ⇒( - 8 ) + ( - 8 ) + ... + ( - 8 ) . ( 51 số - 8 )

                    ⇒- 8 . 51 = - 408

a, \(4\times\left(-\dfrac{1}{2}\right)^3-2\times\left(-\dfrac{1}{2}\right)^2+3\times\left(-\dfrac{1}{2}\right)+1\)

\(=\left(-\dfrac{1}{2}\right)\left[\left(4\times-\dfrac{1}{2}\right)-\left(2\times-\dfrac{1}{2}\right)+3\right]+1\)

\(=\left(-\dfrac{1}{2}\right)\left(-2+1+3\right)+1\)

\(=\left(-\dfrac{1}{2}\right)2+1\)

\(=-1+1\)

\(=0\)

@Trịnh Thị Thảo Nhi

a,

\(A=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}-\frac{\frac{3}{5}+\frac{3}{13}-0,9}{\frac{7}{91}+0,2-\frac{3}{10}}\)

\(A=\frac{155-5\left(\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}{403-13\left(\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}-\frac{\frac{3}{5}+\frac{3}{13}-\frac{9}{10}}{\frac{7}{91}+\frac{2}{10}-\frac{3}{10}}\)

\(A=\frac{155-5}{403-13}-\frac{3\left(\frac{1}{5}+\frac{1}{13}\right)-\frac{9}{10}}{\frac{7}{91}+\left(-\frac{1}{10}\right)}\)

\(A=\frac{5}{13}-\frac{\left(-\frac{9}{130}\right)}{\left(-\frac{3}{130}\right)}=\frac{5}{13}-\frac{\frac{9}{130}}{\frac{3}{130}}\)

\(A=\frac{5}{13}-\frac{9}{130}\cdot\frac{130}{3}\)

\(A=\frac{5}{13}-3=-\frac{34}{13}\)

\(B=\frac{30\cdot4^7\cdot3^{29}-5\cdot14^5\cdot2^{12}}{54\cdot6^{14}\cdot9^7-12\cdot8^5\cdot7^5}\)

\(B=\frac{30\cdot\left(2^2\right)^7\cdot3^{29}-5\cdot\left(2\cdot7\right)^5\cdot2^{12}}{54\cdot\left(2\cdot3\right)^{14}\cdot\left(3^2\right)^7-12\cdot\left(2^3\right)^5\cdot7^5}\)

\(B=\frac{30\cdot2^{14}\cdot3^{29}-5\cdot2^5\cdot7^5\cdot2^{12}}{54\cdot2^{14}\cdot3^{14}\cdot3^{14}-12\cdot2^{15}\cdot7^5}\)

\(B=\frac{30\cdot3^{29}-5\cdot2^{17}\cdot7^5}{54\cdot3^{28}-12\cdot2^{15}\cdot7^5}=\frac{30\cdot3-5\cdot2^2}{54-12}=\frac{5}{3}\)

1: \(A=\dfrac{5\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}{13\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}+\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{13}-\dfrac{3}{10}\right)}{\dfrac{1}{13}+\dfrac{1}{5}-\dfrac{3}{10}}\)

=5/13+3

=5/13+39/13

=44/13

2: \(B=\dfrac{2^{15}\cdot3^{30}\cdot5-5\cdot2^5\cdot7^5\cdot2^{12}}{3^3\cdot2\cdot2^{14}\cdot3^{14}\cdot3^{14}-2^2\cdot3\cdot2^{15}\cdot7^5}\)

\(=\dfrac{5\cdot2^{15}\left(3^{30}-2^2\cdot7^5\right)}{3^{31}\cdot2^{15}-2^{17}\cdot3\cdot7^5}\)

\(=\dfrac{5\cdot2^{15}\left(3^{30}-2^2\cdot7^5\right)}{3\cdot2^{15}\cdot\left(3^{30}-2^2\cdot7^5\right)}=\dfrac{5}{3}\)

a: \(A=\dfrac{5\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}{13\left(31-\dfrac{2}{7}-\dfrac{1}{11}+\dfrac{1}{23}\right)}+\dfrac{3\left(\dfrac{1}{5}+\dfrac{1}{13}-\dfrac{3}{10}\right)}{\left(\dfrac{1}{5}-\dfrac{3}{10}+\dfrac{1}{13}\right)}\)

=5/13+3

=5/13+39/13

=44/13

b: \(B=\dfrac{3^{30}\cdot2^{15}\cdot5-5\cdot2^5\cdot7^5\cdot2^{12}}{2\cdot3^3\cdot2^{14}\cdot3^{14}\cdot3^{14}-3\cdot2^2\cdot2^{15}\cdot7^5}\)

\(=\dfrac{2^{15}\cdot5\left(3^{30}-7^5\cdot2^2\right)}{2^{15}\cdot3^{31}-3\cdot2^{17}\cdot7^5}\)

\(=\dfrac{2^{15}\cdot5\left(3^{30}-7^5\cdot2^2\right)}{2^{15}\cdot3\left(3^{30}-7^5\cdot2^2\right)}=\dfrac{5}{3}\)

\(A=\frac{\left(140\frac{7}{30}-138\frac{5}{12}\right):18\frac{1}{6}}{0,002}\)

\(A=\frac{\left(\frac{4207}{30}-\frac{1661}{12}\right):\frac{109}{6}}{\frac{1}{500}}\)

\(A=\frac{\left(\frac{4207}{30}-\frac{1661}{12}\right)\times\frac{6}{109}}{\frac{1}{500}}\)

\(A=\left(\frac{4207}{30}-\frac{1661}{12}\right)\times\frac{6}{109}\times500\)

\(A=\frac{109}{60}\times\frac{6}{109}\times500\)

\(A=\frac{1}{10}\times500\)

\(A=50\)

\(B=\frac{155-\frac{10}{7}-\frac{5}{11}+\frac{5}{23}}{403-\frac{26}{7}-\frac{13}{11}+\frac{13}{23}}\)

\(B=\frac{5\left(31-\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}{13\left(31-\frac{2}{7}-\frac{1}{11}+\frac{1}{23}\right)}\)

\(B=\frac{5}{13}\)