\(12-10,34.\frac{3}{13}\left(x-1\right)=\left(\frac{1}{21.22}+\frac{1}{22.23}+...+\frac{1}{29.30}\right)280\)

<=> \(12-10,34.\frac{3}{13}\left(x-1\right)=\left(\frac{1}{21}-\frac{1}{22}+\frac{1}{22}-\frac{1}{23}+...+\frac{1}{29}-\frac{1}{30}\right).280\)

<=> \(12-10,34.\frac{3}{13}\left(x-1\right)=\left(\frac{1}{21}-\frac{1}{30}\right)280\)

<=> \(12-10,34.\frac{3}{13}\left(x-1\right)=4\)

<=> \(8=10,34.\frac{3}{13}.\left(x-1\right)\)

<=> \(x-1=\frac{5200}{1551}\)

<=> \(x=\frac{6751}{1551}\)

Ta có:

\(\frac{1}{21.22}+\frac{1}{22.23}+...+\frac{1}{29.30}=\frac{1}{21}-\frac{1}{22}+\frac{1}{22}-\frac{1}{23}+...+\frac{1}{29}-\frac{1}{30}=\frac{1}{21}-\frac{1}{30}\)

phương trình đã cho trở thành

\(12-10,34.\frac{3}{13}\left(x-1\right)=\left(\frac{1}{21}-\frac{1}{30}\right).280\)

\(\Leftrightarrow x-1=\frac{\left(\frac{1}{21}-\frac{1}{30}\right).280-12}{-10,34.\frac{3}{13}}\Leftrightarrow x=\frac{\left(\frac{1}{21}-\frac{1}{30}\right).280-12}{-10,34.\frac{3}{13}}+1\)

\(\Leftrightarrow x=\frac{6751}{1551}\)

Ta có: \(S=20.21+21.22+22.23+....+39.40\)

\(\Rightarrow3S=3.20.21+3.21.22+.....+3.39.40\)

\(\Rightarrow3S=20.21.\left(22-19\right)+22.23.\left(24-21\right)+.....+39.40.\left(41-38\right)\)

\(\Rightarrow3S=39.40.41-19.20.21\)

\(\Rightarrow3S=55980\)

\(\Rightarrow S=18660\)

Cách giải

=(39.40-20.21):0.01+1

=19.19:0.01+1

=1919+1

=1920

3s=20.21.(22-19)+21.22.(23-20)+22.23.(24-21)+...+39.40.(41-38)

=20.21.22 - 20.21.19+21.22.23 - 21.22.20 +22.23.24 -22.23.21 + ...+ 39.40.41-39.40.38

=39.40.41- 20.21.19=63960-7980=55980

s=18660

Đặt \(A=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{21.22}\)

\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{21}-\frac{1}{22}\)

\(A=\frac{1}{5}-\frac{1}{22}\)

\(A=\frac{17}{110}\)

\(\frac{1}{5.6}\)+\(\frac{1}{6.7}\)+...+\(\frac{1}{21.22}\)

=\(\frac{1}{5}\)-\(\frac{1}{6}\)+\(\frac{1}{6}\)-\(\frac{1}{7}\)+...+\(\frac{1}{21}\)-\(\frac{1}{22}\)

=\(\frac{1}{5}\)-\(\frac{1}{22}\)

=\(\frac{17}{110}\)

\(A=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{21}-\frac{1}{22}\)

\(A=\frac{1}{5}-\frac{1}{22}\)

\(A=\frac{22}{110}-\frac{5}{110}\)

\(A=\frac{17}{110}\)

2/1.2*2/20.21

\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{20.21}\)

\(=2\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{20.21}\right)\)

\(=2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{20}-\frac{1}{21}\right)\)

\(=2\left(1-\frac{1}{21}\right)=2.\frac{20}{21}=\frac{40}{21}\)

P>Q chắc chắc đúng ủng hộ mình đi