3)a. Vì xy // zt mà AB vuông góc với xy => AB vuông góc với zt( t/c từ vuông góc đến song song) =>góc ABD = \(^{90^0}\)

(còn lại tự giải ik)

\(\frac{x+1}{x-2}=\frac{5}{4}\)

=>  \(4\left(x+1\right)=5\left(x-2\right)\)

=>  \(4x+4=5x-10\)

=>  \(5x-4x=10+4\)

=>  \(x=14\)

\(\frac{x+1}{x-2}=\frac{5}{4}\)

\(\Leftrightarrow\)\(\left(x+1\right).4=5.\left(x-2\right)\)

\(\Leftrightarrow\)\(4x+4=5x-10\)

\(\Leftrightarrow\)\(5x-4x=10+4\)

\(\Leftrightarrow\)\(x=14\)

\(\frac{-1}{10}\)-\(\frac{2}{5}\)x+\(\frac{7}{20}\)=\(\frac{1}{10}\)

              \(\frac{2}{5}\)x+\(\frac{7}{20}\)=\(\frac{-1}{10}\)-\(\frac{1}{10}\)

              \(\frac{2}{5}\)x+\(\frac{7}{20}\)=\(\frac{-2}{10}\)

              \(\frac{2}{5}\)x              =\(\frac{-4}{20}\)-\(\frac{7}{20}\)

              \(\frac{2}{5}\)x              =\(\frac{-11}{20}\)

                       x               =\(\frac{-11}{20}\).\(\frac{5}{2}\)

                      x                =-1\(\frac{3}{8}\)

vậy...

-1/10-2/5x+7/20=1/10

2/5x+7/10=-1/10-1/10

2/5x+7/10=0

2/5x=0-7/10

2/5x=-3/10

x=-3/10:2/5

x=-3/4

\(\frac{8}{9}-\frac{1}{72}-\frac{1}{56}-...-\frac{1}{2}\)

\(=\frac{8}{9}-\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{72}\right)\)

\(=\frac{8}{9}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{8.9}\right)\)

\(=\frac{8}{9}-\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{9-8}{8.9}\right)\)

\(=\frac{8}{9}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{8}-\frac{1}{9}\right)\)

\(=\frac{8}{9}-\left(1-\frac{1}{9}\right)=\frac{8}{9}-\frac{8}{9}=0\)

3B=3/1.4+3/4.7+3/7.10+...+3/100.103

3B=(4-1)/1.4+(7-4)/4.7+(10-7)/7.10+...+(103-100)/100.103

3B=1-1/4+1/4-1/7+1/7-1/10+...+1/100-1/103=1-1/103=102/103

B=102/(3.103)=34/103

HT

\(\frac{1}{1.4}+\frac{1}{4.7}+.....+\frac{1}{100.103}\)

Đặt :

\(A=\frac{1}{1.4}+\frac{1}{4.7}+....+\frac{1}{100.103}\)

\(3A=\frac{3}{1.4}+\frac{3}{4.7}+.....+\frac{3}{100.103}\)

\(3A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{100}-\frac{1}{103}\)

\(3A=1-\frac{1}{103}\)

\(3A=\frac{102}{103}\)

\(A=\frac{34}{103}\)

giúp mình với ToT

trả lời nhớ k nhé!

=\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{1999\cdot2000}\)

=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+...-\frac{1}{999}+\frac{1}{999}-\frac{1}{2000}\)

=\(\frac{1}{1}-\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+...+\left(-\frac{1}{999}+\frac{1}{999}\right)-\frac{1}{2000}\)

=\(\frac{1}{1}+0+0+...+0-\frac{1}{2000}\)

=\(\frac{1}{1}-\frac{1}{2000}\)

=\(\frac{2000}{2000}-\frac{1}{2000}\)

=\(\frac{1999}{2000}\)