Tính (9,1 + 10,2 + 11,3 + ...+ 17,9) + 19 = (9,1 + 17,9). 9 : 2 + 19 = 121,5 + 19 = 140,5

Tính Mẫu =  \(\frac{40}{42}.\frac{54}{56}.\frac{70}{72}.....\frac{2650}{2652}=\frac{\left(5.8\right).\left(6.9\right).\left(7.10\right)....\left(50.53\right)}{\left(6.7\right).\left(7.8\right).\left(8.9\right)....\left(51.52\right)}=\frac{\left(5.6.7...50\right).\left(8.9.10....53\right)}{\left(6.7.8...51\right).\left(7.8.9...52\right)}=\frac{5.53}{51.7}=\frac{265}{357}\)

Vậy \(P=\frac{140,5}{\frac{265}{357}}.\frac{265}{357}=140,5\)

Xét \(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{123}\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{121}+\frac{1}{123}\right)-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{122}\right)\)

\(=\left(1+\frac{1}{3}+...+\frac{1}{121}+\frac{1}{123}\right)-2\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{61}\right)\)

\(=\frac{1}{62}+\frac{1}{63}+\frac{1}{64}+...+\frac{1}{123}\)

\(Q=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\text{ }\)

\(Q=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)

\(Q=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right).0\)

\(Q=0\)

Q=(1/99+12/999+123/999).(1/2-1/3-1/6) =(1/99+12/999+123/999).0 Q=0

Muốn cho số có hai chữ số giống nhau và chia hết cho 2 thì số đó phải là một trong các số 22, 44, 66, 88. Bây giờ ta tìm trong những số này số mà chia cho 5 thì dư 3.

Đó là số 88.

Xem thêm tại: http://loigiaihay.com/bai-99-trang-39-sgk-toan-6-tap-1-c41a3896.html#ixzz4xczZ4dOb

\(\frac{\frac{4}{17}+\frac{4}{19}-\frac{4}{2111}}{\frac{5}{17}+\frac{5}{19}-\frac{5}{2111}}-\frac{\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}}{-\frac{5}{123}+\frac{5}{19}-\frac{5}{371}+1}\)

\(=\frac{4.\left(\frac{1}{17}+\frac{1}{19}-\frac{1}{2111}\right)}{5.\left(\frac{1}{17}+\frac{1}{19}-\frac{1}{2111}\right)}+\frac{\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}}{5.\left(\frac{1}{123}-\frac{1}{19}+\frac{1}{371}-\frac{1}{5}\right)}=\frac{4}{5}+\frac{1}{5}=1\)

Cho tam giác ABC có đường cao AD .Gọi E là trung điểm  của AB .F đối xứng vs D qua E c/m AB = DF

Ta có: B= (1/99+12/999-123/9999).(1/2-1/3-1/6)

             B= (1/99+12/999-123/9999).(3/6-2/6-1/6)

             B= (1/99+12/999-123/9999).0

             B= 0

B = (1/99+12/999-123/9999).(1/2-1/3-1/6)

B= (1/99+12/999+123/9999).0

B=0

tk mình nha !

=0x(1/99+12/999+123/99990

=0

a: \(A=\dfrac{63-48-70}{84}:\dfrac{-3\cdot84+4\cdot60+5\cdot70}{840}\cdot\dfrac{1-14}{12}\)

\(=\dfrac{-55}{84}\cdot\dfrac{840}{338}\cdot\dfrac{-13}{12}=\dfrac{55}{1}\cdot\dfrac{10}{338}\cdot\dfrac{13}{12}=\dfrac{275}{156}\)

b: \(=-234\cdot123+123\cdot4356-123\cdot2312+234\cdot123-234\cdot2312+2312\cdot234+2312\cdot123\)

\(=123\cdot4356-123\cdot2312+123\cdot2312=123\cdot4356=535788\)