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tính cụm 2 ra 0 từ đó tính đc ra bằng 0 đó bạn. k cho mình nha
\(Q=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\text{ }\)
\(Q=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)
\(Q=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right).0\)
\(Q=0\)
Q=(1/99+12/999+123/999).(1/2-1/3-1/6) =(1/99+12/999+123/999).0 Q=0
\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right).0\)
\(=0\)
\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)=\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right).0=0\)
b)
Số số hạng của B là : ( 58 - 1 ) : 3 + 1 = 20 ( số )
Tổng B là : ( 58 + 1 ) x 20 : 2 = 590
Vậy,......
Học tốt~
a)
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\)
\(A=1-\frac{1}{10}\)
\(A=\frac{9}{10}\)
Học tốt~
a)\(=\frac{-3}{7}+\frac{15}{26}-\frac{2}{13}+\frac{3}{7}\)
\(=\left(\frac{-3}{7}+\frac{3}{7}\right)-\left(\frac{15}{26}+\frac{2}{13}\right)\)
\(=0-\frac{19}{26}\)
\(=-\frac{19}{26}\)
c)\(=\frac{-11}{23}.\left(\frac{6}{7}+\frac{8}{7}\right)-\frac{1}{23}\)
\(=\frac{-11}{23}.2-\frac{1}{23}\)
\(=\frac{-22}{23}-\frac{1}{23}\)
\(=-1\)
\(1\frac{1}{2}.1\frac{1}{3}.1\frac{1}{4}.1\frac{1}{5}.....1\frac{1}{999}\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.\frac{6}{5}....\frac{1000}{999}\)
\(=\frac{1000}{2}\)
\(=500\)
\(B=\frac{1^2}{2}.\frac{2^2}{6}.\frac{3^2}{12}.....\frac{9^2}{90}=\frac{1^2.2^2.3^2.....9^2}{\left(1.2\right).\left(2.3\right).\left(3.4\right).....\left(9.10\right)}\)
\(=\frac{1^2.2^2.3^2.....9^2}{1.2^2.3^2.....9^2.10}=\frac{1}{10}\)
Bài 4:
\(50B=\frac{100}{1.99}+\frac{100}{3.97}+...+\frac{100}{49.51}\)
\(50B=\frac{1+99}{1.99}+\frac{3+97}{3.97}+...+\frac{99+1}{49.51}\)
\(50B=1+\frac{1}{99}+\frac{1}{97}+\frac{1}{3}+...+\frac{1}{49}+\frac{1}{51}=A\)
\(\Rightarrow\frac{A}{B}=50\).
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 !