Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(Q=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\text{ }\)
\(Q=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)
\(Q=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right).0\)
\(Q=0\)
Q=(1/99+12/999+123/999).(1/2-1/3-1/6) =(1/99+12/999+123/999).0 Q=0
tính cụm 2 ra 0 từ đó tính đc ra bằng 0 đó bạn. k cho mình nha
\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right).0\)
\(=0\)
\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)=\(\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right).0=0\)
A=(9/1999+99/999+999/9999).(1/5-1/4+1/20)
A=(9/1999+99/999+999/9999).(-1/20+1/20)
A=(9/1999+99/999+999/9999).0
A=0
Vì mọi số nhân vs 0 thì đều = 0 kể cả phân số
mk nhanh nhất ủng hộ nha
\(A=\left(\frac{9}{1999}+\frac{99}{999}+\frac{999}{9999}\right)\cdot0\)
A=0
\(\left(\frac{99^9}{11^9}-\frac{99^{99}}{11^{99}}-\frac{99^{999}}{11^{999}}\right)\left(\frac{1}{5}-\frac{1}{7}-\frac{2}{35}\right)\)
\(=\left(\frac{99^9}{11^9}-\frac{99^{99}}{11^{99}}-\frac{99^{999}}{11^{999}}\right)\left(\frac{7}{35}-\frac{5}{35}-\frac{2}{35}\right)\)
\(=\left(\frac{99^9}{11^9}-\frac{99^{99}}{11^{99}}-\frac{99^{999}}{11^{999}}\right).0\)
\(=0\)
Bài 1.
b) \(\frac{5+55+555+5555}{9+99+999+9999}\)
= \(\frac{5\left(1+11+111+1111\right)}{9\left(1+11+111+1111\right)}=\frac{5}{9}\)
c) \(39,2\cdot27+39,2\cdot43+78,4\cdot15\)
= \(39,2\cdot27+39,2\cdot43+39,2\cdot2\cdot15\)
= \(39,2\left(27+43+30\right)=39,2\cdot100=3920\)
d) \(\frac{4}{17}\cdot\frac{3}{11}+\frac{8}{11}\cdot\frac{4}{17}-\frac{4}{17}\)
= \(\frac{4}{17}\left(\frac{3}{11}+\frac{8}{11}-1\right)=\frac{4}{17}\cdot0=0\)
Bài 2.
a) \(\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+\frac{1}{9\cdot11}+...+\frac{1}{57\cdot59}\)
= \(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{57}-\frac{1}{59}\)
= \(\frac{1}{5}-\frac{1}{59}=\frac{54}{295}\)
b) \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)-\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)\)
= \(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\)
= \(\frac{1}{2}-\frac{1}{8}=\frac{3}{8}\)
c) \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\left(1-\frac{1}{5}\right)...\left(1-\frac{1}{2012}\right)\)
= \(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\frac{4}{5}\cdot...\cdot\frac{2011}{2012}=\frac{1}{2012}\)
=1/1*2+1/2*3+...+1/999*1000
=1/1-1/2+1/2-1/3+...+1/999-1/1000
=1-1/1000
So sánh A và B biết;
A = \(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{999}{1000}\)
B = \(\frac{2}{3}.\frac{3}{4}.\frac{4}{5}...\frac{998}{999}\)
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 !