A=(1/99+12/999+123/999)x(1/2-1/3-1/6) = ?
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NT
1
18 tháng 4 2019
Q=(1/99+12/999+123/999)x(1/6-1/6)
Q=(1/99+12/999+123/999)x0
Q=0
học tốt!
9 tháng 5 2016
( 1/99 + 12/999 + 123/999 ) . ( 1/2 - 1/3 - 1/6 )
= ( 1/99 + 12/999 + 123/999 ) . 0
= 0 nha bn
25 tháng 5 2021
q = (1/99+12/999+123/999)*(1/2-1/3-1/6)
= (1/99+12/999+123/999) * 0
= 0
12 tháng 6 2016
\(Q=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\))
\(Q=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right).0=0\)
N
1
2 tháng 5 2016
\(Q=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(Q=\left(\frac{1+12+123}{999}\right)\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)
\(Q=\left(\frac{136}{999}\right)\left(\frac{0}{6}\right)\)
\(Q=0\)
AC
0
CT
3
5 tháng 5 2018
Có: \(Q=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(=\left(\frac{1}{2}+\frac{12}{999}+\frac{123}{999}\right).\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)
\(=\left(\frac{1}{2}+\frac{12}{999}+\frac{123}{999}\right).0=0\)
12 tháng 8 2020
\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
=> \(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)\)
=> \(A=2-\frac{1}{2^{2012}}=\frac{2^{2013}-1}{2^{2012}}\)
12 tháng 8 2020
\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
\(2A=3+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}\)
\(2A-A=A\)
\(=\left(3+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)
\(=3+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}-1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{2012}}\)
\(=2-\frac{1}{2012^2}\)
\(B=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(B=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right)\cdot\left(\frac{6}{12}-\frac{4}{12}-\frac{2}{12}\right)\)
\(B=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right)\cdot0=0\)
A=(1/99+12/999+123/999)x(1/2-1/3-1/6)
=(1/99+12/999+123/999)x(3/6-2/6-1/6)
=(1/99+12/999+123/999)x0
=0