Q=(1/99+12/999+123/999)x(1/6-1/6)

Q=(1/99+12/999+123/999)x0

Q=0

học tốt!

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

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

= 0 nha bn

q = (1/99+12/999+123/999)*(1/2-1/3-1/6)

= (1/99+12/999+123/999) * 0

= 0

Q=(1/99+12/999+123/999)x0

Q=0

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

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

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

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

\(Q=\left(\frac{136}{999}\right)\left(\frac{0}{6}\right)\)

\(Q=0\)

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

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

\(=\left(\frac{1}{2}+\frac{12}{999}+\frac{123}{999}\right).0=0\)

1/2-1/3-1/6=0=>A=0

Thank you, bài này cũng dễ nhỉ

\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)

=> \(2A-A=\left(2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}}\right)\)

=>  \(A=2-\frac{1}{2^{2012}}=\frac{2^{2013}-1}{2^{2012}}\)

\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)

\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)

\(2A=3+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}\)

\(2A-A=A\)

\(=\left(3+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\right)\)

\(=3+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}-1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{2012}}\)

\(=2-\frac{1}{2012^2}\)

 \(B=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right)\cdot\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)

\(B=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right)\cdot\left(\frac{6}{12}-\frac{4}{12}-\frac{2}{12}\right)\)

\(B=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{9999}\right)\cdot0=0\)