Chỉ cần tính 1/2-1/3-1/6=6/12-4/12-2/+2=0/12=0

Suy ra (1/99+12/999+123/9999).(1/2-1/3-1/6)=(1/99+12/999+123/9999).0=0

Vậy kết quả là 0.

mk nha

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

=(9/407-123/9999).0

=0

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

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

Ta có \(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}=\dfrac{1}{6}-\dfrac{1}{6}=0\) nên Q = 0.

\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right).\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\) 

\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right).0\) 

\(Q=0\)

đi hỏi toán lớp 3 à 

\(A=\frac{5}{9}.\frac{7}{13}+\frac{5}{9}.\frac{6}{13}\)(( . là dấu nhân nha )

\(A=\frac{5}{9}.\left(\frac{7}{13}+\frac{6}{13}\right)\)

\(A=\frac{5}{9}\)

\(B=\left(\frac{1}{19}+\frac{12}{999}+\frac{123}{9999}\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)

         Mà \(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}=0\)

\(\Rightarrow B=0\)

này nói cho mà bít làm gì có cái luận nào k cho học sinh lớp 6 hỏi toán lớp 3 đâu

\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)

\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right)\left(\dfrac{3}{6}-\dfrac{2}{6}-\dfrac{1}{6}\right)\)

\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right)\cdot\dfrac{0}{6}\)

\(Q=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{9999}\right)\cdot0\)

\(Q=0\)