Có: \(Q=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right).\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
\(=\left(\frac{1}{2}+\frac{12}{999}+\frac{123}{999}\right).\left(\frac{3}{6}-\frac{2}{6}-\frac{1}{6}\right)\)
\(=\left(\frac{1}{2}+\frac{12}{999}+\frac{123}{999}\right).0=0\)
tính giá trị của biểu thức: Q = (1/99+12/999+123/999). (1/2-1/3-1/6)
tính giá trị :\(Q=\left(\frac{1}{99}+\frac{12}{999}+\frac{123}{999}\right)\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\)
(1/99+12/999+123/999).(1/2-1/3-1/6)
Rút gọn biểu thức:
A= 1+1/2+1/2^2+1/2^3+...+1/2^2012
B= ( 1/99+12/999+123/999)(1/2-1/3-1/6)
q=(1/99+12/999+123/999)*(1/2-1/3-1/6)
Q= (1/99+12/999+123/999)(1/2-1/3-1/6)
Q = (1/99 + 12/999 + 123/999) x (1/2 - 1/3 - 1/6)
A=(1/99+12/999+123/999)x(1/2-1/3-1/6) = ?
1.Tính nhanh
A =( \(\dfrac{1}{99}+\dfrac{12}{999}-\dfrac{123}{9999}\)) x \(\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
2.Tìm n ∈ Z để \(\dfrac{n+3}{n-2}\)nhận giá trị nguyên
