Question

1.Tinh gia tri cua bieu thuc

\(\dfrac{1}{2}+\dfrac{1}{3}\).x-\(\dfrac{1}{6}\).x voi x=\(\dfrac{-3}{5}\)

Answer 1

Thay \(x=-\dfrac{3}{5}\) vào biểu thức ta có:

\(\dfrac{1}{2}+\dfrac{1}{3}.\dfrac{-3}{5}-\dfrac{1}{6}.\dfrac{-3}{5}\)

\(=\dfrac{1}{2}+\dfrac{-3}{5}.\left(\dfrac{1}{3}-\dfrac{1}{6}\right)=\dfrac{1}{2}+\dfrac{-3}{5}.\dfrac{1}{6}\)

\(=\dfrac{1}{2}+\dfrac{-1}{10}=\dfrac{2}{5}\)

Chúc bạn học tốt!!!

Answer 2

đóng góp một cách khác:

đặt biểu thức trên là A.

\(A=\dfrac{1}{2}+\dfrac{1}{3}x-\dfrac{1}{6}x=\dfrac{1}{2}+\dfrac{x}{6}\)

Thay \(x=-\dfrac{3}{5}\) vào biểu thức A, ta có:

\(A=\dfrac{1}{2}+\dfrac{-\dfrac{3}{5}}{6}\\ =\dfrac{1}{2}-\dfrac{1}{10}\\ =\dfrac{5}{10}-\dfrac{1}{10}\\ =\dfrac{4}{10}\\ =\dfrac{2}{5}\)

Vậy giá trị biểu thức A tại \(x=-\dfrac{3}{5}\) là \(\dfrac{2}{5}\)

Answer 3

Thay \(x=-\dfrac{3}{5}\) vào biểu thức ta có:

\(\dfrac{1}{2}+\dfrac{1}{3}.\dfrac{-3}{5}-\dfrac{1}{6}.\dfrac{-3}{5}\)

\(=\dfrac{1}{2}+\dfrac{-3}{5}.\left(\dfrac{1}{3}-\dfrac{1}{6}\right)=\dfrac{1}{2}+\dfrac{-3}{5}.\dfrac{1}{6}\)

\(=\dfrac{1}{2}+\dfrac{-1}{10}=\dfrac{2}{5}\)

Answer 4

Thay \(\dfrac{-3}{5}\)vào x ta có:

\(\dfrac{1}{2}+\dfrac{1}{3}x-\dfrac{1}{6}x=\dfrac{1}{2}+\dfrac{1}{3}.\dfrac{-3}{5}-\dfrac{1}{6}.\dfrac{-3}{5}\)

\(=\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{6}\right).\dfrac{-3}{5}=\dfrac{1}{2}+\dfrac{1}{6}.\dfrac{-3}{5}\)

\(=\dfrac{1}{2}+\dfrac{-1}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)