Question

Tính giá trị biểu thức:

`4.D=d-(a+c)+(b+d)-b+(b-c)` với \(a=1\dfrac{1}{3};b=\dfrac{7}{2};c=\dfrac{-5}{6};d=\dfrac{1}{12}\)

Answer 1

\(D=d-a-c+b+d-b+b-c\)

\(=-a+b-2c+2d\)

\(=-\dfrac{4}{3}+\dfrac{7}{2}-2\cdot\dfrac{-5}{6}+2\cdot\dfrac{1}{12}\)

\(=\dfrac{-8+21}{6}+\dfrac{10}{6}+\dfrac{1}{6}=\dfrac{13+10+1}{6}=\dfrac{24}{6}=4\)

Answer 2

`D=d-(a+c)+(b+d)-b+(b-c)`

`D=d-a-c+b+d-b+b-c`

`D=(d+d)-a-(c+c)+(b-b+b)`

`D=2d-a-2c+b`

Thay `a=1 1/3;b=7/2;c=[-5]/6;d=1/12` vào `D` có:

`D=2 . 1/12-1 1/3-2 . [-5]/6+7/2`

`D=1/6-4/3+5/3+7/2`

`D=1/6-8/6+10/6+21/6=24/6=4`