a,
\(3-\left|\dfrac{-1}{2}\right|\\ =3-\dfrac{1}{2}\\ =\dfrac{6}{2}-\dfrac{1}{2}\\ =\dfrac{5}{2}\)
b,
\(\left|\dfrac{-1}{4}\right|+\dfrac{3}{4}-\left|-1\right|\\ =\dfrac{1}{4}+\dfrac{3}{4}-1\\ =1-1\\ =0\)
c,
\(\left|0,25\right|=-\left(-0,25\right)\\ 0,25=0,25\)
a ) 3−|−12|
= 3 - \(\dfrac{1}{2}\)
= \(\dfrac{6}{2}\)- \(\dfrac{1}{2}\)= \(\dfrac{5}{2}\).
b ) |−14|+34−|−1|
= \(\dfrac{1}{4}\)+ \(\dfrac{3}{4}\) - 1.
= 1 - 1= 0.
c )
=> 0,25 = 0,25.
a) \(3-\left|\dfrac{-1}{2}\right|\)
\(=3-\dfrac{1}{2}\)
\(=\dfrac{6}{2}-\dfrac{1}{2}\)
\(=\dfrac{5}{2}\)
b) \(\left|\dfrac{-1}{4}\right|+\dfrac{3}{4}-\left|-1\right|\)
\(=\dfrac{1}{4}+\dfrac{3}{4}-1\)
\(=1-1=0\)
c) Ta có: \(\left|0.25\right|=0.25\)
Áp dụng quy tắc dấu ngoặc ta có: \(0.25=-\left(-0.25\right)\\ \Rightarrow\left|0.25\right|=-\left(-0.25\right)\\ \rightarrowđpcm\)