Tính bằng cách hợp lí.
\(2 \frac{3}{4}\) \(- \frac{3}{5}\) \(- 2 , 25\) \(=\)?
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\(\dfrac{5}{6}\cdot\dfrac{1}{3}+\dfrac{1}{2}\cdot\dfrac{3}{4}\cdot\dfrac{5}{3}\)
\(=\dfrac{5\cdot1}{6\cdot3}+\dfrac{1}{2}\cdot\dfrac{3\cdot5}{4\cdot3}\)
\(=\dfrac{5}{18}+\dfrac{1}{2}\cdot\dfrac{15}{12}\)
\(=\dfrac{5}{18}+\dfrac{1}{2}\cdot\dfrac{5}{4}\)
\(=\dfrac{5}{18}+\dfrac{1\cdot5}{2\cdot8}\)
\(=\dfrac{5}{18}+\dfrac{5}{8}\)
\(=\dfrac{65}{72}\)
\(\dfrac{5}{6}.\dfrac{1}{3}+\dfrac{5}{6}+\dfrac{3}{4}\)
\(=\dfrac{5}{6}.\left(\dfrac{1}{3}+\dfrac{3}{4}\right)=\dfrac{5}{6}.\dfrac{7}{12}=\dfrac{35}{72}\)
= (2/5 + 3/5) + (6/9 + 1/3) + (7/4 + 1/4)
= 1 + (6/9 + 3/9) + 2 = 1 + 1 + 2 = 4
\(\dfrac{2}{5}+\dfrac{6}{9}+\dfrac{7}{4}+\dfrac{3}{5}+\dfrac{1}{3}+\dfrac{1}{4}\\ =\left(\dfrac{2}{5}+\dfrac{3}{5}\right)+\left(\dfrac{7}{4}+\dfrac{1}{4}\right)+\left(\dfrac{2}{3}+\dfrac{1}{3}\right)\\ =1+2+1=4\)
= -5 / 7 x ( 2 / 11 + 9 / 11) + 12/7
= -5/7 x 1 + 12/7
= -5/7 + 12/7
= 1
b, = ( 9/4 x 105 - 9/4 x 101) : 3/2 - 64 x ( 1/4 - 3/4)
= [ 9/4 x ( 105 - 101)] : 3/2 -64 x -1/2
= 9/4 x 4 : 3 /2 - (-32)
= 9 : 3/2 + 32
= 18/3 +32
=
\(A=\left(\frac{5}{3}-\frac{3}{7}+9\right)-\left(2+\frac{5}{7}-\frac{2}{3}\right)+\left(\frac{8}{7}-\frac{4}{3}-10\right)\)
\(=\frac{5}{3}-\frac{3}{7}+9-2-\frac{5}{7}+\frac{2}{3}+\frac{8}{7}-\frac{4}{3}-10\)
\(=\left(\frac{5}{3}+\frac{2}{3}-\frac{4}{3}\right)-\left(\frac{3}{7}+\frac{5}{7}-\frac{8}{7}\right)+\left(9-2-10\right)\)
\(=1-0-3\)
\(=-2\)
5/3-3/7+9-2-5/7+2/3+8/7-4/3-10
(5/3-4/3+2/3)+(8/7-3/7-5/7)+(9-2-10)
1+0-3=-2
=(2,75-2,25)-0.6=0,5-0,6=-0,1