-1,25 . (\(\dfrac{3}{2}\)-0,75)+3,5
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1,25 x 4,2 + 119,6 : 5,2 - 23,19
= 5,25 + 23 - 23,19
= 5,06
14,75 - 7,15 : 5 + 2,6 x 3,5
= 14,75 - 1,43 + 9,1
= 22,42
A = - \(\dfrac{1}{21}\) - \(\dfrac{1}{28}\)
A =- \(\dfrac{4}{84}\)- \(\dfrac{3}{84}\)
A = -\(\dfrac{7}{84}\)
A = - \(\dfrac{1}{12}\)
B = - \(\dfrac{8}{18}\) - \(\dfrac{15}{27}\)
B = - \(\dfrac{4}{9}\) - \(\dfrac{5}{9}\)
B = -1
C = - \(\dfrac{5}{12}\) + 0,75
C = -\(\dfrac{5}{12}\) + \(\dfrac{3}{4}\)
C = - \(\dfrac{5}{12}\) + \(\dfrac{9}{12}\)
C = \(\dfrac{4}{12}\)
C= \(\dfrac{1}{3}\)
D = 3,5 - { - \(\dfrac{2}{7}\))
D = \(\dfrac{7}{2}\) + \(\dfrac{2}{7}\)
D = \(\dfrac{49}{14}\) + \(\dfrac{4}{14}\)
D = \(\dfrac{53}{14}\)
\(a,14,75-7,15:5+2,6\times3,5\\ =14,75-1,43+9,1\\ =13,32+9,1\\ =22,42\\ b,13,5:4,5\times2,4+\left(156,9-80,6\right)\\ =13,5:4,5\times2,4+76,3\\ =3\times2,4+76,3\\ =7,2+76,3\\ =83,5\\ c,1,25\times4,2+119,6:5,2-23,19\\ =5,25+23-23,19\\ =28,25-23,19\\ =5,06\)
1) 2,75 - 5/6 × 2/5 = 2,75 - (5/6) × (2/5) = 2,75 - 1/3 = 2,75 - 0,33 = 2,42
2) 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 3/4) - 3/5 = 1,25 - (5/6 - 9/12) - 3/5 = 1,25 - (10/12 - 9/12) - 3/5 = 1,25 - 1/12 - 3/5 = 1,25 - 0,08 - 0,6 = 1,25 - 0,68 = 0,57
3) 4/9 × 0,75 + 8/5 + 3,125 = (4/9) × 0,75 + 8/5 + 3,125 = 0,44 + 8/5 + 3,125 = 0,44 + 1,6 + 3,125 = 0,44 + 4,725 = 5,165
4) 1,125 - 4/7 - 0,12 = 1,125 - (4/7) - 0,12 = 1,125 - 0,57 - 0,12 = 0,435 - 0,12 = 0,315
5) (1/3 + 0,4) × 3,5 + (1/6 + 0,75) × 6/5
\(A=\frac{0,375-0,3+\frac{3}{10}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}+\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}\)
\(\Rightarrow A=\frac{\frac{3}{8}-\frac{3}{10}+\frac{3}{11}+\frac{3}{12}}{\frac{-5}{8}+\frac{5}{10}-\frac{5}{11}-\frac{5}{12}}+\frac{\frac{3}{2}+\frac{3}{3}-\frac{3}{4}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{4}}\)
\(\Rightarrow A=\frac{-3.\left(-\frac{1}{8}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)}{5\left(-\frac{1}{8}+\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)}+\frac{3.\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}{5\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\right)}\)
\(\Rightarrow A=\frac{-3}{5}+\frac{3}{5}\)
\(\Rightarrow A=0\)
Vậy A = 0
@@ Học tốt @@
# Chiyuki Fujito
Ta có : \(\frac{\left(11,2-10-1,2\right)x\left(3,75-0,75\right)}{2011+3014}=\frac{0x3}{2011+3014}=\frac{0}{2011+3014}=0\)
\(\Rightarrow\left(2000\times7,5+2012:3\right)\times\left(21-3,5\times0,25\right)\times\frac{\left(11,2-10-1,2\right)\times\left(3,75-0,75\right)}{2011+3014}\)
\(=\left(2000\times7,5+2012:3\right)\times\left(21-3,5\times0,25\right)\times0\)
\(=0\)
N=a:2-2:b
N=1,5:2-2:(-0,75)
N=0,75-2:0.75.(-1)
N=0,75-2.(-1):0,75
N=0,75-(-2):0,75
N=0,75+2:0,75
N=75/100+200/75
N=75/100+8/3
N=41/12
hoặc N=(-1,5):2-2:(-0,75)
N=(-0,75)-2.(-1):0,75
N=(-0,75)+2:0,75
N=(-75/100)+200/75
N=(-3/4)+8/3
N=23/12
P=(-2):a2-b.2/3
P=(-2):1,5.1,5-(-0,75).2/3
P=(-0,75).1,5-(-0,75).2/3
P=(-0,75)(1,5-2/3)
P=(-0,75).5/6
P=5/8
`-1,25 . (3/2 - 0,75) + 3,5`
`= -1,25 . (1,5 - 0,75) + 3,5`
`= -1,25 . 0,75 + 3,5`
`= -0,9375 + 3,5`
`= 2,5625`
\(-1,25\cdot\left(\dfrac{3}{2}-0,75\right)+3,5\\ =-\dfrac{5}{4}.\left(\dfrac{6}{4}-\dfrac{3}{4}\right)+\dfrac{7}{2}\\ =-\dfrac{5}{4}\cdot\dfrac{3}{4}+\dfrac{7}{2}\\ =-\dfrac{15}{16}+\dfrac{56}{16}\\ =\dfrac{41}{16}\)