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A= (3,1 - 2,5) - (-2,5 + 3,1)
=3.1-2.5+2.5-3.1
=0
B= (5,3 - 2,8) - (4 + 5,3)
=5.3 - 2.8-4-5.3
= -6.8
C= -(251.3 + 281) + 3.251 - (1-281)
=-251.3-281+3.251-1+281
=-1
D= -(3/5+3/4)−(−3/4+2/5)
= -3/5-3/4+3/4-2/5
= -5/5
=-1
\(A=\left(3,1-2,5\right)-\left(-2,5+3,1\right)=3,1-2,5+2,5-3,1=0\)
\(B=\left(5,3-2,8\right)-\left(4+5,3\right)=5,3-2,8-4-5,3=-6,8\)
\(C=-\left(215\cdot3+281\right)+3\cdot215-\left(1-281\right)=-215\cdot3-281+3\cdot215-1+281=1\)
\(D=-\left(\frac{3}{5}+\frac{3}{4}\right)-\left(-\frac{3}{4}+\frac{2}{5}\right)=-\frac{3}{5}-\frac{3}{4}+\frac{3}{4}-\frac{2}{5}=-1\)
\(A=\left(3,1-2,5\right)-\left(-2,5+3,1\right)\)
\(A=3,1-2,5+2,5-3,1\)
\(A=\left(3,1-3,1\right)-\left(2,5-2,5\right)\)
\(A=0-0\)
\(A=0\)
\(B=\left(5,3-2,8\right)-\left(4+5,3\right)\)
\(B=5,3-2,8-4-5,3\)
\(B=\left(5,3-5,3\right)-\left(2,8+4\right)\\ B=0-6,8\\ B=-6,8\)
Khi bỏ dấu ngoặc, mỗi số trong ngoặc phải đổi dấu :
\(A=3,1-2,5+2,5-3,1=3,1-3,1=0\)
ai tích mình tích lại nhưng phải lên điểm mình tích gấp đôi
A = \(\left(3,1-2,5\right)-\left(-2,5+3,1\right)\)
\(=3,1-2,5+2,5-3,1=6,2\)
B = \(\left(5,3-2,8\right)-\left(4+5,3\right)\)
\(=5,3-2,8-4-5,3=-6,8\)
C = \(-\left(251.3+281\right)+3.251-\left(1-281\right)\)
\(=-251.3-281+3.251-1+281=-1\)
D = \(-\left(\frac{3}{5}+\frac{3}{4}\right)-\left(-\frac{3}{4}+\frac{2}{5}\right)\)
\(=-\frac{3}{5}-\frac{3}{4}+\frac{3}{4}-\frac{2}{5}=-1\)
(Nhớ k cho mình với nhá!)
\(A=\left(3,1-2,5\right)-\left(-2,5+3,1\right)\)
\(=3,1-2,5+2,5-3,1=0\)
\(B=\left(5,3-2,8\right)-\left(4+5,3\right)\)
\(=5,3-2,8-4-5,3=-6,8\)
\(C=-\left(251.3+281\right)+3.215-\left(1-281\right)\)
\(=-753-281+645-1+281\)
\(=-753+654-1=-100\)
\(D=-\left(\dfrac{3}{5}+\dfrac{3}{4}\right)-\left(\dfrac{-3}{4}+\dfrac{2}{5}\right)\)
\(=-\dfrac{3}{5}-\dfrac{3}{4}+\dfrac{3}{4}-\dfrac{2}{5}=\dfrac{-3}{5}-\dfrac{2}{5}=-1\)
\(A=\left(3,1-2,5\right)-\left(-2,5+3,1\right)\)
\(=3,1-2,5+2,5-3,1\)
\(=\left(3,1-3,1\right)-\left(2,5-2,5\right)\)
\(=0\)
Vậy \(A=0.\)
\(B=\left(5,3-2,8\right)-\left(4+5,3\right)\)
\(=5,3-2,8-4-5,3\)
\(=\left(5,3-5,3\right)-\left(2,8+4\right)\)
\(=-\dfrac{34}{5}\)
Vậy \(B=\dfrac{-34}{5}.\)
\(C=-\left(251.3+281\right)+3.215-\left(1-281\right)\)
\(=-251.3-281+3.215-1+281\)
\(=-753-\left(281-281\right)+645\)
\(=-753+645=-108\)
Vậy \(C=-108.\)
\(D=-\left(\dfrac{3}{5}+\dfrac{3}{4}\right)-\left(-\dfrac{3}{4}+\dfrac{2}{5}\right)\)
\(=-\dfrac{3}{5}-\dfrac{3}{4}+\dfrac{3}{4}-\dfrac{2}{5}\)
\(=\left(\dfrac{-3}{5}-\dfrac{2}{5}\right)-\left(\dfrac{3}{4}-\dfrac{3}{4}\right)\)
\(=-1\)
Vậy D = -1.
E\(=5,5.\left(-1,6\right)\)=-8,8
\(E=5,5.2-5,5.3,6\)=11-19,8=-8,8
F= -3,1.(-2,7)=8,37
F=-3,1.3+3,1.5,7=-9,3+17,67=8,37
C1:E=5,5.(2-3,6)
E=5,5.(-1,6)
E=-8,8
C2:E=5,5(2-3,6)
E=5,5.2-5,5.3,6
E=11-19,8
E=-8,8
C1:F=-3,1(3-5,7)
F=-3,1.(-2,7)
F=8,37
C2:F=-3,1(3-5,7)
F=-3,1.3-(-3,1).5,7
F=-9,3-(-17,67)
F=-9,3+17,67
F=8,37
C1:A=(\(\frac{36-4+3}{6}-\frac{30+10-9}{6}-\frac{18-14+15}{6}=\frac{35}{6}-\frac{31}{6}-\frac{19}{6}=-\frac{5}{2}\)
C2:A=\(\left(6-5-3\right)+\left(\frac{-2}{3}-\frac{5}{3}+\frac{7}{3}\right)+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)=-2+0+-\frac{1}{2}=-\frac{5}{2}\)
a) \(A=\left(3,1-2,5\right)-\left(-2,5+3,1\right)\)
\(A=0,6-0,6\)
\(A=0\)
b) \(B=\left(5,3-2,8\right)-\left(4+5,3\right)\)
\(B=2,5-7,3\)
\(B=-6,8\)
c) \(C=-\left(25-3+2,81\right)+3,251-\left(1-2,8\right)\)
\(C=-24,81+3,251+1,8\)
\(C=-19,759\)
`@` `\text {Ans}`
`\downarrow`
`a,`
`A =(3,1 - 2,5) - (-2,5 + 3,1)`
`= 3,1 - 2,5 + 2,5 - 3,1`
`= (3,1 - 3,1) + (-2,5+2,5)`
`= 0`
Vậy, `A = 0`
`b,`
`B = (5,3 - 2,8) - (4+5,3)`
`= 5,3 - 2,8 - 4 - 5,3`
`= (5,3 - 5,3) + (-2,8-4)`
`= -6,8`
Vậy, `B = -6,8`
`c,`
`C = -(25 - 3 + 2,81) + 3,251 - (1 - 2,8)` `?`
`= -25 + 3 - 2,81 + 3,251 - 1 + 2,8`
`= (-25 + 3 - 1) + (-2,81 + 2,8) + 3,251`
`= -23 - 0,01 + 3,251`
`= -19,759`
A = (3,1 – 2,5) – (-2,5 + 3,1)
= 3,1 – 2,5 + 2,5 -3,1
= ( 3,1 - 3,1) +(2,5 – 2,5)
= 0 + 0 = 0