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a,(-35,8)+(-17,2) + 16,4 + 4,6
= [ (-35,8)+(-17,2)]+(16,4 + 4,6 )
= (-53)+21
= -32
b, (5,3 - 2,8) -( 4+ 5,3 )
=5,3 -2,8 -4-5,3
=5,3 +(-2,8)+ (-4)+(-5,3 )
= [5,3 +(-5,3 )]+[(-2,8)+ (-4)]
= 0+(-6,8)
=-6,8
c, (34:72 + 32,28): 5 -(57,25 - 36,05):2
ko biết
d, 2,5.(-4,68)+2,5.(-5,32)
=2,5 .[(-4,68)+(-5,32)]
= 2,5 . (-10)
=-25
e, 5,36.12,34+(-5,36).2,34
= 5,36.12,34+5,36.(-2,34)
= 5,36.[12,34+(-2,34)]
=5,36 . 10
= 53,6
\(a,\left(-38,5\right)+\left(-17,2\right)+16,4+4,6\\ =\left[\left(-35,8\right)+\left(-17,2\right)\right]+\left(16,4+4,6\right)\\ =-53+21\\ =-32\)
\(b,\left(5,3-2,8\right)-\left(4+5,3\right)=2,5-9,3=-6,8\)
\(c,\left(34,72+32,28\right):5-\left(57,25-36,05\right):2\\ =77:5-21,2:2\\ =15,4-10,5\\ =4,9\) (là 34,72 đúng không ah?)
\(d,2,5.\left(-4,68\right)+2,5.\left(-5,32\right)\\ =2,5.\left[\left(-4,68\right)+\left(-5,32\right)\right]\\ =2,5.10\\ =25\)
\(5,36.12,34+\left(-5,36\right).2,34\\ =66,1424+-12,5424\\ =53,6\) (thay vì bê cái khác vào tk thì tôi tự tính tay:'))
a: \(=\left(1.25\right)^{16}\cdot8^{16}\cdot8=8\cdot10^{16}\)
b: \(=\left(\dfrac{5}{2}\right)^{13}\cdot4^{13}\cdot4^2=10^{13}\cdot4^2\)
c: \(=\left(0.25\right)^4\cdot8^4\cdot8^2=2^4\cdot8^2=64\cdot16=1024\)
d: \(=\left(\dfrac{1}{2}\right)^{15}\cdot2^{18}=2^3=8\)
e: \(=\left(\dfrac{1}{3}\cdot6\right)^7\cdot\left(\dfrac{1}{2}\right)^7\cdot\dfrac{1}{2}=2^7\cdot\left(\dfrac{1}{2}\right)^7\cdot\dfrac{1}{2}=\dfrac{1}{2}\)
\(\left(-2,5\right).\left(-4\right)=10\)
\(\left(-2,5\right).0,5.\left(-2\right).2=\left[\left(-2,5\right).\left(-2\right).\left(0,5.2\right)\right]=5.1=5\)
\(\left(-0,5\right).0,5.\left(-2\right).2=\left[\left(-0,5\right).\left(-2\right).\left(0,5.2\right)\right]=1.1=1\)
\(25.\left(-5\right).\left(-0,4\right).\left(-0,2\right)=\left[25.\left(-0,4\right).\left(-5\right).\left(-0,2\right)\right]=-10.1=-10\)
`(2/3-0,25+2)-(2-5/2+1/4)-(2,5-1/3)`
`= 2/3 -1/4 +2-2+ 5/2 -1/4 -5/2 +1/3`
`= (2/3 +1/3) +(-1/4 -1/4) + (2-2) + (5/2-5/2)`
`= 3/3 + (-1/2) + 0 + 0`
`= 1 +(-1/2)`
`= 1/2`
\(\left(\dfrac{2}{3}-0,25+2\right)-\left(2-\dfrac{5}{2}+\dfrac{1}{4}\right)-\left(2,5-\dfrac{1}{3}\right)\\ =\dfrac{2}{3}-0,25+2-2+\dfrac{5}{2}-\dfrac{1}{4}-2,5+\dfrac{1}{3}\\ =\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(\dfrac{5}{2}-2,5\right)+\left(2-2\right)+\left(-\dfrac{1}{4}-0,25\right)\\ =\dfrac{3}{3}+\left(2,5-2,5\right)+0+\left(-\dfrac{1}{4}-\dfrac{1}{4}\right)\\ =1+0+0+\left(-\dfrac{1}{2}\right)=\dfrac{1}{2}\)