a) (2.5)2 và 22.52
b)105/25 và (10/2)5
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1/2×(2×x-3)+105/2=-137/2
1/2×(2×x-3)=-137/2-105/2
1/2×(2×x-3)=-242/2
2×x-3=-242/2:1/2
2×x-3=-242
2.x=(-242)+3
2.x=239
x=239:2
x=239/2
A,-12×25-25×75-25×13
=[(-12)-75-13].25
=(-100).25
-2500
B,-50/7×49/10-35/2×-10/7+ -25/3× -9/5
=(-35)-(-25)+(-13)
=-23
C,-354+265-156+125
=[(-354)-156]+(265+125)
=(-510)+277
=-233
D,-74+40-50+16-35
=(-74)+40+(-50)+16+(-35)
=[(-74)+(-35)]+[40+(-50)+16]
=(-109)+26
=-83
E,-2/3×4/5+-4/5×4/3-0,125
=-2/3.4/5+4/5.(-4/3)-0,125
=4/5.[-2/3+(-4/3)]-0,125
=4/5.(-2)-0,125
=-8/5-0,125
=(-1,6)+(-0,125)
=-1,725
a) 2018 + (– 3) < 2018
b) (– 105) + 5 > (– 105)
c) (– 59) + (– 10) < (–59)
a: =-36-329-71+136-2021
=100-400-2021
=-300-2021
=-2321
a) 58.75+58.50-58.25
=58.(75+50-25)
=58.100
=5800
b) \(20:2^2-5^9:5^8\)
\(20:4-5\)
\(5-5=0\)
c)\(\left(5^{19}:5^{17}-4\right):7\)
\(\left(5^2-4\right):7\)
\(\left(25-4\right):7\)
\(21:7\)
=\(3\)
d) \(84:4+3^9:3^7+5^0\)
\(=21+3^2+1\)
\(=21+9+1\)
\(30+1=31\)
e) \(295-\left(31-2^2.5\right)^2\)
\(295-\left(31-4.5\right)^2\)
\(295-\left(31-20\right)^2\)
\(295-\left(11\right)^2\)
\(295-121\)
\(=174\)
f) \(11^{25}:11^{23}-3^5:\left(1^{10}\right)+2^3-60\)
\(11^2-243:1+8-60\)
\(121-243+8-60\)
\(-122+8-60\)
\(-114-60\)
\(=-174\)
a) 1763 + (–2) = 1763 – 2 = 1761.
Vậy 1763 + (–2) < 1763.
b) (–105) + 5 = –(105 – 5) = –100.
So sánh –100 và –105:
|–100| = 100, |–105| = 105. Mà 100 < 105 nên (–100) > (–105).
Vậy (–105) + 5 > (–105).
c) (–29) + (–11) = – (29 + 11) = –40.
So sánh –40 và –29:
|–40| = 40; |–29| = 29. Mà 40 > 29 nên (–40) < (–29).
Vậy (–29) + (–11) < (–29).
Câu 3:
a) \(\dfrac{12}{36}=\dfrac{12:12}{36:12}=\dfrac{1}{3}\)
\(\dfrac{-16}{20}=\dfrac{-16:4}{20:4}=\dfrac{-4}{5}\)
b) \(\dfrac{21}{105}=\dfrac{21:21}{105:21}=\dfrac{1}{5}\)
\(\dfrac{35}{150}=\dfrac{35:5}{150:5}=\dfrac{7}{30}\)
Câu 4:
a) \(\dfrac{3}{10}+\dfrac{5}{10}=\dfrac{3+5}{10}=\dfrac{8}{10}=\dfrac{4}{5}\)
b) Ta có: \(\left(-27\right)\cdot36+64\cdot\left(-27\right)+23\cdot\left(-100\right)\)
\(=\left(-27\right)\cdot\left(64+36\right)+23\cdot\left(-100\right)\)
\(=-27\cdot100-23\cdot100\)
\(=100\left(-27-23\right)\)
\(=-50\cdot100=-5000\)
c) \(\dfrac{5}{8}+\dfrac{3}{12}=\dfrac{15}{24}+\dfrac{6}{24}=\dfrac{21}{24}=\dfrac{7}{8}\)
d) Ta có: \(\dfrac{-2}{17}+\dfrac{3}{19}+\dfrac{-15}{17}+\dfrac{16}{19}+\dfrac{5}{6}\)
\(=\left(-\dfrac{2}{17}+\dfrac{-15}{17}\right)+\left(\dfrac{3}{19}+\dfrac{16}{19}\right)+\dfrac{5}{6}\)
\(=-1+1+\dfrac{5}{6}\)
\(=\dfrac{5}{6}\)
a) (2*5)^2=2^2/5^2
b) 10^5/2^5=(10/2)^5