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4.(3x^3 + 1^10) = 4.5^2
=> 3x^3 + 1 = 5^2 = 25
=> 3x^3 = 24
=> x^3 = 8 = 2 ^3
=> x = 2
4.(3x^3 + 1^10) = 4.5^2
=> 3x^3 + 1 = 5^2 = 25
=> 3x^3 = 24
=> x^3 = 8 = 2 ^3
=> x = 2
a, 80 - 4 . 5 2 - 3 . 2 3
= 80 – (4.25 – 3.8)
= 80 – 76 = 4
b, 5 6 : 5 4 + 2 3 . 2 2 - 1 2018
= 5 2 + 2 5 - 1
= 25 + 32 – 1 = 56
c, [36.4 – 4. 82 - 7 . 11 2 ]:4 – 2019 0
= (144 – 4. 5 2 ):4 – 1
= (144 – 100):4 – 1
= 11 – 1 = 10
d, 303 – 3.{[655 – (18:2+1). 4 3 +5]}: 10 0
= 303 – 3.(655 – 10.64 + 5):1
= 303 – 10 = 293
\(F=5+5^3+5^5+...+5^{101}\)
=>\(25F=5^3+5^5+...+5^{103}\)
=>\(25F-F=5^3+5^5+...+5^{103}-5-5^3-...-5^{101}\)
=>\(24F=5^{103}-5\)
=>\(F=\dfrac{5^{103}-5}{24}\)
a) 80-(4.52-3.23)=80-100+24=4
b)[36.4-4.(82-7.11)2]:4-20190
={4.[36-(82-7.11)2]}:4-1
=[36-(82-7.11)2]-1
=11-1=10
c)56:54+23.22-12018
=52+25-1
=25+32-1=56
d)303-3.{[655-(18:2+1).43+55]}:100
=303-3.[(655-9-1).43+55]:1
=303-3[655-640+5]
=303-3(20)
=303-60=243
\(B=1+22+24+26+28+...+2200\)
\(=1+\dfrac{\left(2200+22\right).\left[\left(2200-22\right):2+1\right]}{2}\)
\(=1+\dfrac{2222.1090}{2}\)
\(=1+1210990\)
\(=1210991\)
\(C=5+53+55+57+...+5101\)
\(=5+\dfrac{\left(5101+53\right).\left[\left(5101-53\right):2+1\right]}{2}\)
\(=5+\dfrac{5154.2525}{2}\)
\(=5+6506925\)
\(=6506930\)
a) 80- (4.52 - 3.23)
= 80- ( 208 -69 )
=80+139 quy tắc đổi dấu trừ tước dấu ngoặc
= 219
e: Ta có: \(2448:\left[119-\left(23-6\right)\right]\)
\(=2448:\left(119-23+6\right)\)
\(=2448:102=24\)
c) ( 5 103 - 5 102 - 5 101 ) : ( 5 99 . 26 - 5 99 )
= ( 5 103 - 5 102 - 5 101 ) : [ 5 99 . ( 26 – 1)]
= ( 5 103 - 5 102 - 5 101 ) : ( 5 99 . 25 )
= 5 101 ( 52 – 51 – 50) : ( 5 99 . 52 )
= ( 5 101 . 19 ) : 5 101 = 19
a) 80 - (4.52 - 3.23 )
= 80 - ( 4 . 25 - 3. 8 )
= 80 - ( 100 - 24 )
= 80 - 76 = 4
b) 56 : 54 + 23 . 22 - 12017
= 52 + 25 - 1
= 25 + 32 - 1
= 57 -1 = 56
\(4+4\cdot5+4\cdot5^2+...+4\cdot5^{100}\\ =4\left(1+5+5^2+...+5^{100}\right)\left(1\right)\)
Đặt \(A=1+5+5^2+...+5^{100}\)
\(\Leftrightarrow5A=5+5^2+...+5^{101}\\ \Leftrightarrow4A=5^{101}-1\\ \Leftrightarrow A=\dfrac{5^{101}-1}{4}\)
Thay vào (1)
\(\left(1\right)=4\cdot\dfrac{5^{101}-1}{4}=5^{101}-1:5^{101}-1=1\)
Vậy \(4+4\cdot5+4\cdot5^2+...+4\cdot5^{100}:5^{101}-1=1\)