tinh : (12+22+32+........+20122)x(91-273 : 3)
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\(\left(1^2+2^2+3^2+...+1000000^2\right).\left(91-273:3\right)\\ =\left(1^2+2^2+3^2+...+1000000^2\right).0=0\)
(12+22+32+..+10000002).(91-273:3)
=(12+22+32+..+10000002).(91-91)
=(12+22+32+..+10000002).0
=0
(12+22...+10002).(91-273:3)
=(12+22...+10002).(91-91)
=(12+22...+10002).0
=0
a) 19 + (x - 23) = 50
x - 23 = 50 - 19
x - 23 = 31
=> x = 31 + 23
x = 54
b) (273 - x) - 140 = 34
(273 - x) = 34 + 140
(273 - x) = 174
x = 273 - 174
x = 99
c) 56 - (x - 14) = 22
x - 14 = 56 - 22
x - 14 = 34
x = 34 + 14
x = 48
d) 172 - (137 - x) = 93
137 - x = 172 - 93
137 - x = 79
x = 137 - 79
x = 58
e) 13. (x - 14) = 91
x - 14 = 91 : 13
x - 14 = 7
x = 14 + 7
x = 21
g) (13 - x).17 = 51
13 - x = 51 : 17
13 - x = 3
x = 13 - 3
x = 10
chúc bn học giỏi ^^ !
ok mk nha!! ! 5654645645767676576558578779769675634636457567687689978978978978766876765657886567567
1.19+(x-23)=50
<=>x-23=50-19
<=>x-23=31
<=>x=31+23
<=.x=54
vậy x=54
a) \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3^{32}-1\right)< 3^{32}-1=B\)
b) \(A=2011.2013=\left(2012-1\right)\left(2012+1\right)=2012^2-1< 2012^2=B\)
a: =1/8-1/8+2/5=2/5
b: =(-1/15-14/15)+(23/27+31/27)=2-1=1
c: \(=\dfrac{3}{7}\left(\dfrac{22}{21}+\dfrac{5}{21}+\dfrac{15}{21}\right)=\dfrac{3}{7}\cdot2=\dfrac{6}{7}\)
d: \(=\dfrac{-8}{9}\cdot\dfrac{3}{2}+\dfrac{1}{9}\cdot\dfrac{-3}{2}=-\dfrac{3}{2}\)
=(12+22+...+20122)(91-91)
=(12+22+...+2012).0
=0