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c) x.(1+2+3+4+...+100)=0
x.5050=0
x=0:5050=0
Vậy x=0
d) x.(1+2+3+4+5+...+100)=5050
x.5050=5050
x=1
Vậy x=1
e) x+1+x+2+x+3+x+4+...+x+100=5050
(x+x+x+x+...+x)+(1+2+3+4+...+100)=5050
100 số hạng x
x.100+5050=5050
x.100=0
x=0
Vậy x=0
a) 3x2 + 12x =0
<=> 3x( x+ 4)=0
<=> \(\orbr{\begin{cases}3x=0\\x+4=0\end{cases}}\) <=>\(\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)
d) \(4x^3=4x\)
<=> \(4x^3-4x=0\)
<=> 4x( x2 -1) =0
<=> 4x ( x - 1) ( x+ 1) =0
<=> 4x=0 hoac x-1=0 hoac x+1=0
<=> x=0 hoac x=1 hoac x=-1
Bài 1
a) A = 2^0 + 2^1 + 2^2 +...+ 2^50
2A=2^1+2^2+2^3+...+2^51
2A-A=(2^1+2^2+2^3+...+2^51)-(2^0 + 2^1 + 2^2 +...+ 2^50)
A=(2^1-2^1)+(2^2-2^2)+...+(2^50-2^50)+(2^51-2^1)
A=0+0+...+0+(2^51-2^1)
A=2^51-2^1
b)B = 5 + 5^2 + 5^3 +...+ 5^99 + 5^100
5B=5^2+5^3+5^4+...+5^100+5^101
5B-B=(5^2+5^3+5^4+...+5^100+5^101)-( 5 + 5^2 + 5^3 +...+ 5^99 + 5^100)
4B=(5^2-5^2)+(5^3-5^3)+...+(5^100-5^100)+(5^101-5)
4B=0+0+...+0+(5^101-5)
4B=5^101-5
B=(5^101-5)/4
c)C = 3 - 3^2 + 3^3 - 3^4 +...+ 3^2009 - 3 ^2010
3C=3^2-3^3+3^4-3^5+...+3^2010-3^2011
3C-C=(3^2-3^3+3^4-3^5+...+3^2010-3^2011)-(3 - 3^2 + 3^3 - 3^4 +...+ 3^2009 - 3 ^2010)
...............................................!!!!!!!!!!!!!!!!!!!!!!!!
Bài 2
8(mình k0 chắc)
a) 10^30 và 2^100
Ta có: 10^30 = (10^3)^10 = 1000^10
2^100 = (2^10)^10 = 1024^10
Do 1024^10 > 1000^10 => 2^100 > 10^30
b) 333^444 và 444^333
Ta có: 333^444 = 111^444 x 3^444
444^333 = 111^333 x 4^333
Tách: 3^444 = (3^4)^111 =81^111 <=>4^333 = (4^3)^111 = 64^111
Mà: {111^444 > 111^333 (1)
{81^111 > 64^111 hay: (3^4)^111 > (4^3)^111 (2)
Từ (1) và (2) ta có:333^444 > 444^333
c) 3^450 =(3^3)^150 =27^150
5^300=(5^2)^150=25^150
vì 27^150 >25^150 =>3^450 > 5^300
vậy 3^450 > 5^300
a) \(10^{30}=\left(10^3\right)^{10}=1000^{10}\)
\(2^{100}=\left(2^{10}\right)^{10}=1024^{10}\)
Mà \(1000^{10}< 1024^{10}\Rightarrow10^{30}< 2^{100}\)
b) \(3^{400}=\left(3^4\right)^{100}=81^{100}\)
\(5^{300}=\left(5^3\right)^{100}=125^{100}\)
Mà \(81^{100}< 125^{100}\Rightarrow3^{400}< 5^{300}\)
c) \(333^{444}=\left(3.111\right)^{444}=3^{444}.111^{444}=\left(3^4\right)^{111}.111^{444}=81^{111}.111^{444}\)
\(444^{333}=\left(4.111\right)^{333}=4^{333}.111^{333}=\left(4^3\right)^{111}.111^{333}=64^{111}.111^{333}\)
Mà \(81^{111}.111^{444}>64^{111}.111^{333}\Rightarrow333^{444}>444^{333}\)
a) (a-b)-(c-b)-a=a-b-c+b-a=(a-a)+(b-b)-c=0+0-c=-c
1)
a) (a-b) - (c-b) - a= a - b - c + b - a = -c
b) -(300 - 400) + (300 - 400) + 100 = 100
2) a) 4x - 20 = 50 - ( 30 -3x)
4x - 20 = 50 - 30 + 3x
4x - 3x = 50 - 30 + 20= 40
x = 40
b) 100 - (-5x) = 40 - (-4x +10)
100 + 5x = 40 + 4x -10
5x - 4x = 40 - 10 -100
x = -70
c) (-50) + (-7x) = 100 - 8x
-7x + 8x = 100 + 50
x = 150