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Bài 1:
Có : 2009 = 2008 + 1 = x + 1
Thay 2009 = x + 1 vào biểu thức trên,ta có :
x\(^5\)- 2009x\(^4\)+ 2009x\(^3\)- 2009x\(^2\)+ 2009x - 2010
= x\(^5\)- (x + 1)x\(^4\)+ (x + 1)x\(^3\)- (x +1)x\(^2\)+ (x + 1) x - (x + 1 + 1)
= x\(^5\)- x\(^5\)- x\(^4\)+ x\(^4\)- x\(^3\)+ x\(^3\)- x\(^2\)+ x\(^2\)+ x - x -1 - 1
= -2
\(a,20^8.4^8=\left(20.4\right)^8=80^8\)
\(b,10^6:2^6=\left(10:2\right)^6=5^6\)
\(c,5^4.2^8=5^4.\left(2^2\right)^4=5^4.4^4=\left(5.4\right)^4=20^4\)
\(d,7^8.9^4=7^8.\left(3^2\right)^4=7^8.3^8=\left(7.3\right)^8=21^8\)
\(e,27^4:25^6=\left(3^3\right)^4:\left(5^2\right)^6=3^{12}:5^{12}=\left(3:5\right)^{12}=\left(\frac{3}{5}\right)^{12}\)
\(a,10^8.2^8=\left(10.2\right)^8=20^8\)
\(b,10^8:2^8=\left(10:2\right)^8=5^8\)
\(c,25^4.2^8=25^4.\left(2^2\right)^4=25^4.4^4=\left(25.4\right)^4=100^4\)
\(d,15^8.9^4=15^8.\left(3^2\right)^4=15^8.3^8=\left(15.3\right)^8=45^8\)
\(e,27^2.25^3=\left(3^3\right)^2.\left(5^2\right)^3=3^6.5^6=\left(3.5\right)^6=15^6\)
a, 108 . 28
= ( 10 . 2 ) 8
= 208
b, 108 : 28
= ( 10 : 2 )8
= 58
c, 254 . 28
= ( 52 )4 . 28
= 58 . 28
= ( 5 . 2 ) 8
= 108
d, 158 . 94
= 158 . ( 32 )4
= 158 . 38
= ( 15 . 3 ) 8
= 458
e, 272 . 253
= ( 33 )2 . ( 52 ) 3
= 36 . 56
= ( 3 . 5 ) 6
= 15 6
a, (1/5)5x(3)5
=1/3125.243
=243/3125
b, (0,125)^3.512
=1/512.512
=1
c,(0,25)^4.1024
=1/256.1024
=4
2,hầu như chịu hết
a) (x-3)2=0
=> x-3 =0
=> x=3
b) =(x+1+1)(x+1-1)=0
=> x(x+2)=0
=> x=0 hoặc x = -2
c) -x3=8
=>x= \(\sqrt[3]{8}\)=2
1a) \(^{\left(x-3\right)^2=0}\)
=> x-3=0
x = 0+3
x = 3
b)\(^{\left(x+1\right)^2-1=0}\)
(x+1)^2=0+1=1
TH1: x+1=1
x = 1-1 =0
TH2: x+1=-1
x = (-1)-1=-2
c)\(^{-\left(x^3\right)=8}\)
-(x^3)=8
-(x^3)=-(2^3)
=> x=2