Bài 1.41

Ta có: 2⁹<2¹⁰; 2¹⁰<2¹¹

⇒2⁹<2¹⁰<2¹¹

⇒2⁹<2¹¹

Bài 1.42

a) 5⁷.5³=5⁷⁺³=5¹⁰

b) 5⁸:5⁴=5^8-4=5⁴

Bài 1.41

\(2^9=2^{10}:2=1024:2=512\)

\(2^{11}=2^{10}\cdot2=1024\cdot2=2048\)

\(2^9=2^{10-1}=2^{10}:2=1024:2=512\\ 2^{11}=2^{10+1}=2^{10}\cdot2=1024\cdot2=2048\)

Với \(2^{10}=1024\) 

hay \(2.2.2.2.2.2.2.2.2.2=1024\)

⇒ \(2^9=...\) ( do \(9< 10\) nên \(1024:2=512\) )

Vậy \(9^2=512\)

⇒ \(2^{11}=...\) ( do \(11>10\) nên \(1024.2=2048\) )

Vậy \(9^{11}=2048\)

Lời giải:

$A=\frac{2^{10}+2-1}{2^9+1}=\frac{2(2^9+1)-1}{2^9+1}=2-\frac{1}{2^9+1}$

$B=\frac{2^{12}+1}{2^{11}+1}=\frac{2(2^{11}+1)-1}{2^{11}+1}=2-\frac{1}{2^{11}+1}$

Vì $2^9+1< 2^{11}+1\Rightarrow \frac{1}{2^9+1}> \frac{1}{2^{11}+1}$

$\Rightarrow 2-\frac{1}{2^9+1}< 2-\frac{1}{2^{11}+1}$

$\Rightarrow A< B$

1/

Tổng A là tổng các số hạng cách đều nhau 4 đơn vị.

Số số hạng: $(101-1):4+1=26$

$A=(101+1)\times 26:2=1326$

2/

$B=(1+2+2^2)+(2^3+2^4+2^5)+(2^6+2^7+2^8)+(2^9+2^{10}+2^{11})$

$=(1+2+2^2)+2^3(1+2+2^2)+2^6(1+2+2^2)+2^9(1+2+2^2)$

$=(1+2+2^2)(1+2^3+2^6+2^9)$

$=7(1+2^3+2^6+2^9)\vdots 7$

\(119H=\frac{119\left(119^{209}+1\right)}{119^{210}+1}=\frac{119^{210}+119}{119^{210}+1}=1+\frac{118}{119^{210}}\)

\(119K=\frac{119\left(119^{210}+1\right)}{119^{211}+1}=\frac{119^{211}+119}{119^{211}+1}=1+\frac{118}{119^{211}+1}\)

Vì 119²¹¹+1>119²¹⁰+1 nên \(\frac{118}{119^{211}+1}< \frac{118}{119^{210}+1}\)

\(=>119K< 119H\)

\(=>K< H\)

c ) = (1 - 2 ) + ( 3 - 4 ) + .........  + ( 209 - 210 ) + 211

     = -1 + -1 + -1 ........ + -1 + 211 ( có 105 số - 1)

     =  -1 x 105 + 211

    = - 105  + 211

    =  106

b ) = 1152 - 374 - 1152 + -65 + 374

     = ( 1152 - 1152 ) + ( 374 - 374 ) + -65

     = 0 + 0 + -65

     = 0 +-65 

     =  -65

210 = 1024

Vì 1024 > 100 nên 210 > 100

a)  5 8 + 3 17 + 4 18 + 20 − 17 + − 2 9 + 21 56

= 5 8 + 3 17 + 2 9 + − 20 17 + − 2 9 + 3 8 = 5 8 + 3 8 + 3 17 + − 20 17 + 2 9 + − 2 9 = 1 + − 1 + 0 = 0

b)  − 2 11 + 6 7 + 1 2 + − 9 11 + 1 7

= − 2 11 + − 9 11 + 6 7 + 1 7 = − 1 + 1 = 0