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 \(\frac{118}{119^{211}+1}< \frac{118}{119^{210}+1}\)

\(=>119K< 119H\)

\(=>K< H\)

6: =x^2-7xy+5xy-35y^2

=x(x-7y)+5y(x-7y)

=(x-7y)(x+5y)

7: =x^2-2xy-8xy+16y^2

=x(x-2y)-8y(x-2y)

=(x-2y)(x-8y)

8: =3x^2-6xy-4xy+8y^2

=3x(x-2y)-4y(x-2y)

=(x-2y)(3x-4y)

9: =4x^2+4xy+y^2-16y^2

=(2x+y)^2-16y^2

=(2x+y-4y)(2x+y+4y)

=(2x-3y)*(2x+5y)

10: =2(x^2+5xy+4y^2)

=2(x+y)(x+4y)

11: =5x(x+2y+y^2)

( x + 1 ) + ( x + 2 ) + ( x + 3 ) + ... + ( x + 210 ) + ( x + 211 ) = 23632

x + 1 + x + 2 + x +3 + .... + x + 210 + x + 211 = 23632

( x + x + x + .... + x ) + ( 1 + 2 + 3 + ... + 210 + 211 ) = 23632

211x + 22366 = 23632

             211x  = 23632 - 22366

             211x  = 1266

=> x = 1266 : 211 = 6

Vậy x = 6

ta có (x+1) + (x+2) + (x+3) + ... + (x+210) + (x+211) = 23632

     = X + ( 1 + 2 +3 + 4 + .... + 210 + 211 ) = 23632

     = X + ( 1 + 211 ) + ( 2 + 210 ) + .... =23632

     = X + 212 x 105 =23632

     = X + 22260 = 23632

     = X          = 23632 - 22260

     = X          = 1372

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