= 34 . ( 29 - 59 )

= 34 .  -30

= -1020

29.34-34.59

=34.(29-59)

=34.(-30)

=-1020

\(29.34-34.59\)

\(=34.\left(29-59\right)\)

\(=34.\left(-30\right)\)

\(=-1020\)

210+[16+(-210)-16]

=[210+(-210)]+[16+(-16)]

=0+0

=0

k mk nhe

= [ (-210)+210]+(16-16)

= 0 + 0

= 0

0 nhé bạn

Bạn kiểm tra lại giùm mình đi nha:

15.37.4+120.21+21.5.12                                                         16.64+76.34

=15.37.4+(6.7).4.(5.3)+(3.5).(7.3).4                                          =

=15.37.4+42.4.15+15.21.4

=15.4.(37+42+21)

=60.100

=6000

840 : (x : 29) = 210

=> x : 29 = 840 : 210

=> x : 29 = 4

=> x = 4×29

=> x = 116

\(840:\left(x:29\right)=210\)

\(x:29=840:210\)

\(x:29=4\)

\(x=4\times29\)

\(x=116\)

Vậy: x=116 

x=116

$m_{H_2O} = D.V = 1.4,459 = 4,459(gam)$

Gọi $n_X = a(mol)$
$2X + 2H_2O \to 2XOH + H_2$
Theo PTHH : $n_{H_2} = \dfrac{1}{2}n_X = 0,5a(mol)$

$n_{XOH} = n_X = a(mol)$

Sau phản ứng, $m_{dd} = 0,897 + 4,459 - 0,5a.2 = 5,356 - a(gam)$

Suy ra : 

$C\%_{XOH} = \dfrac{a(X + 17)}{5,356 - a}.100\% = 29,34\%$

mà $X.a = 0,897$ nên $a = 0,039 ; X = 23(Natri)$

Đặt \(A=2^0+2^1+...+2^{10}\)

\(\Leftrightarrow2A=2^2+2^3+...+2^{11}\\ \Leftrightarrow A=2^{11}-2^0=2^{11}-1\)

  S = 1 + 2 + 2² + ... + 2⁹ + 2¹⁰

2S = 2 . (2 + 2² + ... + 2⁹ + 2¹⁰)

2S = 2 + 2² + 2³ + ... + 2¹⁰ + 2¹¹  

2S - S = (2 + 2² + 2³ + ... + 2¹⁰ + 2¹¹ ) - (1 + 2 + 2² + ... + 2⁹ + 2¹⁰)

S = 2¹¹ - 1

Đặt A = 2⁰+2¹+...+2¹⁰

=>2A = 2²+2³+...+2¹¹

=>A= 2¹¹ − 2⁰ = 2¹¹ − 1

Cho mik 5 sao nha

Chúc hok tốt

2⁹ =2^10-1 = 2¹⁰: 2¹ = 1024 : 2 = 512

2¹¹ = 2¹⁰⁺¹ = 2¹⁰ . 2¹  = 1024. 2 = 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 $2^9+1< 2^{11}+1\Rightarrow \frac{1}{2^9+1}> \frac{1}{2^{11}+1}$

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

$\Rightarrow A< B$