6) 31 - (17 + x) = 18

17 + x = 31 - 18

17 + x = 13

x = 13 - 17

x = -4

7) (-x + 31) - 39 = -69 + 11

31 - x - 39 = -58

-8 - x = -58

x = -8 - (-58)

x = 50

8) x - (18 - 13) = 5 - (24 - 27)

x - 5 = 5 - (-3)

x - 5 = 8

x = 8 + 5

x = 13

9) 315 - (x + 315) = 43 + (9 - 21)

315 - x - 315 = 43 - 12

-x = 31

x = -31

10) 25 - (25 - x) = 12 + (42 - 65)

25 - 25 + x = 12 - 23

x = -11

a) \(A=2+2^2+2^3+\dots+2^{60}\)

\(2A=2^2+2^3+2^4+\dots+2^{61}\)

\(2A-A=\left(2^2+2^3+2^4+\dots+2^{61}\right)-\left(2+2^2+2^3+\dots+2^{60}\right)\)

\(A=2^{61}-2\)

Vậy: \(A=2^{61}-2\).

b)

+) \(A=2+2^2+2^3+\dots+2^{60}\)

\(=\left(2+2^2\right)+\left(2^3+2^4\right)+\left(2^5+2^6\right)+\dots+\left(2^{59}+2^{60}\right)\)

\(=2\cdot\left(1+2\right)+2^3\cdot\left(1+2\right)+2^5\cdot\left(1+2\right)+\dots+2^{59}\cdot\left(1+2\right)\)

\(=2\cdot3+2^3\cdot3+2^5\cdot3+\dots+2^{59}\cdot3\)

\(=3\cdot\left(2+2^3+2^5+\dots+2^{59}\right)\)

Vì \(3\cdot\left(2+2^3+2^5+\dots+2^{59}\right)⋮3\) nên \(A⋮3\)

+) \(A=2+2^2+2^3+\dots+2^{60}\)

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+\left(2^9+2^{10}+2^{11}+2^{12}\right)+\dots+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)

\(=2\cdot\left(1+2+2^2+2^3\right)+2^5\cdot\left(1+2+2^2+2^3\right)+2^9\cdot\left(1+2+2^2+2^3\right)+\dots+2^{57}\cdot\left(1+2+2^2+2^3\right)\)

\(=2\cdot15+2^5\cdot15+2^9\cdot15+\dots+2^{57}\cdot15\)

\(=15\cdot\left(2+2^5+2^9+\dots+2^{57}\right)\)

Vì \(15⋮5\) nên \(15\cdot\left(2+2^5+2^9+\dots+2^{57}\right)⋮5\)

hay \(A\vdots5\)

+) \(A=2+2^2+2^3+\dots+2^{60}\)

\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+\left(2^7+2^8+2^9\right)+\dots+\left(2^{58}+2^{59}+2^{60}\right)\)

\(=2\cdot\left(1+2+2^2\right)+2^4\cdot\left(1+2+2^2\right)+2^7\cdot\left(1+2+2^2\right)+\dots+2^{58}\cdot\left(1+2+2^2\right)\)

\(=2\cdot7+2^4\cdot7+2^7\cdot7+\dots+2^{58}\cdot7\)

\(=7\cdot\left(2+2^4+2^7+\dots+2^{58}\right)\)

Vì \(7\cdot\left(2+2^4+2^7+\dots+2^{58}\right)⋮7\) nên \(A⋮7\)

$Toru$

a) �=2+22+23+⋯+260A=2+22+23+⋯+260

2�=22+23+24+⋯+2612A=22+23+24+⋯+261

2�−�=(22+23+24+⋯+261)−(2+22+23+⋯+260)2A−A=(22+23+24+⋯+261)−(2+22+23+⋯+260)

�=261−2A=261−2

Vậy: �=261−2A=261−2.

b)

+) �=2+22+23+⋯+260A=2+22+23+⋯+260

=(2+22)+(23+24)+(25+26)+⋯+(259+260)=(2+22)+(23+24)+(25+26)+⋯+(259+260)

=2⋅(1+2)+23⋅(1+2)+25⋅(1+2)+⋯+259⋅(1+2)=2⋅(1+2)+23⋅(1+2)+25⋅(1+2)+⋯+259⋅(1+2)

=2⋅3+23⋅3+25⋅3+⋯+259⋅3=2⋅3+23⋅3+25⋅3+⋯+259⋅3

=3⋅(2+23+25+⋯+259)=3⋅(2+23+25+⋯+259)

Vì 3⋅(2+23+25+⋯+259)⋮33⋅(2+23+25+⋯+259)⋮3 nên $A⋮3$

+) �=2+22+23+⋯+260A=2+22+23+⋯+260

=(2+22+23+24)+(25+26+27+28)+(29+210+211+212)+⋯+(257+258+259+260)=(2+22+23+24)+(25+26+27+28)+(29+210+211+212)+⋯+(257+258+259+260)

=2⋅(1+2+22+23)+25⋅(1+2+22+23)+29⋅(1+2+22+23)+⋯+257⋅(1+2+22+23)=2⋅(1+2+22+23)+25⋅(1+2+22+23)+29⋅(1+2+22+23)+⋯+257⋅(1+2+22+23)

=2⋅15+25⋅15+29⋅15+⋯+257⋅15=2⋅15+25⋅15+29⋅15+⋯+257⋅15

=15⋅(2+25+29+⋯+257)=15⋅(2+25+29+⋯+257)

Vì $15⋮5$ nên 15⋅(2+25+29+⋯+257)⋮515⋅(2+25+29+⋯+257)⋮5

hay $A⋮5$

+) �=2+22+23+⋯+260A=2+22+23+⋯+260

=(2+22+23)+(24+25+26)+(27+28+29)+⋯+(258+259+260)=(2+22+23)+(24+25+26)+(27+28+29)+⋯+(258+259+260)

=2⋅(1+2+22)+24⋅(1+2+22)+27⋅(1+2+22)+⋯+258⋅(1+2+22)=2⋅(1+2+22)+24⋅(1+2+22)+27⋅(1+2+22)+⋯+258⋅(1+2+22)

=2⋅7+24⋅7+27⋅7+⋯+258⋅7=2⋅7+24⋅7+27⋅7+⋯+258⋅7

=7⋅(2+24+27+⋯+258)=7⋅(2+24+27+⋯+258)

Vì 7⋅(2+24+27+⋯+258)⋮77⋅(2+24+27+⋯+258)⋮7 nên $A⋮7$

0,8 : 2,5

= 0,32

a, \(N_2+3H_2⇌2NH_3\)

b, Ta có: \(K_c=\dfrac{\left[NH_3\right]^2}{\left[N_2\right]\left[H_2\right]^3}=\dfrac{0,62^2}{0,45.0,14^3}\approx311,31\)

c, - Tăng nhiệt độ → giảm hiệu suất.

- Tăng áp suất → tăng hiệu suất.

- Thêm bột sắt (xúc tác) → không làm thay đổi hiệu suất.

Ta có: \(n_{KOH}=0,05.0,1=0,005\left(mol\right)\)

\(n_{HNO_3}=0,052.0,1=0,0052\left(mol\right)\)

PT: \(KOH+HNO_3\rightarrow KNO_3+H_2O\)

Xét tỉ lệ: \(\dfrac{0,005}{1}< \dfrac{0,0052}{1}\), ta được HNO₃ dư.

Theo PT: \(n_{HNO_3\left(pư\right)}=n_{KOH}=0,005\left(mol\right)\)

\(\Rightarrow n_{H^+}=n_{HNO_3\left(dư\right)}=0,0002\left(mol\right)\)

\(\Rightarrow\left[H^+\right]=\dfrac{0,0002}{0,05+0,052}=\dfrac{1}{510}\left(M\right)\)

⇒ pH = -log[H⁺] ≃ 2,71

Gọi x (cuốn) là số sách cần tìm (x ∈ ℕ* và 99 < x < 1000)

Do khi xếp thành từng bó 18 cuốn, 22 cuốn, 24 cuốn đều thừa ra 5 cuốn nên x - 5 ∈ BC(18; 22; 24)

Ta có:

18 = 2.3²

22 = 2.11

24 = 2³.3

⇒ BCNN(18; 22; 24) = 2³.3².11 = 792

⇒ x - 5 ∈ BC(18; 22; 24) = B(792) = {0; 792; 1584; ...}

⇒ x ∈ {5; 797; 1589; ...}

Mà 99 < x < 1000

⇒ x = 797

Vậy số cuốn sách cần tìm là 797 cuốn

cứuuuuu ai chơi lq ko ?

2\(xy\) + ? = 8 thế em

\(\dfrac{5x+y^2}{x^2y}-\dfrac{5y-x^2}{xy^2}\left(dkxd:x,y\ne0\right)\)

\(=\dfrac{y\left(5x+y^2\right)}{x^2y^2}-\dfrac{x\left(5y-x^2\right)}{x^2y^2}\)

\(=\dfrac{5xy+y^3-5xy+x^3}{x^2y^2}\)

\(=\dfrac{x^3+y^3}{x^2y^2}\)

They are speaking to some Japanese tourists.

They are speaking to some Japanese tourists.