17+17+17=17x3=41

\(\frac{5.7.8.9.10}{7.8.9.10.11}=\frac{5}{11}\)

\(\frac{5\times7\times8\times9\times10}{7\times8\times9\times10\times11}=\frac{5\times̸7\times̸8\times̸9\times̸10}{̸7\times̸8\times̸9\times̸10\times11}=\frac{5}{11}\)

5 /11 là đúng cảm ơn các bạn

\(\frac{x+5}{13}+\frac{x+6}{12}+\frac{x+7}{11}=\frac{x+8}{10}+\frac{x+9}{9}+\frac{x+10}{8}\)

\(\Leftrightarrow\left(\frac{x+5}{13}+1\right)+\left(\frac{x+6}{12}+1\right)+\left(\frac{x+7}{11}+1\right)=\left(\frac{x+8}{10}+1\right)+\left(\frac{x+9}{9}+1\right)+\left(\frac{x+10}{8}\right)\)

\(\Leftrightarrow\frac{x+18}{13}+\frac{x+18}{12}+\frac{x+18}{11}=\frac{x+18}{10}+\frac{x+18}{9}+\frac{x+18}{8}\)

ta chuyển về vế trái được 

\(\Leftrightarrow\left(x+18\right)\left(\frac{1}{13}+\frac{1}{122}+\frac{1}{11}-\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\right)=0\)

\(\Leftrightarrow x+2018=0\)(do cái còn lại khác 0)

\(\Leftrightarrow x=-2018\)

mình nghĩ đề cậu viết thiếu mình sửa rồi

Ta có:

\(\frac{x+5}{13}+\frac{x+6}{12}+\frac{x+7}{11}=\frac{x+8}{10}+\frac{x+9}{9}+\frac{x+10}{8}\)

\(\Rightarrow\left(\frac{x+5}{13}+1\right)+\left(\frac{x+6}{12}+1\right)+\left(\frac{x+7}{11}+1\right)=\left(\frac{x+8}{10}+1\right)+\left(\frac{x+9}{9}+1\right)+\left(\frac{x+10}{8}+1\right)\)

\(\Rightarrow\frac{x+18}{13}+\frac{x+18}{12}+\frac{x+18}{11}=\frac{x+18}{10}+\frac{x+18}{9}+\frac{x+18}{8}\)

\(\Rightarrow\frac{x+18}{13}+\frac{x+18}{12}+\frac{x+18}{11}-\frac{x+18}{10}-\frac{x+18}{9}-\frac{x+18}{8}=0\)

\(\Rightarrow\left(x+18\right)\times\left(\frac{1}{13}+\frac{1}{12}+\frac{1}{11}-\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\right)=0\)

Vì \(\frac{1}{13}+\frac{1}{12}+\frac{1}{11}-\frac{1}{10}-\frac{1}{9}-\frac{1}{8}\ne0\)

\(\Rightarrow x+18=0\)

\(\Rightarrow x=-18\)

Vậy phương trình có nghiệm là x = -18

xin lỗi vì tớ nhầm một chút ở dòng cuối cho mình xin lỗi nhé

Hãy k tui đi

Nếu bạn k tui

Tui không k lại đâu

THANKS

k là gì vậy bạn, tui ko hiểu bạn nói gì.

ý bạn là gì vây?

Bài 11:

a, Ta có: \(n_{H_2}=\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

PT: \(A+H_2SO_4\rightarrow ASO_4+H_2\)

___0,2________________0,2 (mol)

\(\Rightarrow M_A=\dfrac{4,8}{0,2}=24\left(g/mol\right)\)

Vậy: A là Mg.

b, Theo PT: \(n_{H_2SO_4}=n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow V_{ddH_2SO_4}=\dfrac{0,2}{0,5}=0,4\left(l\right)\)

c, Theo PT: \(n_{MgSO_4}=n_{H_2}=0,2\left(mol\right)\)

\(\Rightarrow C_{M_{MgSO_4}}=\dfrac{0,2}{0,4}=0,5M\)

Bạn tham khảo nhé!

Bài 9:

Giả sử KL cần tìm là A.

PT: \(A+2HCl\rightarrow ACl_2+H_2\)

____0,3___0,6 (mol)

\(\Rightarrow M_A=\dfrac{7,2}{0,3}=24\left(g/mol\right)\)

Vậy: A là Magie (Mg)

Bạn tham khảo nhé!

Bài 10:

Ta có: \(n_{H_2}=\dfrac{8,96}{22,4}=0,4\left(mol\right)\)

PT: \(X+2HCl\rightarrow XCl_2+H_2\)

___0,4________________0,4 (mol)

\(\Rightarrow M_X=\dfrac{9,6}{0,4}=24\left(g/mol\right)\)

Vậy: X là Magie (Mg)

Bạn tham khảo nhé!

Bài 9

a; A = (1 + \(\dfrac{1}{3}\))\(\times\)(1 + \(\dfrac{1}{8}\))\(\times\)(1 + \(\dfrac{1}{15}\))\(\times\)...\(\times\)(1 + \(\dfrac{1}{9999}\))

    A = \(\dfrac{3+1}{3}\)\(\times\)\(\dfrac{8+1}{8}\)\(\times\)\(\dfrac{15+1}{15}\)\(\times\)...\(\times\)\(\dfrac{9999+1}{9999}\)

   A  = \(\dfrac{4}{3}\)\(\times\)\(\dfrac{9}{8}\)\(\times\)\(\dfrac{16}{15}\)\(\times\)...\(\times\)\(\dfrac{10000}{9999}\)

A = \(\dfrac{2\times2\times3\times3\times4\times4\times...\times100\times100}{1\times3\times2\times4\times3\times5\times...\times99\times101}\)

A = \(\dfrac{2\times100}{1\times101}\)

A = \(\dfrac{200}{101}\)

b; B = (1 - \(\dfrac{1}{4}\))\(\times\) (1 - \(\dfrac{1}{9}\)) \(\times\) (1 - \(\dfrac{1}{16}\))\(\times\)...\(\times\)(1 - \(\dfrac{1}{10000}\))

    B = \(\dfrac{4-1}{4}\) \(\times\) \(\dfrac{9-1}{9}\) \(\times\) \(\dfrac{16-1}{16}\) \(\times\) \(\dfrac{10000-1}{10000}\)

    B = \(\dfrac{3}{4}\)\(\times\) \(\dfrac{8}{9}\)\(\times\)\(\dfrac{15}{16}\)\(\times\)...\(\times\) \(\dfrac{9999}{10000}\)

   B = \(\dfrac{1\times3}{2\times2}\)\(\times\)\(\dfrac{2\times4}{3\times3}\)\(\times\)\(\dfrac{3\times5}{4\times4}\)\(\times\)...\(\times\)\(\dfrac{99\times101}{100\times100}\)

  B = \(\dfrac{1\times101}{2\times100}\)

  B = \(\dfrac{101}{200}\)

Câu 7 :

a, Tham khảo : https://olm.vn/hoi-dap/detail/43902845942.html

b, Tham khảo : https://olm.vn/hoi-dap/detail/7963533510.html

Bài 8:

Bài 7: 

b: Ta có: \(\dfrac{2}{1\cdot2}+\dfrac{2}{2\cdot3}+\dfrac{2}{3\cdot4}+...+\dfrac{2}{18\cdot19}+\dfrac{2}{19\cdot20}\)

\(=2\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{18}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{20}\right)\)

\(=2\cdot\dfrac{19}{20}=\dfrac{19}{10}\)

11 * x - 11 * 6 = 4 * x + 11

11 * x - 66 = 4 * x + 11

11 * x - 4 * x = 66 + 11

7 * x = 77

x = 77 : 7

x = 11

cảm ơn nguyen thi bao hoan nhiều nhé