Cấu hình e lớp ngoài cùng của ion X 2 + là 3 s 2 3 p 6 3 d 6 .  Cấu hình e của X là

A. 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 8

B.  1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 4

C.  1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 6 4 s 2  

D. 1 s 2 2 s 2 2 p 6 3 s 2 3 p 6 3 d 4 4 s 1  

1) 2155-(174+2155)+(-68+174-7)-1911-(1234-1911)

=2155-174-2155+(-68)+174-7-1911-1234+1911

=(2155-2155)+(174-174)+(1911-1911)-(68+7+1234)

= 0 + 0 + 0 - 1309

= -1309

\(A=\dfrac{2^4.3^3+2^3.3^4}{2^5.3^4-2^6.3^3}=\dfrac{2^3.3^3.\left(2+3\right)}{2^5.3^3.\left(3-2\right)}=\dfrac{2^3.3^3.5}{2^5.3^3.1}\)

\(=\dfrac{5}{2^2}=\dfrac{5}{4}\)

bài 1:

a) \(4\dfrac{1}{2}x:\dfrac{5}{12}=0,5\) ; b)\(1,5+1\dfrac{1}{4}x=\dfrac{2}{3}\)

\(\dfrac{9}{2}x:\dfrac{5}{12}=\dfrac{1}{2}\) \(\dfrac{3}{2}+\dfrac{5}{4}x=\dfrac{2}{3}\)

\(\dfrac{9}{2}x\) \(=\dfrac{1}{2}.\dfrac{5}{12}\) \(\dfrac{5}{4}x=\dfrac{2}{3}-\dfrac{3}{2}\)

\(\dfrac{9}{2}x\) \(=\dfrac{5}{24}\) \(\dfrac{5}{4}x=\dfrac{-5}{6}\)

\(x\) \(=\dfrac{5}{24}:\dfrac{9}{2}\) \(x=\dfrac{-5}{6}:\dfrac{5}{4}\)

\(x\) \(=\dfrac{5}{108}\) \(x=\dfrac{-2}{3}\)

c) Cho mình hỏi x ở đâu vậy ???

d)\(\left(x-5\right):\dfrac{1}{3}=\dfrac{2}{5}\) e)\(\left(4,5-2x\right):\dfrac{3}{4}=1\dfrac{1}{3}\)

\(\left(x-5\right)\) \(=\dfrac{2}{5}.\dfrac{1}{3}\) \(\left(\dfrac{9}{2}-2x\right):\dfrac{3}{4}=\dfrac{4}{3}\)

\(x-5\) \(=\dfrac{2}{15}\) \(\dfrac{9}{2}-2x\) =\(\dfrac{4}{3}.\dfrac{3}{4}\)

\(x\) \(=\dfrac{2}{15}+5\) \(\dfrac{9}{2}-2x=1\)

\(x\) \(=\dfrac{77}{15}\) \(2x=\dfrac{9}{2}-1\)

f) \(\left(2,7x-1\dfrac{1}{2}x\right):\dfrac{2}{7}=\dfrac{-21}{7}\) \(2x=\dfrac{7}{2}\)

\(\left(\dfrac{27}{10}x-\dfrac{3}{2}x\right):\dfrac{2}{7}=-3\) \(x=\dfrac{7}{2}:2\)

\(\left[x\left(\dfrac{27}{10}-\dfrac{3}{2}\right)\right]=-3.\dfrac{2}{7}\) \(x=\dfrac{7}{4}\)

\(x.\dfrac{6}{5}=\dfrac{-6}{7}\)

\(x=\dfrac{-6}{7}:\dfrac{6}{5}\)

\(x=\dfrac{-5}{7}\)

bài 2:

Theo bài ra ta có :\(\dfrac{a}{27}=\dfrac{-5}{9}=\dfrac{-45}{b}\)

\(\Rightarrow9a=27.\left(-5\right)\Rightarrow a=\dfrac{27.\left(-5\right)}{9}=-15\)

\(\Rightarrow\left(-5\right)b=\left(-45\right).9\Rightarrow b=\dfrac{\left(-45\right).9}{-5}=81\)

Vậy \(a=-15;b=81\)

hình như là sai đề nếu đúng thì phải: \(\dfrac{6}{1.3}+\dfrac{6}{3.5}+\dfrac{6}{5.7}+\dfrac{6}{7.9}\)

222³³³= (222³)¹¹¹=[(2.111)³]¹¹¹=(2³)¹¹¹.(111¹¹¹)³

=8¹¹¹.111¹¹¹.(111¹¹¹)²=(8.111)¹¹¹.(111¹¹¹)²=888¹¹¹.(111¹¹¹)²

333²²²=(333²)¹¹¹=[(3.111)²]¹¹¹=(3²)¹¹¹.(111¹¹¹)²=9¹¹¹.(111¹¹¹)²

Vì 888¹¹¹>9¹¹¹ nên 222³³³>333²²²

Ta thấy: \(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)

\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)

\(...\)

\(\dfrac{1}{100^2}< \dfrac{1}{99.100}\)

\(\Rightarrow\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)

\(\Rightarrow\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{1}{2}-\dfrac{1}{100}\)

\(\Rightarrow\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{100^2}< \dfrac{49}{100}< \dfrac{50}{100}< \dfrac{1}{2}\) (ĐPCM)

ta có : \(\dfrac{3}{2}\)A= \(\dfrac{3}{4}+\)\(\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+\)\(...+\left(\dfrac{3}{2}\right)^{2013}\) (1)

A= \(\dfrac{1}{2}+\dfrac{3}{2}\)\(+\left(\dfrac{3}{2}\right)^2+...+\)\(\left(\dfrac{3}{2}\right)^{2012}\) (2)

Lấy (1) trừ đi (2) vế theo vế:

\(\dfrac{3}{2}A-A=\dfrac{3}{4}-\dfrac{1}{2}-\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^{2013}\)

\(\dfrac{1}{2}A=\left(\dfrac{3}{2}\right)^{2013}-\dfrac{5}{4}\Rightarrow A=\dfrac{3^{2013}}{2^{2012}}-\dfrac{5}{2}\)

ta có : \(B=\left(\dfrac{3}{2}\right)^{2013}:2=\dfrac{3^{2013}}{2^{2013}}.\dfrac{1}{2}=\dfrac{3^{2013}}{2^{2014}}\)

Vậy \(A-B=\dfrac{3^{2013}}{2^{2014}}-\left(\dfrac{3^{2013}}{2^{2012}}-\dfrac{5}{2}\right)\)

a) B= 5.6.7+6.7.8+7.8.9+...+2015.2016.2017

4B=4.(5.6.7+6.7.8+7.8.9+...+2015.2016.2017)

4B=5.6.7.4+6.7.8.4+7.8.9.4+...+2015.2016.2017.4

4B=5.6.7.(8-4)+6.7.8.(9-5)+7.8.9.(10-6)+...+2015.2016.2017.(2018-2014)

4B=5.6.7.8-4.5.6.7+6.7.8.9-5.6.7.8+7.8.9.10-6.7.8.9+...+2015.2016.2017.2018-2014.2015.2016.2017

4B= -4.5.6.7

B=\(\dfrac{-4.5.6.7}{4}\)

B= -210

b) D=7.9.11+9.11.13+11.13.15+...+101.103.105

8D= 8(7.9.11+9.11.13+11.13.15+...+101.103.105)

8D=7.9.11.8+9.11.13.8+11.13.15.8+...+101.103.105.8

8D=7.9.11.(13-5)+9.11.13.(15-7)+11.13.15.(17-9)+...+101.103.105.(107-99)

8D=7.9.11.13-5.7.9.11+9.11.13.15-7.9.11.13+11.13.15.17-9.11.13.15+...+101.103.105.107-99.101.103.105

8D=-5.7.9.11

D= \(\dfrac{-5.7.9.11}{8}\)

D= \(\dfrac{3465}{8}\)

Bài này mình làm rồi 

x⁴⁸ = x

=> x = {0;1}