\(A=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{99}}\)

\(\Rightarrow\dfrac{A}{3}=\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\)

\(\Rightarrow A-\dfrac{A}{3}=\dfrac{2A}{3}=\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\right)-\left(\dfrac{1}{3^2}+\dfrac{1}{3^3}+\dfrac{1}{3^4}+...+\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\dfrac{2A}{3}=\left(\dfrac{1}{3^2}-\dfrac{1}{3^2}\right)+\left(\dfrac{1}{3^3}-\dfrac{1}{3^3}\right)+...+\left(\dfrac{1}{3^{99}}-\dfrac{1}{3^{99}}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)=\dfrac{1}{3}-\dfrac{1}{3^{100}}\)

\(\Rightarrow2A=3\cdot\left(\dfrac{1}{3}-\dfrac{1}{3^{100}}\right)\)

\(\Rightarrow\text{A}=\dfrac{1-\dfrac{1}{3^{99}}}{2}\)

\(\Rightarrow A=\dfrac{1}{2}-\dfrac{1}{2.3^{99}}< \dfrac{1}{2}\)

a)

Dãy trên có số số hạng là:

( 20 - 1 ) : 1 + 1 = 20 ( số hạng )

Tổng của dãy trên là:

( 20 + 1 ) x 20 : 2 = 210 

Đáp số: 210

b)

Dãy trên có số số hạng là:

( 21 - 1 ) : 2 + 1 = 11 ( số hạng )

Tổng của dãy trên là:

( 21 + 1 ) x 11 : 2 = 121

Đáp số: 121

c) ( 2x - 1 ) x 2 = 13

2x - 1 = \(\dfrac{13}{2}\)

2x = \(\dfrac{15}{2}\)
\(x=\dfrac{15}{4}\)

32 x ( x - 10 ) = 32

( x - 10 ) = 1

x = 11

\(A=1+2+3+...+20\)

Số hạng:

\(\left(20-1\right):1+1=20\) (số hạng)

Tổng: \(\left(20+1\right)\cdot20:2=210\)

\(B=1+3+5+...+21\)

Số hạng:

\(\left(21-1\right):2+1=11\) (số hạng) 

Tổng: \(\left(21+1\right)\cdot11:2=121\)

\(\left(2x-1\right)\cdot2=13\)

\(\Rightarrow2x-1=\dfrac{13}{2}\)

\(\Rightarrow2x=\dfrac{15}{2}\)

\(\Rightarrow x=\dfrac{15}{4}\)

\(32\cdot\left(x-10\right)=32\)

\(\Rightarrow x-10=1\)

\(\Rightarrow x=11\)

`a,` Khoảng cách là : `1`

Số số hạng là : \(\dfrac{20-1}{1}+1=20\)

Tổng là : \(\dfrac{20+1\times20}{2}=\)`210`

`b,` Khoảng cách là : `2`

Số số hạng là : \(\dfrac{21-1}{2}+1=11\)

Tổng là : \(\dfrac{21+1\times11}{2}=121\)

\(c,\left(2x-1\right).2=13\\ \Rightarrow2x-1=\dfrac{13}{2}\\ \Rightarrow2x=\dfrac{13}{2}+1\\ \Rightarrow2x=\dfrac{15}{2}\\ \Rightarrow x=\dfrac{15}{2}:2\\ \Rightarrow x=\dfrac{15}{4}\)

\(32.\left(x-10\right)=32\\ \Rightarrow x-10=32:32\\ \Rightarrow x-10=1\\ \Rightarrow x=1+10\\ \Rightarrow x=11\)

a, (4 - x )⁵ +(x - 2)⁵ =32

(=) 1024  -  x⁵ + x⁵  - 32 = 32

(=) -x⁵ + x⁵ = 32 + 32 - 1024

(=) 0x =  -960

=) phương trình vô nghiệm

sauriengroblox

3/2+5/4+9/8/+17/16+33/32-6+x-1/x+1=31/32-2/2015

=(1+1/2)+(1+1/4)+(1+1/8)+(1+1/16)+(1+1/32-6+x-1/x+1=31/32-2/2015

=(1/2+1/4+1/8+1/16+1/32)+(1+1+1+1+1)-6+x-1/x+1=31/32-2/2015

=31/32+5-6+x-1/x+1=31/32-2/2015

=5-6+x-1/x+1=31/32-2/2015-31/32

=-1+x-1/x+1=-2/2015

=x-1/x+1=-2/2015- -1

=x-1/x+1=2013/2015

=>x=2014

g)

Áp dụng tính chất của dãy tỉ số bằng nhau : 

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{x+y}{3+5}=\dfrac{32}{8}=4\)

\(+)\)\(\dfrac{x}{3}=4\Rightarrow x=4\times3=12\)

\(+)\)\(\dfrac{y}{5}=4\Rightarrow y=4\times5=20\)

Vậy \(x=12;y=20\)

h)

\(\dfrac{1}{2}-\left|x-\dfrac{1}{3}\right|=\left|-\dfrac{1}{5}\right|\)

\(\dfrac{1}{2}-\left|x-\dfrac{1}{3}\right|=\dfrac{1}{5}\)

\(\left|x-\dfrac{1}{3}\right|=\dfrac{1}{2}-\dfrac{1}{5}\)

\(\left|x-\dfrac{1}{3}\right|=\dfrac{5}{10}-\dfrac{2}{10}\)

\(\left|x-\dfrac{1}{3}\right|=\dfrac{3}{10}\)

\(\Rightarrow x-\dfrac{1}{3}=\dfrac{3}{10}\) hoặc \(x-\dfrac{1}{3}=\dfrac{-3}{10}\)

\(\Rightarrow\left[{}\begin{matrix}x-\dfrac{1}{3}=\dfrac{3}{10}\\x-\dfrac{1}{3}=\dfrac{-3}{10}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{10}+\dfrac{1}{3}\\x=\dfrac{-3}{10}+\dfrac{1}{3}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{30}+\dfrac{10}{30}\\x=\dfrac{-9}{30}+\dfrac{10}{30}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{19}{30}\\x=\dfrac{1}{30}\end{matrix}\right.\)

4 + x = 7 - 2x

4 + x + 2x = 7

4 + 3x = 7

3x = 7 - 4

3x = 3

x = 3 : 3

x =1

5 + x = 8,3 + 4,7 - x

5 + x + x = 8,3 + 4,7

5 + 2x = 13

2x = 13 -  5

2x = 8

x = 8 : 2

x= 4

Làm đc thì mình k cho 5 cái luôn

d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)

\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)

hay \(x=\dfrac{25}{372}\)

Vậy: \(x=\dfrac{25}{372}\)

e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)

\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)

f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)

\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)

\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)

\(\Leftrightarrow3x=\dfrac{1}{9}\)

hay \(x=\dfrac{1}{27}\)

g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)

\(\Leftrightarrow\dfrac{4}{3}x=2\)

hay \(x=\dfrac{3}{2}\)

Vậy: \(x=\dfrac{3}{2}\)

h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)

  Vậy ...

i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)

  Vậy ...

\(\dfrac{x}{4}=-\dfrac{5}{3}\)\(\Rightarrow x.3=-5.4\Rightarrow3x=-20\Rightarrow x=\dfrac{-20}{3}\)

\(\dfrac{1}{2}+x=1,5.\dfrac{4}{3}\)\(\Rightarrow\dfrac{1}{2}+x=2\Rightarrow x=2-\dfrac{1}{2}\Rightarrow x=\dfrac{3}{2}\)

\(\dfrac{2}{3}-\dfrac{1}{3}.x=\dfrac{1}{5}\Rightarrow\)\(\dfrac{1}{3}.x=\dfrac{2}{3}-\dfrac{1}{5}\Rightarrow\dfrac{1}{3}.x=\dfrac{7}{15}\Rightarrow x=\dfrac{7}{15}:\dfrac{1}{3}\)\(\Rightarrow x=\dfrac{7}{5}\)

\(\left(x-5,2\right).\dfrac{1}{3}=15\Rightarrow x-5,2=15:\dfrac{1}{3}\Rightarrow x-5,2=45\)\(\Rightarrow x=45+5,2\Rightarrow x=50,2\)

P/s Nhớ tick cho mình nha. Thanks bạn