a) Tổng \({S_n}\) là tổng của cấp số nhân có số hạng đầu \({u_1} = 1\) và công bội \(q = \frac{1}{3}\) nên ta có:

\({S_n} = \frac{{{u_1}\left( {1 - {q^n}} \right)}}{{1 - q}} = \frac{{1\left( {1 - {{\left( {\frac{1}{3}} \right)}^n}} \right)}}{{1 - \frac{1}{3}}} = \frac{{1 - {{\left( {\frac{1}{3}} \right)}^n}}}{{\frac{2}{3}}} = \frac{3}{2}\left( {1 - \frac{1}{{{3^n}}}} \right) = \frac{3}{2} - \frac{1}{{{{2.3}^{n - 1}}}}\)

b) Ta có:

\(\begin{array}{l}{S_n} = 9 + 99 + 999 + ... + \underbrace {99...9}_{n\,\,chu\,\,so\,\,9} = \left( {10 - 1} \right) + \left( {100 - 1} \right) + \left( {1000 - 1} \right) + ... + \left( {\underbrace {100...0}_{n\,\,chu\,\,so\,\,0} - 1} \right)\\ = \left( {10 + 100 + 1000 + ... + \underbrace {100...0}_{n\,\,chu\,\,so\,\,0}} \right) - n\end{array}\)

Tổng \(10 + 100 + 1000 + ... + \underbrace {100...0}_{n\,\,chu\,\,so\,\,0}\) là tổng của cấp số nhân có số hạng đầu \({u_1} = 10\) và công bội \(q = 10\) nên ta có:

\(10 + 100 + 1000 + ... + \underbrace {100...0}_{n\,\,chu\,\,s\^o \,\,0} = \frac{{10\left( {1 - {{10}^n}} \right)}}{{1 - 10}} = \frac{{10 - {{10}^{n + 1}}}}{{ - 9}} = \frac{{{{10}^{n + 1}} - 10}}{9}\)

Vậy \({S_n} = \frac{{{{10}^{n + 1}} - 10}}{9} - n = \frac{{{{10}^{n + 1}} - 10 - 9n}}{9}\)

a, \(\frac{x}{9}-\frac{3}{y}=\frac{1}{18}\)

\(\Rightarrow\frac{xy}{9y}-\frac{27}{9y}=\frac{1}{18}\Rightarrow y=2\)

\(\Rightarrow\frac{xy}{9y}-\frac{27}{9y}=\frac{1}{18}=\frac{2x}{18}-\frac{27}{18}=\frac{1}{18}\)

\(\Rightarrow2x-27=1\)

\(\Rightarrow2x=28\Rightarrow x=14\)

vậy x = 14

a, \(\frac{x}{9}-\frac{3}{y}=\frac{1}{18}\)

\(\Rightarrow\frac{xy}{9y}-\frac{27}{9y}=\frac{1}{9.2}\)

\(\Rightarrow9y=9.2\Rightarrow y=2\)

thay y = 2 vào ta có :

\(\frac{2x}{18}-\frac{27}{18}=\frac{1}{18}\)

\(\Rightarrow2x-27=1\Rightarrow2x=28\Rightarrow x=14\)

b, \(\frac{1}{x}=\frac{y}{2}-\frac{1}{3}\)

\(\Rightarrow\frac{1}{x}=\frac{3y}{6}-\frac{2}{6}\)

\(\Rightarrow\frac{1}{x}=\frac{3y-2}{6}\)

\(\Rightarrow x=6\)

2. \(B=\frac{10n-3}{4n-10}=\frac{\frac{5}{2}.\left(4n-10\right)+22}{4n-10}=\frac{5}{2}+\frac{22}{4n-10}\)

để \(B\) có giá trị lớn nhất thì \(\frac{22}{4n-10}\) là số dương lớn nhất 

=> 4n - 10 là số dương nhỏ nhất ( n thuộc N )

\(\Rightarrow4n-10=2\Rightarrow4n=12\Rightarrow n=3\)

ta có : 

\(B=\frac{10n-3}{4n-10}=\frac{30-3}{12-10}=\frac{27}{2}\)

Vậy để \(B\) có giá trị lớn nhất thì \(n=3\)

giá trị lớn nhất của \(B=\frac{27}{2}\)

Ta có : 

\(\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\) ( cái đề hình như có 1 phân số \(\frac{2}{9}\) đúng không bạn ) 

\(\Leftrightarrow\)\(\frac{2}{42}+\frac{2}{56}+\frac{2}{72}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{9}\)

\(\Leftrightarrow\)\(2\left(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2}{9}\)

\(\Leftrightarrow\)\(\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)

\(\Leftrightarrow\)\(\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+...+\frac{1}{x\left(x+1\right)}=\frac{1}{9}\)

\(\Leftrightarrow\)\(\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{1}{9}\)

\(\Leftrightarrow\)\(\frac{1}{6}-\frac{1}{x+1}=\frac{1}{9}\)

\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{6}-\frac{1}{9}\)

\(\Leftrightarrow\)\(\frac{1}{x+1}=\frac{1}{18}\)

\(\Leftrightarrow\)\(x+1=1:\frac{1}{18}\)

\(\Leftrightarrow\)\(x+1=18\)

\(\Leftrightarrow\)\(x=18-1\)

\(\Leftrightarrow\)\(x=17\)

Vậy \(x=17\)

Chúc bạn học tốt ~ 

fuck you

\(\left(\frac{99^9}{11^9}-\frac{99^{99}}{11^{99}}-\frac{99^{999}}{11^{999}}\right)\left(\frac{1}{5}-\frac{1}{7}-\frac{2}{35}\right)\)

\(=\left(\frac{99^9}{11^9}-\frac{99^{99}}{11^{99}}-\frac{99^{999}}{11^{999}}\right)\left(\frac{7}{35}-\frac{5}{35}-\frac{2}{35}\right)\)

\(=\left(\frac{99^9}{11^9}-\frac{99^{99}}{11^{99}}-\frac{99^{999}}{11^{999}}\right).0\)

\(=0\)

bài dễ thế không ai làm được hay thật

a) \(\frac{1}{81}\times\left(\frac{1}{3}\right)^{-2}\times9\times3^3\)
\(=\frac{3^7}{3^4}\)
\(=3^3\)
 

b) \(\left(2^5\times4\right)\div\left(2^3\times\frac{1}{16}\right)\)
\(=2^7\div\frac{2^3}{2^{\text{4}}}\)
\(=2^7\div\frac{1}{2}\)
=\(2^6\)
 

c) \(2^5\times3^2\left(\frac{3}{2}\right)^{-2}\)
\(=2^5\times3^2\times\frac{2^2}{3^2}\)
\(=2^3\)

a/ ĐKXĐ: ...

\(\frac{5x-3}{2\left(5x-1\right)\left(5x+1\right)}-\frac{5x-9}{12x\left(5x-1\right)}+\frac{1}{12x}=\frac{8x-5}{16x\left(5x+1\right)}\)

\(\Leftrightarrow\frac{5x-3}{\left(5x-1\right)\left(5x+1\right)}+\frac{1}{6x}\left(1-\frac{5x-9}{5x-1}\right)=\frac{8x-5}{8x\left(5x+1\right)}\)

\(\Leftrightarrow\frac{5x-3}{\left(5x-1\right)\left(5x+1\right)}+\frac{4}{3x\left(5x-1\right)}-\frac{8x-5}{8x\left(5x+1\right)}=0\)

\(\Leftrightarrow24x\left(5x-3\right)+32\left(5x+1\right)-3\left(5x-1\right)\left(8x-5\right)=0\)

\(\Leftrightarrow-120x^2+379x-55=0\)

Bạn có nhầm đề chỗ nào ko nhỉ? Con số thật khủng khiếp (nghiệm ko hề đẹp)

b/ ĐKXĐ:...

\(\Leftrightarrow\frac{1}{3}\left(\frac{1}{x}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+9}\right)=\frac{1}{3}\left(27-\frac{1}{x+9}\right)\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+9}=27-\frac{1}{x+9}\)

\(\Leftrightarrow\frac{1}{x}=27\Rightarrow x=\frac{1}{27}\)

a, \(\frac{5x-3}{50x^2-2}+\frac{5x-9}{12x-60x^2}+\frac{1}{12x}=\frac{8x-5}{80x^2+16x}\) (ĐKXĐ: x \(\ne\) \(\pm\)\(\frac{1}{5}\); x \(\ne\) 0)

\(\Leftrightarrow\) \(\frac{5x-3}{2\left(5x-1\right)\left(5x+1\right)}+\frac{-5x+9}{12x\left(5x-1\right)}+\frac{1}{12x}=\frac{8x-5}{16x\left(5x+1\right)}\)

\(\Leftrightarrow\) \(\frac{24x\left(5x-3\right)\left(5x+1\right)}{48x\left(5x-1\right)\left(5x+1\right)}+\frac{-4\left(5x+1\right)\left(5x-9\right)}{48x\left(5-1x\right)\left(5x+1\right)}+\frac{4\left(5x-1\right)\left(5x+1\right)}{48x\left(5x-1\right)\left(5x+1\right)}=\frac{3\left(8x-5\right)\left(5x-1\right)}{48x\left(5x-1\right)\left(5x+1\right)}\)

\(\Leftrightarrow\) 24x(5x - 3) - 4(5x + 1)(5x - 9) + 4(5x - 1)(5x + 1) = 3(8x - 5)(5x - 1)

\(\Leftrightarrow\) 120x² - 72x - 100x² + 160x + 36 + 100x² - 4 = 120x² - 99x + 15

\(\Leftrightarrow\) 120x² - 120x² - 100x² + 100x² - 72x + 160x + 99x = 15 - 36 + 4

\(\Leftrightarrow\) 187x = -17

\(\Leftrightarrow\) x = \(\frac{-1}{11}\) (TM ĐKXĐ)

Vậy S = {\(\frac{-1}{11}\)}

Chúc bn học tốt!! (Đã được kiểm chứng không sai :)

b, \(\frac{1}{x\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+9\right)}=\frac{1}{3}\left(27-\frac{1}{x+9}\right)\) (ĐKXĐ: x \(\ne\) 0; x \(\ne\) -3; x \(\ne\) -6; x \(\ne\) -9)

\(\Leftrightarrow\) \(\frac{1}{3}\)(\(\frac{1}{x}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+9}\)) = \(\frac{1}{3}\)(27 - \(\frac{1}{x+9}\))

\(\Leftrightarrow\) \(\frac{1}{3}\)(\(\frac{1}{x}-\frac{1}{x+9}\)) = \(\frac{1}{3}\)(27 - \(\frac{1}{x+9}\))

\(\Leftrightarrow\) \(\frac{1}{3}\)(\(\frac{1}{x}-\frac{1}{x+9}\)) - \(\frac{1}{3}\)(27 - \(\frac{1}{x+9}\))

\(\Leftrightarrow\) \(\frac{1}{3}\)(\(\frac{1}{x}-\frac{1}{x+9}-27+\frac{1}{x+9}\)) = 0

\(\Leftrightarrow\) \(\frac{1}{3}\)(\(\frac{1}{x}-27\)) = 0

\(\Leftrightarrow\) \(\frac{1}{x}-27\) = 0

\(\Leftrightarrow\) x = \(\frac{1}{27}\) (TM ĐKXĐ)

Vậy S = {\(\frac{1}{27}\)}

Chúc bn học tốt!!

A=(9/1999+99/999+999/9999).(1/5-1/4+1/20)

A=(9/1999+99/999+999/9999).(-1/20+1/20)

A=(9/1999+99/999+999/9999).0

A=0

Vì mọi số nhân vs 0 thì đều = 0 kể cả phân số

mk nhanh nhất ủng hộ nha

\(A=\left(\frac{9}{1999}+\frac{99}{999}+\frac{999}{9999}\right)\cdot0\)

A=0

1/21 = 2/42 = 1/7-1/8

1/28 = 2/56 = 1/8-1/9

.....

1/(x*(x+1)) = 1/x -1/(x+1)

cộng lại với nhau ta sẽ được 1/7 - 1/(x+1) = 2/9

suy ra 1/(x+1) = -5/62 :D

Đặt \(A=\frac{1}{21}+\frac{1}{28}+\frac{1}{36}+...+\frac{2}{x\left(x+1\right)}\)

=> \(A=\frac{2}{6.7}+\frac{2}{7.8}+\frac{2}{8.9}+...+\frac{2}{x\left(x+1\right)}\)

\(\frac{A}{2}=\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+...+\frac{1}{x}-\frac{1}{x+1}\)

=> \(\frac{A}{2}=\frac{1}{6}-\frac{1}{x+1}=\frac{x+1-6}{6\left(x+1\right)}=\frac{x-5}{6\left(x+1\right)}\) => \(A=\frac{x-5}{3\left(x+1\right)}=\frac{2}{9}\)

<=> 3(x-5)=2(x+1)  <=> 3x-15=2x+2  <=> x=17

Đáp số: x=17