Bn tham khảo ở đây nhé, mk lm r`: Câu hỏi của Su su - Toán lớp 7 | Học trực tuyến

tớ vào xem ko dk

quy luật ở giữ chưa rõ rằng 

$\textbf{a)}$

$A=\dfrac{\left(\dfrac23\right)^3\cdot\left(-\dfrac34\right)^2\cdot(-1)^{2019}}{36\cdot\dfrac15\cdot\left(\dfrac25\right)^2\cdot\left(-\dfrac5{12}\right)^3}$

$=\dfrac{\dfrac8{27}\cdot\dfrac9{16}\cdot(-1)}{36\cdot\dfrac15\cdot\dfrac4{25}\cdot\left(-\dfrac{125}{1728}\right)}$

$=\dfrac{-\dfrac16}{-\dfrac5{12}}$

$=\dfrac16\cdot\dfrac{12}5$

$=\dfrac25.$

$\textbf{b)}$

$B=\dfrac1{19}+\dfrac9{19\cdot29}+\dfrac9{29\cdot39}+\cdots+\dfrac9{2009\cdot2019}$

$=\dfrac1{19}+\left(\dfrac1{19}-\dfrac1{29}\right)+\left(\dfrac1{29}-\dfrac1{39}\right)+\cdots+\left(\dfrac1{2009}-\dfrac1{2019}\right)$

$=\dfrac1{19}+\dfrac1{19}-\dfrac1{2019}$

$=\dfrac2{19}-\dfrac1{2019}$

$=\dfrac{2\cdot2019-19}{19\cdot2019}$

$=\dfrac{4019}{38361}.$

\(3.\)

\(\frac{x-1}{2011}+\frac{x-2}{2010}+\frac{x-3}{2009}=\frac{x-4}{2008}\)

\(\Rightarrow\)\(\frac{x-1}{2011}-1+\frac{x-2}{2010}-1+\frac{x-3}{2009}-1-\frac{x-4}{2008}+1+2=0\)

\(\Rightarrow\)\(\frac{x-1}{2011}-\frac{2011}{2011}+\frac{x-2}{2010}-\frac{2010}{2010}+\frac{x-3}{2009}-\frac{2009}{2009}-\frac{x-4}{2008}+\frac{2008}{2008}=0\)

\(\Rightarrow\)\(\frac{x-2012}{2011}+\frac{x-2012}{2010}+\frac{x-2012}{2009}-\frac{x-2012}{2008}=0\)

\(\Rightarrow\)\(x-2012\left(\frac{1}{2011}+\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}\right)=0\)

\(\Rightarrow\)\(x=2012\)

Vì \(\frac{9^{1006}-1}{4}\) là số chẵn nên x là số lẻ

\(\Rightarrow\left(-3\right)^x=-3^x\)

Đặt A=1-3+3²-3³+...-3^x

3A=3-3²+3³-3⁴+...+3^x+1

3A+A=[3-3²+3³-3⁴+...+3^x+1] -[1-3+3²-3³+...-3^x]

4A=3^x+1-1

\(A=\frac{3^{x+1}-1}{4}=\frac{9^{1006}-1}{4}=\frac{\left(3^2\right)^{1006}-1}{4}=\frac{3^{1012}-1}{4}\)

=>x+1=2012

=>x=2012-1=2011

vậy x=2011