a) 

( 4x - 9 ) ( 2,5 + (-7/3) . x ) = 0

\(\Rightarrow\orbr{\begin{cases}4x-9=0\\2,5+\frac{-7}{3}x=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{9}{4}\\x=\frac{15}{14}\end{cases}}\)

P/s: đợi xíu làm câu b

b) \(\frac{1}{x\left(x+1\right)}\cdot\frac{1}{\left(x+1\right)\left(x+2\right)}\cdot\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2015}\)

\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2015}\)

\(\frac{-1}{x+3}=\frac{1}{2015}\)

\(\Leftrightarrow x+3=-2015\)

\(\Leftrightarrow x=-2018\)

Vậy,.........

\(\frac{\left(\frac{2}{3}\right)^3\cdot\left(-\frac{3}{4}^2\right)\cdot\left(-1\right)^{2003}}{\left(\frac{2}{5}\right)^2\cdot\left(-\frac{5}{12}\right)^3}\)

\(=\frac{\frac{8}{27}\cdot\frac{9}{16}\cdot\left(-1\right)}{\frac{4}{25}\cdot\left(-\frac{125}{1728}\right)}\)

\(=\frac{-\frac{1}{6}}{-\frac{5}{432}}=-\frac{1}{6}:\left(-\frac{5}{432}\right)=\frac{72}{5}\)

\(\left[6.\left(\frac{-1}{3}\right)^2-3.\left(\frac{-1}{3}\right)+1\right]:\left(\frac{-1}{3}-1\right)\)

\(=\left[6.\frac{1}{9}-\left(-1\right)+1\right]:\frac{-4}{3}\)

\(=\left[\frac{2}{3}-\left(-1\right)+1\right]:\frac{-4}{3}\)

\(=\frac{8}{3}:\frac{-4}{3}=\frac{-24}{12}=-2\)

~ Hok tốt ~

#)Giải :

a)\(2009^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-15^3\right)}=2009^{\left(1000-1^3\right)...\left(1000-10^3\right)...\left(1000-15^3\right)}=2009^0=1\)

b)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)

$\textbf{A)}$

$A=2009^{(1000-1^3)}\cdot(1000-2^3)\cdots(1000-15^3).$

$\text{Vì }1000-10^3=1000-1000=0.$

$\Rightarrow A=0.$

a) \(\frac{\left(-1\right)}{4}^2+\frac{3}{8}.\left(\frac{-1}{6}\right)-\frac{3}{16}:\left(\frac{-1}{2}\right)=\left(\frac{-1}{4}\right)^2+\left(\frac{-3}{68}\right)-\left(\frac{-3}{8}\right)=\left(\frac{1}{16}\right)+\left(\frac{-3}{68}\right)-\left(\frac{-3}{8}\right)=\frac{5}{272}-\left(\frac{-3}{8}\right)=\frac{107}{272}\)

$\textbf{a)}$

$A=\left(-\dfrac14\right)^2+\dfrac38\cdot\left(-\dfrac16\right)-\dfrac3{16}:\left(-\dfrac12\right)$

$=\dfrac1{16}-\dfrac1{16}+\dfrac38$

$=\dfrac38.$

$\textbf{a)}$

$A=\left(1-\dfrac12\right)\left(1-\dfrac13\right)\left(1-\dfrac14\right)\cdots\left(1-\dfrac1n\right)$

$=\dfrac12\cdot\dfrac23\cdot\dfrac34\cdots\dfrac{n-1}{n}$

$=\dfrac12\cdot\dfrac23\cdot\dfrac34\cdots\dfrac{n-1}{n}$

$=\dfrac1n.$

$\textbf{b)}$

$B=\left(1-\dfrac1{2^2}\right)\left(1-\dfrac1{3^2}\right)\cdots\left(1-\dfrac1{n^2}\right)$

$=\dfrac{(2-1)(2+1)}{2^2}\cdot\dfrac{(3-1)(3+1)}{3^2}\cdots\dfrac{(n-1)(n+1)}{n^2}$

$=\left(\dfrac12\cdot\dfrac23\cdot\dfrac34\cdots\dfrac{n-1}{n}\right)\left(\dfrac32\cdot\dfrac43\cdot\dfrac54\cdots\dfrac{n+1}{n}\right)$

$=\dfrac1n\cdot\dfrac{n+1}{2}$

$=\dfrac{n+1}{2n}.$

$B=\left(\dfrac1{2^2}-1\right)\left(\dfrac1{3^2}-1\right)\cdots\left(\dfrac1{99^2}-1\right)$

$=\left(-\dfrac{2^2-1}{2^2}\right)\left(-\dfrac{3^2-1}{3^2}\right)\cdots\left(-\dfrac{99^2-1}{99^2}\right)$

$=(-1)^{98}\prod_{k=2}^{99}\dfrac{(k-1)(k+1)}{k^2}$

$=\left(\prod_{k=2}^{99}\dfrac{k-1}{k}\right)\left(\prod_{k=2}^{99}\dfrac{k+1}{k}\right)$

$=\left(\dfrac12\cdot\dfrac23\cdot\dfrac34\cdots\dfrac{98}{99}\right)\left(\dfrac32\cdot\dfrac43\cdot\dfrac54\cdots\dfrac{100}{99}\right)$

$=\dfrac1{99}\cdot\dfrac{100}{2}$

$=\dfrac{50}{99}.$

$\textbf{a)}$

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

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

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

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

$=\dfrac25.$

$\textbf{b)}$

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

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

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

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

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

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