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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,.........
#)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)\(\left(\frac{1}{125}-\frac{1}{1^3}\right)\left(\frac{1}{125}-\frac{1}{2^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...\left(\frac{1}{125}-\frac{1}{5^3}\right)...\left(\frac{1}{125}-\frac{1}{25^3}\right)=\left(\frac{1}{125}-\frac{1}{1^3}\right)...0...\left(\frac{1}{125}-\frac{1}{25^3}\right)=0\)
\(A=\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right)\cdot\left(\frac{1}{4^2}-1\right)\cdot...\cdot\left(\frac{1}{2014^2}-1\right)\)
\(A=\frac{-3}{2^2}\cdot\frac{-8}{3^2}\cdot\frac{-15}{4^2}\cdot...\cdot\frac{-2014^2+1}{2014^2}\)
\(A=\frac{1\cdot\left(-3\right)}{2^2}\cdot\frac{2\cdot\left(-4\right)}{3^2}\cdot\frac{3\cdot\left(-5\right)}{4^2}\cdot...\cdot\frac{2013\cdot\left(-2015\right)}{2014^2}\)
\(A=\frac{1\cdot2\cdot3\cdot...\cdot2013}{2\cdot3\cdot4\cdot...\cdot2014}\cdot\frac{\left(-3\right)\cdot\left(-4\right)\cdot\left(-5\right)\cdot...\cdot\left(-2015\right)}{2\cdot3\cdot4\cdot...\cdot2014}\)
\(A=\frac{1}{2014}\cdot\frac{-2015}{2}\)
\(A=\frac{-2015}{4028}\)
