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a) \(\dfrac{15x}{7y^3}.\dfrac{2y^2}{x^2}=\dfrac{15x.2y^2}{7y^3.x^2}=\dfrac{30}{7xy}\)
b) \(\dfrac{4y^2}{11x^4}.\left(-\dfrac{3x^2}{8y}\right)=\dfrac{-4y^2.3x^2}{11x^4.8y}=\dfrac{-3y}{22x^2}\)
c) \(\dfrac{x^3-8}{5x+20}.\dfrac{x^2+4x}{x^2+2x+4}\\ =\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{5\left(x+4\right)}.\dfrac{x\left(x+4\right)}{x^2+2x+4}\\ =\dfrac{x^2-2x}{5}\)
1.
a) \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
b) \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
Bài 1:
a, \(x\left(x+4\right)+x+4=0\)
\(\Leftrightarrow x\left(x+4\right)+\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-1\end{matrix}\right.\)
Vậy \(x=-4\) hoặc \(x=-1\)
b, \(x\left(x-3\right)+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-2\)
a) 1/x(x + 1) + 1/(x + 1)(x + 2) + 1/(x + 2)(x + 3) + 1/(x + 3)(x + 4)
( 1/x - 1/x+1) + (1/x+1 - 1/x+2) + (1/x+2 - 1/ x+3) + 1/(x+3 - 1/x+4)
(1/x +1/x+4) - ( 1/x+2 - 1/x+2) - ( 1/x+3 - 1/x+3)
1/x +1/x+4
2x+4/x(x+4)
a) \(\dfrac{4y^2}{11x^4}:\left(-\dfrac{8y}{33x^2}\right)\)
\(=\dfrac{4y^2}{11x^4}.\left(-\dfrac{33x^2}{8y}\right)\)
\(=-\dfrac{4y^2.33x^2}{11x^4.8y}\)
\(=-\dfrac{3y}{2x^2}\)
b) \(\dfrac{x^2-4}{3x+12}.\dfrac{x+4}{2x-4}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}.\dfrac{x+4}{2\left(x-2\right)}\)
\(=\dfrac{\left(x-2\right)\left(x+2\right)\left(x+4\right)}{3\left(x+4\right).2\left(x-2\right)}\)
\(=\dfrac{x+2}{6}\).
\(\dfrac{\left(x-2\right)\left(x+2\right)\left(x+4\right)}{3\left(x+4\right).2\left(x-2\right)}=\dfrac{x+2}{6}\)