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) \(\dfrac{x^3+8y^3}{2y+x}\)
\(=\dfrac{x^3+\left(2y\right)^3}{x+2y}\)
\(=\dfrac{\left(x+2y\right)\left[x^2+x.2y+\left(2y\right)^2\right]}{x+2y}\)
\(=x^2+2xy+4y^2\)
b) \(\dfrac{a-1}{2\left(a-4\right)}+\dfrac{a}{a-4}\) MTC: \(2\left(a-4\right)\)
\(=\dfrac{a-1}{2\left(a-4\right)}+\dfrac{2a}{2\left(a-4\right)}\)
\(=\dfrac{a-1+2a}{2\left(a-4\right)}\)
\(=\dfrac{3a-1}{2\left(a-4\right)}\)
c) \(\dfrac{x^3+3x^2y+3xy^2+y^3}{2x+2y}\)
\(=\dfrac{\left(x+y\right)^3}{2\left(x+y\right)}\)
\(=\dfrac{\left(x+y\right)^2}{2}\)
d) \(\left(x-5\right)^2+\left(7-x\right)\left(x+2\right)\)
\(=\left(x^2-2.x.5+5^2\right)+\left(7x+14-x^2-2x\right)\)
\(=x^2-10x+25+7x+14-x^2-2x\)
\(=39-5x\)
e) \(\dfrac{3x}{x-2}-\dfrac{2x+1}{2-x}\)
\(=\dfrac{3x}{x-2}+\dfrac{2x+1}{x-2}\)
\(=\dfrac{3x+2x+1}{x-2}\)
\(=\dfrac{5x+1}{x-2}\)
h) \(\dfrac{1}{3x-2}-\dfrac{1}{3x+2}-\dfrac{3x+6}{4-9x^2}\)
\(=\dfrac{1}{3x-2}-\dfrac{1}{3x+2}+\dfrac{3x+6}{9x^2-4}\)
\(=\dfrac{1}{3x-2}-\dfrac{1}{3x+2}+\dfrac{3x+6}{\left(3x-2\right)\left(3x+2\right)}\) MTC: \(\left(3x-2\right)\left(3x+2\right)\)
\(=\dfrac{3x+2}{\left(3x-2\right)\left(3x+2\right)}-\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}+\dfrac{3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{\left(3x+2\right)-\left(3x-2\right)+\left(3x+6\right)}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+2-3x+2+3x+6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+10}{\left(3x-2\right)\left(3x+2\right)}\)
\(\frac{x^2}{9}+\frac{1}{x^2}=\frac{5}{3}\left(\frac{x}{3}-\frac{1}{x}\right)\)
đặt \(\left(\frac{x}{3}-\frac{1}{x}\right)=t\Rightarrow t^2=\left(\frac{x^2}{9}-\frac{2}{3}+\frac{1}{x^2}\right)\Rightarrow\frac{x^2}{9}+\frac{1}{x^2}=t^2+\frac{2}{3}\)
\(\Leftrightarrow t^2-\frac{5}{3}t+\frac{2}{3}=0\Leftrightarrow t^2-2.\frac{5}{6}t+\left(\frac{5}{6}\right)^2=\frac{25}{36}-\frac{24}{36}=\frac{1}{36}=\left(\frac{1}{6}\right)^2\)
\(\Rightarrow\left[\begin{matrix}t-\frac{5}{6}=\frac{1}{6}\\t-\frac{5}{6}=-\frac{1}{6}\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}t=1\\t=\frac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}\frac{x}{3}-\frac{1}{x}=1\\\frac{x}{3}-\frac{1}{x}=\frac{2}{3}\end{matrix}\right.\)\(\Leftrightarrow\left\{\begin{matrix}x\ne0\\x^2-x-3=0\\x^2-2x-3=0\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x\ne0\\\left[\begin{matrix}x=\frac{1-\sqrt{13}}{2}\\x=\frac{1+\sqrt{13}}{2}\end{matrix}\right.\\\left[\begin{matrix}x=-1\\x=3\end{matrix}\right.\end{matrix}\right.\)
Kết luận: \(\left[\begin{matrix}x=\frac{1-\sqrt{13}}{2}\\x=\frac{1+\sqrt{13}}{2}\\x=-1\\x=3\end{matrix}\right.\)