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a) \(=x\left(3x^3-x^2+5\right)\)
b) \(=\left(2x+3y\right)\left(x-y\right)\)
c) \(=\left(x^2-3x\right)-\left(4x-12\right)=x\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x-4\right)\)
a, = x.(3x3 - x2 + 5)
b, = 2x.(x - y) + 3y.(x - y) = (x - y).(2x + 3y)
c, = x2 - 3x - 4x + 12 = (x2 - 3x) - (4x - 12) = x.(x - 3) - 4.(x - 3) = (x - 3).(x - 4)
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\(1,\Leftrightarrow x^2-8x+16-x^2+x+12=7\\ \Leftrightarrow-7x=-21\\ \Leftrightarrow x=3\\ 2,\Leftrightarrow\left(x-4\right)^2-\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
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\(a,\Leftrightarrow2x\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
\(b,\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3+x>0\\2x-5>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+3< 0\\2x-5< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>\dfrac{5}{2}\\x< -3\end{matrix}\right.\)
\(c,\Leftrightarrow x\left(x+3\right)< 0\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 0\\x+3>0\end{matrix}\right.\\\left\{{}\begin{matrix}x>0\\x+3< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow-3< x< 0\)
\(d,\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+3>0\\x+5>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+3< 0\\x+5< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>-3\\x< -5\end{matrix}\right.\)
\(e,\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3-2x\ge0\\x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}3-2x\le0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x\le\dfrac{3}{2}\\x>1\end{matrix}\right.\\\left\{{}\begin{matrix}x\ge\dfrac{3}{2}\\x< 1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow1< x\le\dfrac{3}{2}\)
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\(n_Fe=\dfrac{11,2}{56}=0,2(mol)\\ PTHH:Fe+2HCl\to FeCl_2+H_2\\ a,n_{H_2}=n_{Fe}=0,2(mol)\\ \Rightarrow V_{H_2(đktc)}=0,2.22,4=4,48(l)\\ b,n_{HCl}=2n_{Fe}=0,4(mol)\\ \Rightarrow m_{HCl}=0,4.36,5=14,6(g)\\ c,n_{FeCl_2}=n_{Fe}=0,2(mol)\\ \Rightarrow m_{FeCl_2}=0,2.127=25,4(g)\)