a,
\(\dfrac{7x-1}{6}+2x=\dfrac{16-x}{5}\)
\(\Rightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
\(\Leftrightarrow35x-5+60x=96-6x\)
\(\Leftrightarrow101x=101\)
\(\Leftrightarrow x=1\)
\(Vậy:S=\left\{1\right\}\)
b,
\(2x^3-7x^2=3x^2-12x\)
\(\Leftrightarrow2x^3-10x^2-12x=0\)
\(\Leftrightarrow2x\left(x^2-5x-6\right)=0\)
\(\Leftrightarrow2x\left(x^2+6x-1x-6\right)=0\)
\(\Leftrightarrow2x\left[x\left(x+6\right)-1\left(x+6\right)\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\\x=1\end{matrix}\right.\)
\(Vậy:S=\left\{0;-6;1\right\}\)
câu còn lại tí nữa lm :")
d) \(4x^2-1=x^2\left(2x+1\right)\)
\(\Leftrightarrow\) \(\left(4x^2-1\right)-x^2\left(2x+1\right)=0\)
\(\Leftrightarrow\) \(\left(2x-1\right)\left(2x+1\right)-x^2\left(2x+1\right)=0\)
\(\Leftrightarrow\) \(\left(2x+1\right)\left(2x-1-x^2\right)=0\)
\(\Leftrightarrow\) \(-\left(2x+1\right)\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\end{matrix}\right.\)
Vậy S = {\(-\dfrac{1}{2};1\)}
c) \(\dfrac{x+1}{x+2}+\dfrac{4}{4-x^2}=1-\dfrac{5x^2+10x+20}{x^3-8}\)
\(\Leftrightarrow\) \(\dfrac{x+1}{x+2}+\dfrac{4}{x^2-4}=1-\dfrac{5\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\)
\(\Leftrightarrow\) \(\dfrac{x+1}{x+2}-\dfrac{4}{\left(x-2\right)\left(x+2\right)}=1-\dfrac{5}{x-2}\)
\(\Rightarrow\) \(\left(x+1\right)\left(x-2\right)-4=\left(x-2\right)\left(x+2\right)-5x\left(x+2\right)\)
\(\Leftrightarrow\) \(x^2-x-2-4=x^2-4-5x-10\)
\(\Leftrightarrow\) \(x^2-x-6=x^2-5x-14\)
\(\Leftrightarrow x^2-x^2-x+5x=\left(-14\right)+6\)
\(\Leftrightarrow4x=-8\)
\(\Rightarrow x=-2\)
b) \(2x^3-7x^2=3x^2-12x\)
\(\Leftrightarrow\) \(2x^3-7x^2-3x^2+12x=0\)
\(\Leftrightarrow\) \(2x^3-10x^2+12x=0\)
\(\Leftrightarrow2x\left(x^2-5x+6\right)=0\)
\(\Leftrightarrow2x\left(x^2-2x-3x+6\right)=0\)
\(\Leftrightarrow2x [ x\left(x-2\right)-3\left(x-2\right)]=0\)
\(\Leftrightarrow2x\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-2=0\\x-3=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=3\end{matrix}\right.\)
Vậy S = {0;2;3}
