tìm x biết:a)3x^4-9x^3=-9x^2+27x.c)x^2(x+8)+x^2=-8x.d)(x+3)(x^2-3x+5)=x^2+3x.Cần gấp ạ

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
3{x^4} – 9{x^3} =  – 9{x^2} + 27x\\
 \Leftrightarrow 3{x^4} – 9{x^3} + 9{x^2} – 27x = 0\\
 \Leftrightarrow {x^4} – 3{x^3} + 3{x^2} – 9x = 0\\
 \Leftrightarrow \left( {{x^4} – 3{x^3}} \right) + \left( {3{x^2} – 9x} \right) = 0\\
 \Leftrightarrow {x^3}\left( {x – 3} \right) + 3x\left( {x – 3} \right) = 0\\
 \Leftrightarrow \left( {x – 3} \right)\left( {{x^3} + 3x} \right) = 0\\
 \Leftrightarrow \left( {x – 3} \right).x\left( {{x^2} + 3} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x – 3 = 0\\
x = 0\\
{x^2} + 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 3\\
x = 0
\end{array} \right.\\
c,\\
{x^2}\left( {x + 8} \right) + {x^2} =  – 8x\\
 \Leftrightarrow {x^2}\left( {x + 8} \right) + {x^2} + 8x = 0\\
 \Leftrightarrow {x^2}\left( {x + 8} \right) + x\left( {x + 8} \right) = 0\\
 \Leftrightarrow \left( {x + 8} \right)\left( {{x^2} + x} \right) = 0\\
 \Leftrightarrow \left( {x + 8} \right).x.\left( {x + 1} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 8 = 0\\
x = 0\\
x + 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x =  – 8\\
x = 0\\
x =  – 1
\end{array} \right.\\
d,\\
\left( {x + 3} \right)\left( {{x^2} – 3x + 5} \right) = {x^2} + 3x\\
 \Leftrightarrow \left( {x + 3} \right)\left( {{x^2} – 3x + 5} \right) = x.\left( {x + 3} \right)\\
 \Leftrightarrow \left( {x + 3} \right).\left[ {\left( {{x^2} – 3x + 5} \right) – x} \right] = 0\\
 \Leftrightarrow \left( {x + 3} \right).\left( {{x^2} – 4x + 5} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 3 = 0\\
{x^2} – 4x + 5 = 0
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x =  – 3\\
{\left( {x – 2} \right)^2} + 1 = 0\,\,\,\,\,\left( {vn} \right)
\end{array} \right. \Leftrightarrow x =  – 3
\end{array}\)