Đáp án:
2g) \(\left[ \begin{array}{l}
x = \dfrac{1}{9}\\
x = - \dfrac{7}{9}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
1h) - {x^3} - 8{x^2} - 10x + 20 = 0\\
\to \left[ \begin{array}{l}
x = 1,034041836\\
x = - 3,486535223\\
x = - 5,547506613
\end{array} \right.\\
B2:\\
g)9{\left( {3x + 1} \right)^2} - 16 = 0\\
\to {\left( {3x + 1} \right)^2} = \dfrac{{16}}{9}\\
\to \left| {3x + 1} \right| = \dfrac{4}{3}\\
\to \left[ \begin{array}{l}
3x + 1 = \dfrac{4}{3}\\
3x + 1 = - \dfrac{4}{3}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{9}\\
x = - \dfrac{7}{9}
\end{array} \right.\\
h)9{\left( {3x + 1} \right)^2} - 16{\left( {2x - 1} \right)^2} = 0\\
\to 9{\left( {3x + 1} \right)^2} = 16{\left( {2x - 1} \right)^2}\\
\to 3\left| {3x + 1} \right| = 4\left| {2x - 1} \right|\\
\to \left[ \begin{array}{l}
3\left( {3x + 1} \right) = 4\left( {2x - 1} \right)\left( {DK:x \ge - \dfrac{1}{3}} \right)\\
3\left( {3x + 1} \right) = - 4\left( {2x - 1} \right)\left( {DK:x < - \dfrac{1}{3}} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
9x + 3 = 8x - 4\\
9x + 3 = - 8x + 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 7\\
17x = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = - 7\left( l \right)\\
x = \dfrac{1}{{17}}\left( l \right)
\end{array} \right.\\
\to x \in \emptyset
\end{array}\)
( 1h nghiệm xấu bạn nha bạn xem lại đề nhé )
