9. (X + 0,7)^3 = -27
10. ( 2/3x -1/3)^5 = 1/243
11. ( (2/5-3X)^2 = 9/25
12. (2x-1)^10 = 49^5
13. X : 5^2 = (3/5)^2 : 3^2
14. (x – 3/5)^2 = 4
15. (x – 1/4)^2 = 4/9
16. (2x – 5)^4 = -81
9. (X + 0,7)^3 = -27
10. ( 2/3x -1/3)^5 = 1/243
11. ( (2/5-3X)^2 = 9/25
12. (2x-1)^10 = 49^5
13. X : 5^2 = (3/5)^2 : 3^2
14. (x – 3/5)^2 = 4
15. (x – 1/4)^2 = 4/9
16. (2x – 5)^4 = -81
Đáp án:
\(\begin{array}{l}
9)x = – \dfrac{{37}}{{10}}\\
10)x = 1\\
11)\left[ \begin{array}{l}
x = – \dfrac{1}{{15}}\\
x = \dfrac{1}{3}
\end{array} \right.\\
12)x = 4\\
13)x = 1\\
14)\left[ \begin{array}{l}
x = \dfrac{{13}}{5}\\
x = – \dfrac{7}{5}
\end{array} \right.\\
15)\left[ \begin{array}{l}
x = \dfrac{{11}}{{12}}\\
x = – \dfrac{5}{{12}}
\end{array} \right.\\
16)x \in \emptyset
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
9){\left( {x + \dfrac{7}{{10}}} \right)^3} = – 27\\
\to x + \dfrac{7}{{10}} = – 3\\
\to x = – \dfrac{{37}}{{10}}\\
10){\left( {\dfrac{2}{3}x – \dfrac{1}{3}} \right)^5} = \dfrac{1}{{243}}\\
\to \dfrac{2}{3}x – \dfrac{1}{3} = \sqrt[5]{{\dfrac{1}{{243}}}}\\
\to \dfrac{2}{3}x – \dfrac{1}{3} = \dfrac{1}{3}\\
\to \dfrac{2}{3}x = \dfrac{2}{3}\\
\to x = 1\\
11){\left( {\dfrac{2}{5} – 3x} \right)^2} = \dfrac{9}{{25}}\\
\to \left| {\dfrac{2}{5} – 3x} \right| = \dfrac{3}{5}\\
\to \left[ \begin{array}{l}
\dfrac{2}{5} – 3x = \dfrac{3}{5}\\
\dfrac{2}{5} – 3x = – \dfrac{3}{5}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = – \dfrac{1}{{15}}\\
x = \dfrac{1}{3}
\end{array} \right.\\
12){\left( {2x – 1} \right)^{10}} = {7^{2.5}}\\
\to 2x – 1 = 7\\
\to x = 4\\
13)\dfrac{x}{{{5^2}}} = \dfrac{{{3^2}}}{{{5^2}}}:{3^2}\\
\to \dfrac{x}{{{5^2}}} = \dfrac{{{3^2}}}{{{5^2}}}.\dfrac{1}{{{3^2}}}\\
\to \dfrac{x}{{{5^2}}} = \dfrac{1}{{{5^2}}}\\
\to x = 1\\
14){\left( {x – \dfrac{3}{5}} \right)^2} = 4\\
\to \left| {x – \dfrac{3}{5}} \right| = 2\\
\to \left[ \begin{array}{l}
x – \dfrac{3}{5} = 2\\
x – \dfrac{3}{5} = – 2
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{13}}{5}\\
x = – \dfrac{7}{5}
\end{array} \right.\\
15){\left( {x – \dfrac{1}{4}} \right)^2} = \dfrac{4}{9}\\
\to \left| {x – \dfrac{1}{4}} \right| = \dfrac{2}{3}\\
\to \left[ \begin{array}{l}
x – \dfrac{1}{4} = \dfrac{2}{3}\\
x – \dfrac{1}{4} = – \dfrac{2}{3}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{11}}{{12}}\\
x = – \dfrac{5}{{12}}
\end{array} \right.\\
16)Do:{\left( {2x – 5} \right)^4} \ge 0\forall x\\
\to {\left( {2x – 5} \right)^4} = – 81\left( {KTM} \right)\\
\to x \in \emptyset
\end{array}\)