Bài 8. Tìm x € Z biết:
a) -12+5(2x-1)2 = 33
b) -12+5(2x-1)2 = 63
c) -12+4(2x-1)2 = 52
d) (3x-1)2 = 25 e) (2x-3)3= 125
f) (2x-3)3= -125
g) -11+3.(2x-1)2 =64
h) -12+5(2x-1)2 = 113
i) -12+4(2x-1)2 = 52
Bài 8. Tìm x € Z biết:
a) -12+5(2x-1)2 = 33
b) -12+5(2x-1)2 = 63
c) -12+4(2x-1)2 = 52
d) (3x-1)2 = 25 e) (2x-3)3= 125
f) (2x-3)3= -125
g) -11+3.(2x-1)2 =64
h) -12+5(2x-1)2 = 113
i) -12+4(2x-1)2 = 52
Đáp án:
i) \(\left[ \begin{array}{l}
x = \dfrac{5}{2}\\
x = – \dfrac{3}{2}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
a) – 12 + 5{\left( {2x – 1} \right)^2} = 33\\
\to 5{\left( {2x – 1} \right)^2} = 45\\
\to {\left( {2x – 1} \right)^2} = 9\\
\to \left| {2x – 1} \right| = 3\\
\to \left[ \begin{array}{l}
2x – 1 = 3\\
2x – 1 = – 3
\end{array} \right. \to \left[ \begin{array}{l}
x = 2\\
x = – 1
\end{array} \right.\\
b) – 12 + 5{\left( {2x – 1} \right)^2} = 63\\
\to 5{\left( {2x – 1} \right)^2} = 75\\
\to {\left( {2x – 1} \right)^2} = 15\\
\to \left| {2x – 1} \right| = \sqrt {15} \\
\to \left[ \begin{array}{l}
2x – 1 = \sqrt {15} \\
2x – 1 = – \sqrt {15}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{\sqrt {15} + 1}}{2}\\
x = \dfrac{{\sqrt {15} – 1}}{2}
\end{array} \right.\\
c) – 12 + 4{\left( {2x – 1} \right)^2}{\rm{ = }}52\\
\to 4{\left( {2x – 1} \right)^2}{\rm{ = 64}}\\
\to {\left( {2x – 1} \right)^2}{\rm{ = }}16\\
\to \left| {2x – 1} \right| = 4\\
\to \left[ \begin{array}{l}
2x – 1 = 4\\
2x – 1 = – 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{5}{2}\\
x = – \dfrac{3}{2}
\end{array} \right.\\
d){\left( {3x – 1} \right)^2} = 25\\
\to \left| {3x – 1} \right| = 5\\
\to \left[ \begin{array}{l}
3x – 1 = 5\\
3x – 1 = – 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 2\\
x = – \dfrac{4}{3}
\end{array} \right.\\
e){\left( {2x – 3} \right)^3} = 125\\
\to 2x – 3 = 5\\
\to x = 4\\
f){\left( {2x – 3} \right)^3} = – 125\\
\to 2x – 3 = – 5\\
\to x = – 1\\
g) – 11 + 3{\left( {2x – 1} \right)^2} = 64\\
\to 3{\left( {2x – 1} \right)^2} = 75\\
\to {\left( {2x – 1} \right)^2} = 25\\
\to \left| {2x – 1} \right| = 5\\
\to \left[ \begin{array}{l}
2x – 1 = 5\\
2x – 1 = – 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = – 2
\end{array} \right.\\
h) – 12 + 5{\left( {2x – 1} \right)^2} = 113\\
\to 5{\left( {2x – 1} \right)^2} = 125\\
\to {\left( {2x – 1} \right)^2} = 25\\
\to \left| {2x – 1} \right| = 5\\
\to \left[ \begin{array}{l}
2x – 1 = 5\\
2x – 1 = – 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 3\\
x = – 2
\end{array} \right.\\
i) – 12 + 4{\left( {2x – 1} \right)^2} = 52\\
\to 4{\left( {2x – 1} \right)^2} = 64\\
\to {\left( {2x – 1} \right)^2}{\rm{ = }}16\\
\to \left| {2x – 1} \right| = 4\\
\to \left[ \begin{array}{l}
2x – 1 = 4\\
2x – 1 = – 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{5}{2}\\
x = – \dfrac{3}{2}
\end{array} \right.
\end{array}\)