by Tìm \(x\) biết:
\(a)\ {{\left( x-1 \right)}^{2}}=x-1\) \(b)\ 7{{x}^{2}}+2x=0\)
\(c)\ 7{{x}^{2}}\left( x-7 \right)+5x\left( 7-x \right)=0\) \(d)\ {{x}^{3}}-3{{x}^{2}}+3-x=0\)
A. \(\begin{array}{l}
a)\,\,\,x \in \left\{ {1;\,\,2} \right\}\\
b)\,\,x \in \left\{ { – \frac{2}{7};\,\,\,0} \right\}\\
c)\,\,x \in \left\{ {0;\,\,\frac{5}{7};\,\,7} \right\}\\
d)\,\,x \in \left\{ { – 1;\,\,1;\,\,3} \right\}
\end{array}\)
B. \(\begin{array}{l}
a)\,\,\,x \in \left\{ {1} \right\}\\
b)\,\,x \in \left\{ { – \frac{2}{7}} \right\}\\
c)\,\,x \in \left\{ {0;\,\,\frac{5}{7};\,\,7} \right\}\\
d)\,\,x \in \left\{ { – 1;\,\,1;\,\,3} \right\}
\end{array}\)
C. \(\begin{array}{l}
a)\,\,\,x \in \left\{ {2} \right\}\\
b)\,\,x \in \left\{ { – \frac{2}{7}} \right\}\\
c)\,\,x \in \left\{ {0;\,\,\frac{5}{7};\,\,7} \right\}\\
d)\,\,x \in \left\{ { – 1;\,\,1} \right\}
\end{array}\)
D. \(\begin{array}{l}
a)\,\,\,x \in \left\{ {1} \right\}\\
b)\,\,x \in \left\{ { – \frac{2}{7}} \right\}\\
c)\,\,x \in \left\{ {0;\,\,\frac{5}{7};\,\,7} \right\}\\
d)\,\,x \in \left\{ { – 1;\,\,1;\,\,3} \right\}
\end{array}\)
Hướng dẫn Chọn đáp án là: A
Lời giải chi tiết:
Hướng dẫn chi tiết:
\(\begin{array}{l}a){\left( {x – 1} \right)^2} = x – 1\\ \Leftrightarrow \left( {x – 1} \right)\left( {x – 1} \right) – \left( {x – 1} \right) = 0\\ \Leftrightarrow \left( {x – 1} \right)\left( {x – 1 – 1} \right) = 0\\ \Leftrightarrow \left( {x – 1} \right)\left( {x – 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x – 1 = 0\\x – 2 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 1\\x = 2\end{array} \right.\end{array}\)
Vậy \(x=1\) hoặc \(x=2\)
\(\begin{array}{l}b)\;7{x^2} + 2x = 0\\ \Leftrightarrow 7x.x + 2.x = 0\\ \Leftrightarrow x\left( {7x + 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = 0\\x = – \frac{2}{7}\end{array} \right.\end{array}\)
Vậy \(x=0\) hoặc \(x = \frac{{ – 2}}{7}\)
\(\begin{array}{l}c)7{x^2}\left( {x – 7} \right) + 5x\left( {7 – x} \right) = 0\\ \Leftrightarrow 7x.x\left( {x – 7} \right) – 5.x\left( {x – 7} \right) = 0\\ \Leftrightarrow \left( {7x.x – 5.x} \right)\left( {x – 7} \right) = 0\\ \Leftrightarrow x\left( {7x – 5} \right)\left( {x – 7} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = 0\\7x – 5 = 0\\x – 7 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 0\\x = \frac{5}{7}\\x = 7\end{array} \right.\end{array}\)
Vậy \(x=0\) hoặc \(x=7\) hoặc \(x=\frac{5}{7}\)
\(\begin{array}{l}d)\;{x^3} – 3{x^2} + 3 – x = 0\\ \Leftrightarrow {x^2}.x – 3.{x^2} + \left( {3 – x} \right) = 0\\ \Leftrightarrow {x^2}\left( {x – 3} \right) – \left( {x – 3} \right) = 0\\ \Leftrightarrow \left( {{x^2} – 1} \right)\left( {x – 3} \right) = 0\\ \Leftrightarrow \left( {x – 1} \right)\left( {x + 1} \right)\left( {x – 3} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x – 1 = 0\\x + 1 = 0\\x – 3 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 1\\x = – 1\\x = 3\end{array} \right.\end{array}\)
Vậy \(x=1\) hoặc \(x=-1\) hoặc \(x=3\).
