Toán tìm x thỏa mãn : a/x^2-x-6=0 b/x^2-4x-12=0 c/(x+2)(2x-1)+1=4x^2 02/08/2021 By Raelynn tìm x thỏa mãn : a/x^2-x-6=0 b/x^2-4x-12=0 c/(x+2)(2x-1)+1=4x^2
a) x² -x-6 = 0 <=> x = 3 và x= -2 b) x² -4x -12 = 0 <=> x= 6 và x=-2 c) (x+2)(2x-1)+1=4x² <=> 2x² – x + 4x -2 +1 -4x² = 0 <=> -2x² +3x -1 = o <=> x = 1 và x= 1/2 ————————————THE END————————- P/s : ko hiểu cứ hỏi nhé !! Trả lời
\(\begin{array}{l}a)\,\,{x^2} – x – 6 = 0\\ \Leftrightarrow {x^2} + 2x – 3x – 6 = 0\\ \Leftrightarrow x\left( {x + 2} \right) – 3\left( {x + 2} \right) = 0\\ \Leftrightarrow \left( {x + 2} \right)\left( {x – 3} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x + 2 = 0\\x – 3 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = – 2\\x = 3\end{array} \right.\\Vay\,\,S = \left\{ { – 2;3} \right\}\\b)\,\,{x^2} – 4x – 12 = 0\\ \Leftrightarrow {x^2} + 2x – 6x – 12 = 0\\ \Leftrightarrow x\left( {x + 2} \right) – 6\left( {x + 2} \right) = 0\\ \Leftrightarrow \left( {x + 2} \right)\left( {x – 6} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x + 2 = 0\\x – 6 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = – 2\\x = 6\end{array} \right.\\Vay\,\,S = \left\{ { – 2;6} \right\}\\c)\,\,\left( {x + 2} \right)\left( {2x – 1} \right) + 1 = 4{x^2}\\ \Leftrightarrow 2{x^2} – x + 4x – 2 + 1 – 4{x^2} = 0\\ \Leftrightarrow – 2{x^2} + 3x – 1 = 0\\ \Leftrightarrow 2{x^2} – 3x + 1 = 0\\ \Leftrightarrow 2{x^2} – 2x – x + 1 = 0\\ \Leftrightarrow 2x\left( {x – 1} \right) – \left( {x – 1} \right) = 0\\ \Leftrightarrow \left( {x – 1} \right)\left( {2x – 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x – 1 = 0\\2x – 1 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 1\\x = \dfrac{1}{2}\end{array} \right.\\Vay\,\,S = \left\{ {1;\dfrac{1}{2}} \right\}\end{array}\) Trả lời
a) x² -x-6 = 0
<=> x = 3 và x= -2
b) x² -4x -12 = 0
<=> x= 6 và x=-2
c) (x+2)(2x-1)+1=4x²
<=> 2x² – x + 4x -2 +1 -4x² = 0
<=> -2x² +3x -1 = o
<=> x = 1 và x= 1/2
————————————THE END————————-
P/s : ko hiểu cứ hỏi nhé !!
\(\begin{array}{l}
a)\,\,{x^2} – x – 6 = 0\\
\Leftrightarrow {x^2} + 2x – 3x – 6 = 0\\
\Leftrightarrow x\left( {x + 2} \right) – 3\left( {x + 2} \right) = 0\\
\Leftrightarrow \left( {x + 2} \right)\left( {x – 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 2 = 0\\
x – 3 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = – 2\\
x = 3
\end{array} \right.\\
Vay\,\,S = \left\{ { – 2;3} \right\}\\
b)\,\,{x^2} – 4x – 12 = 0\\
\Leftrightarrow {x^2} + 2x – 6x – 12 = 0\\
\Leftrightarrow x\left( {x + 2} \right) – 6\left( {x + 2} \right) = 0\\
\Leftrightarrow \left( {x + 2} \right)\left( {x – 6} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 2 = 0\\
x – 6 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = – 2\\
x = 6
\end{array} \right.\\
Vay\,\,S = \left\{ { – 2;6} \right\}\\
c)\,\,\left( {x + 2} \right)\left( {2x – 1} \right) + 1 = 4{x^2}\\
\Leftrightarrow 2{x^2} – x + 4x – 2 + 1 – 4{x^2} = 0\\
\Leftrightarrow – 2{x^2} + 3x – 1 = 0\\
\Leftrightarrow 2{x^2} – 3x + 1 = 0\\
\Leftrightarrow 2{x^2} – 2x – x + 1 = 0\\
\Leftrightarrow 2x\left( {x – 1} \right) – \left( {x – 1} \right) = 0\\
\Leftrightarrow \left( {x – 1} \right)\left( {2x – 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x – 1 = 0\\
2x – 1 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = \dfrac{1}{2}
\end{array} \right.\\
Vay\,\,S = \left\{ {1;\dfrac{1}{2}} \right\}
\end{array}\)