giải phương trình a) (2x-1)(x^2-x+1) b)3x-15=2x(x-5) c)x^-3x=0 d)(x+1)(x+2)=(2-x)(x+2) 24/08/2021 Bởi Ariana giải phương trình a) (2x-1)(x^2-x+1) b)3x-15=2x(x-5) c)x^-3x=0 d)(x+1)(x+2)=(2-x)(x+2)
Đáp án: $\begin{array}{l}a)\left( {2x – 1} \right)\left( {{x^2} – x + 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}2x – 1 = 0\\{x^2} – x + 1 = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{1}{2}\\{x^2} – 2.x.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4} = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = \dfrac{1}{2}\\{\left( {x – \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} = 0\left( {vn} \right)\end{array} \right.\\Vậy\,x = \dfrac{1}{2}\\b)3x – 15 = 2x\left( {x – 5} \right)\\ \Leftrightarrow 3\left( {x – 5} \right) – 2x\left( {x – 5} \right) = 0\\ \Leftrightarrow \left( {x – 5} \right)\left( {3 – 2x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x – 5 = 0\\3 – 2x = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = 5\\x = \dfrac{3}{2}\end{array} \right.\\Vậy\,x = 5;x = \dfrac{3}{2}\\c){x^2} – 3x = 0\\ \Leftrightarrow x\left( {x – 3} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = 0\\x – 3 = 0\end{array} \right.\\ \Leftrightarrow \left[ \begin{array}{l}x = 0\\x = 3\end{array} \right.\\Vậy\,x = 0;x = 3\\d)\left( {x + 1} \right)\left( {x + 2} \right) = \left( {2 – x} \right)\left( {x + 2} \right)\\ \Leftrightarrow \left( {x + 1} \right)\left( {x + 2} \right) + \left( {x + 2} \right)\left( {x – 2} \right) = 0\\ \Leftrightarrow \left( {x + 2} \right)\left( {x + 1 + x – 2} \right) = 0\\ \Leftrightarrow \left( {x + 2} \right).\left( {2x – 1} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = – 2\\x = \dfrac{1}{2}\end{array} \right.\\Vậy\,x = – 2;x = \dfrac{1}{2}\end{array}$ Bình luận
Đáp án:
$\begin{array}{l}
a)\left( {2x – 1} \right)\left( {{x^2} – x + 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
2x – 1 = 0\\
{x^2} – x + 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
{x^2} – 2.x.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4} = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
{\left( {x – \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} = 0\left( {vn} \right)
\end{array} \right.\\
Vậy\,x = \dfrac{1}{2}\\
b)3x – 15 = 2x\left( {x – 5} \right)\\
\Leftrightarrow 3\left( {x – 5} \right) – 2x\left( {x – 5} \right) = 0\\
\Leftrightarrow \left( {x – 5} \right)\left( {3 – 2x} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x – 5 = 0\\
3 – 2x = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 5\\
x = \dfrac{3}{2}
\end{array} \right.\\
Vậy\,x = 5;x = \dfrac{3}{2}\\
c){x^2} – 3x = 0\\
\Leftrightarrow x\left( {x – 3} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x – 3 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 3
\end{array} \right.\\
Vậy\,x = 0;x = 3\\
d)\left( {x + 1} \right)\left( {x + 2} \right) = \left( {2 – x} \right)\left( {x + 2} \right)\\
\Leftrightarrow \left( {x + 1} \right)\left( {x + 2} \right) + \left( {x + 2} \right)\left( {x – 2} \right) = 0\\
\Leftrightarrow \left( {x + 2} \right)\left( {x + 1 + x – 2} \right) = 0\\
\Leftrightarrow \left( {x + 2} \right).\left( {2x – 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = – 2\\
x = \dfrac{1}{2}
\end{array} \right.\\
Vậy\,x = – 2;x = \dfrac{1}{2}
\end{array}$
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