(x-1) ×(x+2)-x-2=0 x(3x+2)+(x+1) ² – (2x+5)(2x-5)=0 (4x+1)(x-2) -(2x-3)(2x+1)=7 19/08/2021 Bởi Reagan (x-1) ×(x+2)-x-2=0 x(3x+2)+(x+1) ² – (2x+5)(2x-5)=0 (4x+1)(x-2) -(2x-3)(2x+1)=7
$\begin{array}{l} + )\,\,\left( {x – 1} \right).\left( {x + 2} \right) – x – 2 = 0\\ \Leftrightarrow \left( {x – 1} \right).\left( {x + 2} \right) – \left( {x + 2} \right) = 0\\ \Leftrightarrow \left( {x + 2} \right).\left( {x – 1 – 1} \right) = 0\\ \Leftrightarrow \left( {x + 2} \right)\left( {x – 2} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l}x + 2 = 0\\x – 2 = 0\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = – 2\\x = 2\end{array} \right.\\ + )\,\,x\left( {3x + 2} \right) + {\left( {x + 1} \right)^2} – \left( {2x + 5} \right)\left( {2x – 5} \right) = 0\\ \Leftrightarrow 3{x^2} + 2x + {x^2} + 2x + 1 – \left( {4{x^2} – 25} \right) = 0\\ \Leftrightarrow 4{x^2} + 4x + 1 – 4{x^2} + 25 = 0\\ \Leftrightarrow 4x + 26 = 0\\ \Leftrightarrow 4x = – 26\\ \Leftrightarrow x = \frac{{ – 13}}{2}\\ + )\,\left( {4x + 1} \right)\left( {x – 2} \right) – \left( {2x – 3} \right)\left( {2x + 1} \right) = 7\\ \Leftrightarrow \left( {4{x^2} – 8x + x – 2} \right) – \left( {4{x^2} + 2x – 6x – 3} \right) – 7 = 0\\ \Leftrightarrow \left( {4{x^2} – 7x – 2} \right) – \left( {4{x^2} – 4x – 3} \right) – 7 = 0\\ \Leftrightarrow 4{x^2} – 7x – 2 – 4{x^2} + 4x + 3 – 7 = 0\\ \Leftrightarrow – 3x – 6 = 0\\ \Leftrightarrow – 3x = 6\\ \Leftrightarrow x = – 2\end{array}$ Bình luận
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$\begin{array}{l}
+ )\,\,\left( {x – 1} \right).\left( {x + 2} \right) – x – 2 = 0\\
\Leftrightarrow \left( {x – 1} \right).\left( {x + 2} \right) – \left( {x + 2} \right) = 0\\
\Leftrightarrow \left( {x + 2} \right).\left( {x – 1 – 1} \right) = 0\\
\Leftrightarrow \left( {x + 2} \right)\left( {x – 2} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x + 2 = 0\\
x – 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = – 2\\
x = 2
\end{array} \right.\\
+ )\,\,x\left( {3x + 2} \right) + {\left( {x + 1} \right)^2} – \left( {2x + 5} \right)\left( {2x – 5} \right) = 0\\
\Leftrightarrow 3{x^2} + 2x + {x^2} + 2x + 1 – \left( {4{x^2} – 25} \right) = 0\\
\Leftrightarrow 4{x^2} + 4x + 1 – 4{x^2} + 25 = 0\\
\Leftrightarrow 4x + 26 = 0\\
\Leftrightarrow 4x = – 26\\
\Leftrightarrow x = \frac{{ – 13}}{2}\\
+ )\,\left( {4x + 1} \right)\left( {x – 2} \right) – \left( {2x – 3} \right)\left( {2x + 1} \right) = 7\\
\Leftrightarrow \left( {4{x^2} – 8x + x – 2} \right) – \left( {4{x^2} + 2x – 6x – 3} \right) – 7 = 0\\
\Leftrightarrow \left( {4{x^2} – 7x – 2} \right) – \left( {4{x^2} – 4x – 3} \right) – 7 = 0\\
\Leftrightarrow 4{x^2} – 7x – 2 – 4{x^2} + 4x + 3 – 7 = 0\\
\Leftrightarrow – 3x – 6 = 0\\
\Leftrightarrow – 3x = 6\\
\Leftrightarrow x = – 2
\end{array}$