Tìm x biết : (3x+2)(2x+9)-(x+2)(x+1)=(x+1)-(x-6) 25/09/2021 Bởi Hadley Tìm x biết : (3x+2)(2x+9)-(x+2)(x+1)=(x+1)-(x-6)
Đáp án: \(\left[ \begin{array}{l}x = \dfrac{{ – 14 + \sqrt {151} }}{5}\\x = \dfrac{{ – 14 – \sqrt {151} }}{5}\end{array} \right.\) Giải thích các bước giải: \(\begin{array}{l}\left( {3x + 2} \right)\left( {2x + 9} \right) – \left( {x + 2} \right)\left( {x + 1} \right) = \left( {x + 1} \right) – \left( {x – 6} \right)\\ \to 6{x^2} + 31x + 18 – {x^2} – 3x – 2 = x + 1 – x + 6\\ \to 5{x^2} + 28x + 9 = 0\\ \to {\left( {x\sqrt 5 } \right)^2} + 2.x\sqrt 5 .\dfrac{{14}}{{\sqrt 5 }} + {\left( {\dfrac{{14}}{{\sqrt 5 }}} \right)^2} – \dfrac{{151}}{5} = 0\\ \to {\left( {x\sqrt 5 + \dfrac{{14}}{{\sqrt 5 }}} \right)^2} = \dfrac{{151}}{5}\\ \to \left[ \begin{array}{l}x\sqrt 5 + \dfrac{{14}}{{\sqrt 5 }} = \sqrt {\dfrac{{151}}{5}} \\x\sqrt 5 + \dfrac{{14}}{{\sqrt 5 }} = – \sqrt {\dfrac{{151}}{5}} \end{array} \right.\\ \to \left[ \begin{array}{l}x = \dfrac{{ – 14 + \sqrt {151} }}{5}\\x = \dfrac{{ – 14 – \sqrt {151} }}{5}\end{array} \right.\end{array}\) Bình luận
Đáp án:
\(\left[ \begin{array}{l}
x = \dfrac{{ – 14 + \sqrt {151} }}{5}\\
x = \dfrac{{ – 14 – \sqrt {151} }}{5}
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
\left( {3x + 2} \right)\left( {2x + 9} \right) – \left( {x + 2} \right)\left( {x + 1} \right) = \left( {x + 1} \right) – \left( {x – 6} \right)\\
\to 6{x^2} + 31x + 18 – {x^2} – 3x – 2 = x + 1 – x + 6\\
\to 5{x^2} + 28x + 9 = 0\\
\to {\left( {x\sqrt 5 } \right)^2} + 2.x\sqrt 5 .\dfrac{{14}}{{\sqrt 5 }} + {\left( {\dfrac{{14}}{{\sqrt 5 }}} \right)^2} – \dfrac{{151}}{5} = 0\\
\to {\left( {x\sqrt 5 + \dfrac{{14}}{{\sqrt 5 }}} \right)^2} = \dfrac{{151}}{5}\\
\to \left[ \begin{array}{l}
x\sqrt 5 + \dfrac{{14}}{{\sqrt 5 }} = \sqrt {\dfrac{{151}}{5}} \\
x\sqrt 5 + \dfrac{{14}}{{\sqrt 5 }} = – \sqrt {\dfrac{{151}}{5}}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{ – 14 + \sqrt {151} }}{5}\\
x = \dfrac{{ – 14 – \sqrt {151} }}{5}
\end{array} \right.
\end{array}\)