Toán Lớp 8: (2x-3)(3x+2)-x(x-1)=3(x-1)^2
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Toán Lớp 8: (2x-3)(3x+2)-x(x-1)=3(x-1)^2
Mn giúp mình vs
Comments ( 1 )
x = \dfrac{{ – 1 + \sqrt {19} }}{2}\\
x = \dfrac{{ – 1 – \sqrt {19} }}{2}
\end{array} \right.\)
(2x – 3)(3x + 2) – x(x – 1) = 3{(x – 1)^2}\\
\to 6{x^2} – 5x – 6 – {x^2} + x = 3\left( {{x^2} – 2x + 1} \right)\\
\to 5{x^2} – 4x – 6 = 3{x^2} – 6x + 3\\
\to 2{x^2} + 2x – 9 = 0\\
\to 2{x^2} + 2.x\sqrt 2 .\dfrac{1}{{\sqrt 2 }} + \dfrac{1}{2} – \dfrac{{19}}{2} = 0\\
\to {\left( {x\sqrt 2 + \dfrac{1}{{\sqrt 2 }}} \right)^2} = \dfrac{{19}}{2}\\
\to \left| {x\sqrt 2 + \dfrac{1}{{\sqrt 2 }}} \right| = \sqrt {\dfrac{{19}}{2}} \\
\to \left[ \begin{array}{l}
x\sqrt 2 + \dfrac{1}{{\sqrt 2 }} = \sqrt {\dfrac{{19}}{2}} \\
x\sqrt 2 + \dfrac{1}{{\sqrt 2 }} = – \sqrt {\dfrac{{19}}{2}}
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{{ – 1 + \sqrt {19} }}{2}\\
x = \dfrac{{ – 1 – \sqrt {19} }}{2}
\end{array} \right.
\end{array}\)