Toán Lớp 9: Giai PT dùng công thức nghiệm
-3$x^{2}$ + 5x = 0
3$x^{2}$ -8 = 0
9$x^{2}$ – 11= 0
x – 2$\sqrt{x}$ =8
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Toán Lớp 9: Giai PT dùng công thức nghiệm
-3$x^{2}$ + 5x = 0
3$x^{2}$ -8 = 0
9$x^{2}$ – 11= 0
x – 2$\sqrt{x}$ =8
Comments ( 2 )
Giải đáp:
$\begin{array}{l}
a) – 3{x^2} + 5x = 0\\
\Leftrightarrow 3{x^2} – 5x = 0\\
\Leftrightarrow \Delta = {5^2} – 4.3.0 = 25\\
\Leftrightarrow \left\{ \begin{array}{l}
{x_1} = \dfrac{{5 + \sqrt {25} }}{{2.3}} = \dfrac{5}{3}\\
{x_2} = \dfrac{{5 – \sqrt {25} }}{{2.3}} = 0
\end{array} \right.\\
Vậy\,x = 0;x = \dfrac{5}{3}\\
b)3{x^2} – 8 = 0\\
\Leftrightarrow \Delta = {0^2} – 4.3.\left( { – 8} \right) = 96\\
\Leftrightarrow \left\{ \begin{array}{l}
{x_1} = \dfrac{{\sqrt {96} }}{{2.3}} = \dfrac{{4\sqrt 6 }}{6} = \dfrac{{2\sqrt 6 }}{3}\\
{x_2} = – \dfrac{{2\sqrt 6 }}{3}
\end{array} \right.\\
Vậy\,x = \dfrac{{2\sqrt 6 }}{3};x = – \dfrac{{2\sqrt 6 }}{3}\\
c)9{x^2} – 11 = 0\\
\Leftrightarrow \Delta = 0 – 4.9.\left( { – 11} \right) = 36.11\\
\Leftrightarrow \left\{ \begin{array}{l}
{x_1} = \dfrac{{\sqrt {36.11} }}{{2.9}} = \dfrac{{6\sqrt {11} }}{{18}} = \dfrac{{\sqrt {11} }}{3}\\
{x_2} = – \dfrac{{\sqrt {11} }}{3}
\end{array} \right.\\
Vậy\,x = \dfrac{{\sqrt {11} }}{3};x = – \dfrac{{\sqrt {11} }}{3}\\
d)Dkxd:x \ge 0\\
x – 2\sqrt x = 8\\
\Leftrightarrow x – 2\sqrt x – 8 = 0\\
\Leftrightarrow {t^2} – 2t – 8 = 0\left( {t = \sqrt x \ge 0} \right)\\
\Leftrightarrow \Delta ‘ = 1 – \left( { – 8} \right) = 9\\
\Leftrightarrow \left\{ \begin{array}{l}
{t_1} = \dfrac{{1 + \sqrt 9 }}{1} = 4\\
{t_2} = \dfrac{{1 – \sqrt 9 }}{1} = – 2\left( {ktm} \right)
\end{array} \right.\\
\Leftrightarrow \sqrt x = 4\\
\Leftrightarrow x = 16\left( {tm} \right)\\
Vậy\,x = 16
\end{array}$
Đây nhé
