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

Đáp án:

$\begin{array}{l}
a) - 3{x^2} + 5x = 0\\
 \Leftrightarrow 3{x^2} - 5x = 0\\
 \Leftrightarrow \Delta  = {5^2} - 4.3.0 = 25\\
 \Leftrightarrow \left\{ \begin{array}{l}
{x_1} = \dfrac{{5 + \sqrt {25} }}{{2.3}} = \dfrac{5}{3}\\
{x_2} = \dfrac{{5 - \sqrt {25} }}{{2.3}} = 0
\end{array} \right.\\
Vậy\,x = 0;x = \dfrac{5}{3}\\
b)3{x^2} - 8 = 0\\
 \Leftrightarrow \Delta  = {0^2} - 4.3.\left( { - 8} \right) = 96\\
 \Leftrightarrow \left\{ \begin{array}{l}
{x_1} = \dfrac{{\sqrt {96} }}{{2.3}} = \dfrac{{4\sqrt 6 }}{6} = \dfrac{{2\sqrt 6 }}{3}\\
{x_2} =  - \dfrac{{2\sqrt 6 }}{3}
\end{array} \right.\\
Vậy\,x = \dfrac{{2\sqrt 6 }}{3};x =  - \dfrac{{2\sqrt 6 }}{3}\\
c)9{x^2} - 11 = 0\\
 \Leftrightarrow \Delta  = 0 - 4.9.\left( { - 11} \right) = 36.11\\
 \Leftrightarrow \left\{ \begin{array}{l}
{x_1} = \dfrac{{\sqrt {36.11} }}{{2.9}} = \dfrac{{6\sqrt {11} }}{{18}} = \dfrac{{\sqrt {11} }}{3}\\
{x_2} =  - \dfrac{{\sqrt {11} }}{3}
\end{array} \right.\\
Vậy\,x = \dfrac{{\sqrt {11} }}{3};x =  - \dfrac{{\sqrt {11} }}{3}\\
d)Dkxd:x \ge 0\\
x - 2\sqrt x  = 8\\
 \Leftrightarrow x - 2\sqrt x  - 8 = 0\\
 \Leftrightarrow {t^2} - 2t - 8 = 0\left( {t = \sqrt x  \ge 0} \right)\\
 \Leftrightarrow \Delta ' = 1 - \left( { - 8} \right) = 9\\
 \Leftrightarrow \left\{ \begin{array}{l}
{t_1} = \dfrac{{1 + \sqrt 9 }}{1} = 4\\
{t_2} = \dfrac{{1 - \sqrt 9 }}{1} =  - 2\left( {ktm} \right)
\end{array} \right.\\
 \Leftrightarrow \sqrt x  = 4\\
 \Leftrightarrow x = 16\left( {tm} \right)\\
Vậy\,x = 16
\end{array}$