Đáp án:
Giải thích các bước giải:
a). $x^{2}$ - 11x + 3 = 0
sử dụng : $\frac{-(b)±\sqrt[]{(b)^2-4ac}}{2.a}$
a = 1
b = -11
c = 3
=> $\frac{-(-11)±\sqrt[]{(-11)^2-4.1.3}}{2.1}$
=> $\frac{-(-11)±\sqrt[]{121-12}}{2}$
=> $\frac{11±\sqrt[]{109}}{2}$
Vậy x = $\frac{11±\sqrt[]{109}}{2}$
b). 2$x^{2}$ + 3x - 27 = 0
$\Leftrightarrow$ 2$x^{2}$ - 6x + 9x - 27 = 0
$\Leftrightarrow$ 2x ( x - 3) + 9 ( x - 3) = 0
$\Leftrightarrow$ (2x + 9 ) (x - 3) = 0
$\Leftrightarrow$ \(\left[ \begin{array}{l}2x + 9 = 0\\x-3 = 0\end{array} \right.\)
$\Leftrightarrow$ \(\left[ \begin{array}{l}x=\frac{-9}{2} \\x=3\end{array} \right.\)
c). $x^{2}$ - 2x - 3 = 0
$\Leftrightarrow$ $x^{2}$ + x - 3x - 3 = 0
$\Leftrightarrow$ x (x + 1) - 3 (x + 1) = 0
$\Leftrightarrow$ (x - 3) (x + 1) = 0
$\Leftrightarrow$ \(\left[ \begin{array}{l}x=3\\x=-1\end{array} \right.\)
d). 2$x^{2}$ + 5x - 3 = 0
$\Leftrightarrow$ 2$x^{2}$ - x + 6x - 3 = 0
$\Leftrightarrow$ x (2x - 1) + 3 (2x - 1) = 0
$\Leftrightarrow$ (x + 3)(2x - 1) = 0
$\Leftrightarrow$ \(\left[ \begin{array}{l}x=-3\\x=\frac{1}{2}\end{array} \right.\)
e). 5x (x - 1) = x - 1
$\Leftrightarrow$ 5$x^{2}$ - 5x - x + 1 = 0
$\Leftrightarrow$ 5x (x - 1) - (x - 1) = 0
$\Leftrightarrow$ (5x - 1) (x - 1) = 0
$\Leftrightarrow$ \(\left[ \begin{array}{l}x=\frac{1}{5}\\x=1\end{array} \right.\)
f). 2 (x + 5) - $x^{2}$ - 5x = 0
$\Leftrightarrow$ 2x + 10 - $x^{2}$ - 5x = 0
$\Leftrightarrow$ - ($x^{2}$ - 2x + 5x - 10 ) = 0
$\Leftrightarrow$ $x^{2}$ - 2x + 5x - 10 = 0
$\Leftrightarrow$ x ( x - 2) + 5 (x - 2 ) = 0
$\Leftrightarrow$ (x + 5) (x - 2) = 0
$\Leftrightarrow$\(\left[ \begin{array}{l}x=5\\x=2\end{array} \right.\)
