Giải câu 81 bài: Ôn tập chương 1 sgk Toán Đại 8 tập 1 Trang 33.
a) \({2 \over 3}x\left( {{x^2} - 4} \right) = 0\) hay
\(\Leftrightarrow {2 \over 3}x\left( {{x^2} - {2^2}} \right) = 0\)
\(\Leftrightarrow {2 \over 3}x\left( {x - 2} \right)\left( {x + 2} \right) = 0\)
$\Leftrightarrow \left[ \begin{array}{ll} x = 0 \\ x + 2 = 0 \\x - 2 = 0 \\\end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{ll} x = 0 \\ x = -2 \\x = 2 \\\end{array} \right.$
Vậy \(x = 0,x = - 2,x = 2\)
\(b){\left( {x + 2} \right)^2} - \left( {x - 2} \right)\left( {x + 2} \right) = 0\)
\(\Leftrightarrow\left( {x + 2} \right)\left[ {\left( {x + 2} \right) - \left( {x - 2} \right)} \right] = 0\)
\(\Leftrightarrow\left( {x + 2} \right)\left( {x + 2 - x + 2} \right) = 0\)
\(\Leftrightarrow\left( {x + 2} \right).4 = 0\)
\(\Leftrightarrowx + 2 = 0\)
\(\Leftrightarrowx = - 2\)
Vậy x = - 2
c) \(x + 2\sqrt 2 {x^2} + 2{x^3} = 0\)
\(\Leftrightarrowx\left( {1 + 2\sqrt 2 x + 2{x^2}} \right) = 0\)
\(\Leftrightarrowx(1 + 2\sqrt 2 x + {\left( {\sqrt 2 x} \right)^2} = 0\)
\(\Leftrightarrowx{\left( {1 + \sqrt 2 x} \right)^2} = 0\)
$\Leftrightarrow \left[ \begin{array}{ll} x = 0 \\ (1 + \sqrt{2}x) ^{2}= 0 \\\end{array} \right.$
$\Leftrightarrow \left[ \begin{array}{ll} x = 0 \\ 1 + \sqrt{2}x = 0 \\\end{array} \right.$
$\Leftrightarrow\left[ \begin{array}{ll} x = 0 \\ x = -\frac{1}{\sqrt{2}} \\\end{array} \right.$
Vậy \(x = 0,x = - {1 \over {\sqrt 2 }}\)