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[thivao10]

Xem đáp án tại: http://thivao10.vn/d...ng-2017-69.html

[tuaneee111]

Bài 5: Ta có: $$ \displaystyle \begin{array}{l}5{{x}^{3}}+6{{x}^{2}}+12x+8=0\Leftrightarrow {{x}^{3}}+6{{x}^{2}}+12x+8=-4{{x}^{3}}\\\Leftrightarrow {{\left( {x+2} \right)}^{3}}={{\left( {-\sqrt[3]{4}x} \right)}^{3}}\Leftrightarrow x+2=-\sqrt[3]{4}x\Leftrightarrow x=\frac{{-2}}{{1+\sqrt[3]{4}}}\end{array}$$