Tìm x:
7x^3 + 3x^2 - 3x + 1 = 0
x^3 + 3x^2 + 3x + 28 = 0
a, [tex]7x^3+3x^2-3x+1=0\\\Leftrightarrow 7x^3+7x^2-4x-4x+x+1=0\\\Leftrightarrow 7x^2(x+1)-4x(x+1)+(x+1)=0\\\Leftrightarrow (x+1)(7x^2-4x+1)=0[/tex]
Ta có: [tex]7x^2-4x+1=7\left ( x^2-\frac{4}{7}x+\frac{1}{7} \right )\\=7\left ( x^2-2.\frac{2}{7}x+\frac{4}{49}+\frac{3}{49} \right )\\=7\left [ \left ( x-\frac{2}{7} \right )^2+\frac{3}{49} \right ]> 0[/tex]
Do đó [tex]x+1=0\Leftrightarrow x=-1[/tex]
Vậy..............
b, [tex]x^3+3x^2+3x+28=0\\\Leftrightarrow x^3+4x^2-x^2-4x+7x+28=0\\\Leftrightarrow x^2(x+4)-x(x+4)+7(x+4)=0\\\Leftrightarrow (x+4)(x^2-x+7)=0[/tex]
Ta có: [tex]x^2-x+7=x^2-2.\frac{1}{2}x+\frac{1}{4}+\frac{27}{4}=\left ( x-\frac{1}{2} \right )^2+\frac{27}{4}> 0[/tex]
Do đó [tex]x+4=0\Leftrightarrow x=-4[/tex]
Vậy...............