a) Với \(x\ne0\) , ta rút gọn :
\(A=\left(6x^3+12x^2\right):2x-2x\left(x+1\right)+5\)
\(A=3x^2+6x-2-2x+5\)
\(A=3x^2+6x+3\)
\(A=3\left(x^2+2x+1\right)\)
\(A=3\left(x+1\right)^2\)
Vậy sau khi rút gọn kết quả là : \(A=3\left(x+1\right)^2\)
b) Ta thấy \(x\ne0\Rightarrow x+1\ne1\)
\(\Rightarrow\left(x+1\right)^2\ge1;\forall x\ne0\)
\(\Rightarrow3\left(x+1\right)^2\ge3>1;\forall x\ne0\)
Vậy \(3\left(x+1\right)^2>1\Leftrightarrow A>1\) với \(\forall x\ne0\) \(\left(ĐPCM\right)\)
a) Với \(x\ne0\) , ta rút gọn :
\(A=\left(6x^3+12x^2\right):2x-2\left(x+1\right)+5\)
\(A=3x^2+6x-2x-2+5\)
\(A=3x^2+4x+3\)
Vậy sau khi rút gọn kết quả là : \(A=3x^2+4x+3\)
b) \(A=3x^2+4x+3\)
\(A=3x^2+4x+\frac{4}{3}+\frac{5}{3}\)
\(A=3\left(x^2+\frac{4}{3}x+\frac{4}{9}\right)+\frac{5}{3}\)
\(A=3\left(x+\frac{2}{3}\right)^2+\frac{5}{3}\)
Ta thấy : \(\left(x+\frac{2}{3}\right)^2\ge0;\forall x\)
\(\Rightarrow3\left(x+\frac{2}{3}\right)^2\ge0;\forall x\)
\(\Rightarrow3\left(x+\frac{2}{3}\right)^2+\frac{5}{3}\ge\frac{5}{3}>0\)
\(\Leftrightarrow A>0\)
Vậy \(A>0\) với \(\forall x\ne0\) \(\left(ĐPCM\right)\)