a) Kết quả là: \(x^2-\frac{7}{2}x+3\)
b) Ta có: \(x^2-2x+1-y^2=\left(x-1\right)^2-y^2=\left(x-1-y\right)\left(x-1+y\right)\)
c) Ta có: \(\frac{\left(3^{10}+x^9\right)}{x^9}=2011\)
\(\Leftrightarrow\frac{3^{10}}{x^9}+\frac{x^9}{x^9}=2011\)
\(\Leftrightarrow\frac{3}{x}+1=2011\)
\(\Leftrightarrow\frac{3}{x}=2011-1\Leftrightarrow x=\frac{2011-1}{3}=670\)
Vậy: x=670
d) Ta có: \(A=x^2+4x+20\)
\(=x^2+4x+4+16=\left(x+2\right)^2+16\)
Ta có: \(\left(x+2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+2\right)^2+16\ge16\forall x\)
Dấu '=' xảy ra khi
\(\left(x+2\right)^2=0\Leftrightarrow x+2=0\Leftrightarrow x=-2\)
Vậy: GTNN của biểu thức \(A=x^2+4x+20\) là 16 khi x=-2
