1) \(a^3+2a^2-13a+10=a^3-a^2+3a^2-3a-10a+10=\)
\(=a^2\left(a-1\right)+3a\left(a-1\right)-10\left(a-1\right)=\left(a-1\right)\left(a^2+3a-10\right)\)
\(=\left(a-1\right)\left(a^2-2a+5a-10\right)=\left(a-1\right)\left[a\left(a-2\right)+5\left(a-2\right)\right]=\)
\(=\left(a-1\right)\left(a-2\right)\left(a+5\right)\)
b) \(\left(a^2+4b^2-5\right)^2-16\left(ab+1\right)^2=\left(a^2+4b^2-5+4ab+4\right)\left(a^2+4b^2-5-4ab-4\right)\)
\(=\left(a^2+4ab+4b^2-1\right)\left(a^2-4ab+4b^2-9\right)=\left[\left(a+2b\right)^2-1\right]\left[\left(a-2b\right)^2-9\right]=\)
\(=\left(a+2b+1\right)\left(a+2b-1\right)\left(a-2b+3\right)\left(a-2b-3\right)\)
2) \(6a-5b=1\Rightarrow5b=6a-1\Rightarrow25b^2=36a^2-12a+1\)
\(\Rightarrow4a^2+25b^2=40a^2-12a+1=40\left(a^2-2\cdot a\cdot\frac{3}{20}+\left(\frac{3}{20}\right)^2\right)+1-\frac{9}{10}\)
\(=40\left(a-\frac{3}{20}\right)^2+\frac{1}{10}\)
Vậy GTNN của \(4a^2+25b^2\)= 1/10. Xảy ra khi a = 3/20 và b = -1/50.
