Bài 1 :
a ) \(A=3x^2-5x+2000\)
\(A=3\left(x^2-\dfrac{5}{3}x+\dfrac{2000}{3}\right)\)
\(A=3\left[\left(x^2-\dfrac{5}{3}x+\dfrac{25}{36}\right)+\dfrac{23975}{36}\right]\)
\(A=3\left[\left(x-\dfrac{5}{6}\right)^2+\dfrac{23975}{36}\right]\)
Vì : \(\left(x-\dfrac{5}{6}\right)^2\ge0\Rightarrow\left(x-\dfrac{5}{6}\right)^2+\dfrac{23975}{36}\ge\dfrac{23975}{35}\Rightarrow3\left[\left(x-\dfrac{5}{6}\right)^2+\dfrac{23975}{36}\right]\ge\dfrac{23975}{12}\)
Vậy GTNN của A là \(\dfrac{23975}{12}\) khi \(\left(x-\dfrac{5}{6}\right)^2=0\Rightarrow x=\dfrac{5}{6}\)
b ) \(B=-2x^2+6x+2018\)
\(B=-2\left(x^2-3x-1009\right)\)
\(B=-2\left[\left(x^2-3x+\dfrac{9}{4}\right)-\dfrac{4045}{4}\right]\)
\(B=-2\left[\left(x-\dfrac{3}{2}\right)^2-\dfrac{4045}{4}\right]\le\dfrac{4045}{2}\)
Vậy GTLN của B là \(\dfrac{4045}{2}\) khi \(\left(x-\dfrac{3}{2}\right)^2=0\Leftrightarrow x=\dfrac{3}{2}\)
Chúc bạn học tốt !!
2)
\(x^9-x^7+x^6-x^5-x^4+x^3-x^2+1\)
\(=x^7\left(x^2-1\right)+x^4\left(x^2-1\right)+x^3\left(x^2-1\right)-1\left(x^2-1\right)\)
\(=\left(x^7+x^4+x^3-1\right)\left(x-1\right)\left(x+1\right)\)
\(\left(x-3\right)\left(x-1\right)\left(x+1\right)\left(x+3\right)+15\)
\(=\left(x^2-1\right)\left(x^2-9\right)+15\)
\(=\left(x^2-5+4\right)\left(x^2-5-4\right)+15\)
\(=\left(x^2-5\right)^2-16+15=\left(x^2-5\right)^2-1\)
\(=\left(x^2-5+1\right)\left(x^2-5-1\right)=\left(x^2-4\right)\left(x^2-6\right)=\left(x-2\right)\left(x+2\right)\left(x^2-6\right)\)
\(x^7+x^5+1\)
\(=x^7-x^6+x^5-x^3+x^2+x^6-x^5+x^4-x^2+x+x^5-x^4+x^3-x+1\)
\(=\left(x^2+x+1\right)\left(x^5-x^4+x^3-x+1\right)\)