Bài 1a: \(A=x^2-5x+1\) \(=x^2-5x+\frac{25}{4}-\frac{21}{4}\) \(=\left(x-\frac52\right)^2-\frac{21}{4}>=-\frac{21}{4}\forall x\) Dấu '=' xảy ra khi \(x-\frac52=0\) =>\(x=\frac52\) b: \(B=\left(x^2+x-6\right)\left(x^2+x+2\right)\) \(=\left(x^2+x\right)^2-4\left(x^2+x\right)-12\) \(=\left(x^2+x\right)^2-4\left(x^2+x\right)+4-16\) \(=\left(x^2+x-2\right)^2-16\ge-16\forall x\) Dấu '=' xảy ra khi \(x^2+x-2=0\) =>(x+2)(x-1)=0=>x+2=0 hoặc x-1=0=>x=-2 hoặc x=1Bài 2:a: \(A=1-x^2+3x\) \(=-x^2+3x-\frac94+\frac{13}{4}\) \(=-\left(x-\frac32\right)^2+\frac{13}{4}\le\frac{13}{4}\forall x\) Dấu '=' xảy ra khi \(x-\frac32=0\) =>\(x=\frac32\)Câu trả lời: Bài 1a: \(A=x^2-5x+1\) \(=x^2-5x+\frac{25}{4}-\frac{21}{4}\) \(=\left(x-\frac52\right)^2-\frac{21}{4}>=-\frac{21}{4}\forall x\) Dấu '=' xảy ra khi \(x-\frac52=0\) =>\(x=\frac52\) b: \(B=\left(x^2+x-6\right)\left(x^2+x+2\right)\) \(=\left(x^2+x\right)^2-4\left(x^2+x\right)-12\) \(=\left(x^2+x\right)^2-4\left(x^2+x\right)+4-16\) \(=\left(x^2+x-2\right)^2-16\ge-16\forall x\) Dấu '=' xảy ra khi \(x^2+x-2=0\) =>(x+2)(x-1)=0=>x+2=0 hoặc x-1=0=>x=-2 hoặc x=1Bài 2:a: \(A=1-x^2+3x\) \(=-x^2+3x-\frac94+\frac{13}{4}\) \(=-\left(x-\frac32\right)^2+\frac{13}{4}\le\frac{13}{4}\forall x\) Dấu '=' xảy ra khi \(x-\frac32=0\) =>\(x=\frac32\)