Từ xy+yz+zx=0 => 2(xy+yz+zx)=0

Từ x+y+z=0 => (x+y+z)²=0

=>x²+y²+z²+2(xy+yz+zx)=0

=>x²+y²+z²=0

Mà \(x^2\ge0,y^2\ge0,z^2\ge0\Rightarrow x^2+y^2+z^2\ge0\)

=>x=y=z=0

Thay x=y=z=0 vào A

=>A=-1+1-1=-1

nhàm dòng cuối 

=>A=-1+0+1=0

Ta có : \(4x^2-5x+2\)

\(=\left(2x\right)^2-2.2x.\frac{5}{4}+\frac{25}{16}+\frac{7}{16}\)

\(=\left(2x-\frac{5}{4}\right)^2+\frac{7}{16}\)

Do \(\left(2x-\frac{5}{4}\right)^2\ge0\forall x\)

\(\Rightarrow\left(2x-\frac{5}{4}\right)^2+\frac{7}{16}\ge\frac{7}{16}>0\forall x\left(đpcm\right)\)

Học tốt nha bạn 

a) \(A=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)+1\)

\(A=\left[\left(x-1\right)\left(x-4\right)\right]\left[\left(x-2\right)\left(x-3\right)\right]+1\)

\(A=\left(x^2-5x+4\right)\left(x^2-5x+6\right)+1\)

Đặt \(a=x^2-5x+5\)

\(\Leftrightarrow A=\left(a-1\right)\left(a+1\right)+1\)

\(\Leftrightarrow A=a^2-1^2+1\)

\(\Leftrightarrow A=a^2\)

Thay \(a=x^2-5x+5\)vào A ta có :

\(A=\left(x^2-5x+5\right)^2\)

b) \(B=\left(x^2+3x+2\right)\left(x^2+7x+12\right)+1\)

\(B=\left(x^2+x+2x+2\right)\left(x^2+3x+4x+12\right)+1\)

\(B=\left[x\left(x+1\right)+2\left(x+1\right)\right]\left[x\left(x+3\right)+4\left(x+3\right)\right]+1\)

\(B=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)

Làm tương tự câu a)

c) \(12x^2-3xy-8xz+2yz\)

\(=3x\left(4x-y\right)-2z\left(4x-y\right)\)

\(=\left(4x-y\right)\left(3x-2z\right)\)

Ta có : \(x+y=1\)

\(\Leftrightarrow\left(x+y\right)^2=1\)

\(\Leftrightarrow x^2+y^2+2xy=1\)

\(\Leftrightarrow x^2+y^2+2m=1\)

\(\Leftrightarrow x^2+y^2=1-2m\)

\(B=\left(x^2-x\right)\left(x^2-x-6\right)+9\)

\(=\left(x^2-x\right)^2-6\left(x^2-x\right)+9=\left(x^2-x-3\right)^2\ge0\)

Vậy GTLN của B là 0

Xin lỗi bạn, GTNN của A là 0 nhé.