Question

Tìm giá trị nhỏ nhất của biểu thức:

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

Answer 1

Đặt x²-2x+1=t, ta có:

\(A=\left(t-1\right)\left(t+1\right)=t^2-1=\left(x^2-2x+1\right)^2-1\ge-1\)

Dấu "=" xảy ra khi \(x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)

Answer 2

Đặt \(\left(x^2-2x\right)\left(x^2-2x=2\right)=k.\left(k+2\right)=A\)

\(\Rightarrow A=k.\left(k+2\right)=k^2+2k\)

\(\Rightarrow A=k^2+k+k+1-1=k\left(k+1\right)+\left(k+1\right)-1\)

\(\Rightarrow A=\left(k+1\right)^2-1\)

\(\Rightarrow A=\left(x^2-2x+1\right)^2-1\)

\(\Rightarrow A=\left(x^2-x-x+1\right)^2-1=\left[x.\left(x-1\right)-\left(x-1\right)\right]^2-1\)

\(\Rightarrow A=\left(x-1\right)^2-1\ge-1\)

( Dấu "=" xảy ra <=> x=1 )