Question

tìm min cuả biẻu thức A = ( x- 1)( x + 2 )(x + 3)(x+6) + 2045

Answer 1

lớp 8?

\(A=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)+2045\)

\(=\left(x^2+6x-x-6\right)\left(x^2+3x+2x+6\right)+2045\)

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)+2045\)

\(=\left(x^2+5x\right)^2-6^2+2045\)

\(=\left(x^2+5x\right)^2+2009\ge2009\)

Dấu "=" xày ra khi x²+5x=0  <=> x=0 hoặc x=-5

Vậy MinA=2009 khi x=0 hoặc x=-5

Answer 2

\(A=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)+2045\)

\(A=\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]+2045\)

\(A=\left(x^2+5x-6\right)\left(x^2+5x+6\right)+2045\)

\(A=\left(x^2+5x\right)^2-36+2045\)

\(A=\left(x^2+5x\right)^2+2009\)

Vì \(\left(x^2+5x\right)^2\ge0\Rightarrow\left(x^2+5x\right)^2+2009\ge2009\)

\(\Rightarrow A\ge2009\)

=> GTNN của A bằng 2009 

Dấu '=' xảy ra khi \(\left(x^2+5x\right)^2=0\Leftrightarrow x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\)

=> x = 0 hoặc x + 5 = 0 <=>  x = -5

Vậy GTNN của A bằng 2009