Question

Tìm giá trị nhỏ nhất của biểu thức E=(x+3)(1-x-x^2)

Answer 1

A=(x−1)(x−2)(x−3)(x−4)+5A=(x−1)(x−2)(x−3)(x−4)+5

⇔A=[(x−1)(x−4)][(x−2)(x−3)]+5⇔A=[(x−1)(x−4)][(x−2)(x−3)]+5

⇔A=(x2−4x−x+4)(x2−3x−2x+6)+5⇔A=(x2−4x−x+4)(x2−3x−2x+6)+5

⇔A=(x2−5x+4)(x2−5x+6)+5⇔A=(x2−5x+4)(x2−5x+6)+5

⇔A=(x2−5x+4)[(x2−5x+4)+2]+5⇔A=(x2−5x+4)[(x2−5x+4)+2]+5

⇔A=(x2−5x+4)2+2(x2−5x+4)+5⇔A=(x2−5x+4)2+2(x2−5x+4)+5

⇔A=(x2−5x+4)2+2x2−10x+8+5⇔A=(x2−5x+4)2+2x2−10x+8+5

⇔A=(x2−5x+4)2+2x2−10x+13⇔A=(x2−5x+4)2+2x2−10x+13

⇔A=(x2−5x+4)2+2x2−10x+252+12⇔A=(x2−5x+4)2+2x2−10x+252+12

⇔A=(x2−5x+4)2+(2x2−10x+252)+12⇔A=(x2−5x+4)2+(2x2−10x+252)+12

⇔A=(x2−5x+4)2+2(x2−5x+254)+12⇔A=(x2−5x+4)2+2(x2−5x+254)+12

⇔A=(x2−5x+4)2+2[x2−2.x.52+(52)2]+12⇔A=(x2−5x+4)2+2[x2−2.x.52+(52)2]+12

⇔A=(x2−5x+4)2+2(x−52)2+12⇔A=(x2−5x+4)2+2(x−52)2+12

Vậy GTNN của A=12A=12 khi ⎧⎩⎨x2−5x+4=0x−52=0{x2−5x+4=0x−52=0 ⇔⎧⎩⎨x2−5x+4=0(loai)x=52