Question

[Nguyễn Lê Đức Thành]

Tìm GTNN 
a A=x²-6x+15

b B=x²+5x+7

c C= (x+1)(x+2)(x+3)(x+4)+10

Answer 1

[Quỳnh Anh]

Trả lời:

a, \(A=x^2-6x+15=\left(x^2-6x+9\right)+6=\left(x-3\right)^2+6\ge6\forall x\)\(6\forall x\)

Dấu "=" xảy ra khi x - 3 = 0 <=> x = 3

Vậy GTNN của A = 6 khi x = 3

b, \(B=x^2+5x+7=\left(x^2+2.x.\frac{5}{2}+\frac{25}{4}\right)+\frac{3}{4}=\left(x+\frac{5}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)

Dấu "=" xảy ra khi \(x+\frac{5}{2}=0\Leftrightarrow x=-\frac{5}{2}\)

Vậy GTNN của B = 3/4 khi x = - 5/2

c, \(C=\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+10\)

\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]+10\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+10\)

Đặt \(x^2+5x+4=t\)

\(\Rightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)+10=t\left(t+2\right)+10\)

\(=t^2+2t+10=\left(t^2+2t+1\right)+9=\left(t+1\right)^2+9\ge9\forall t\)

Dấu "=" xảy ra khi \(t+1=0\Leftrightarrow t=-1\)

hay \(x^2+5x+4=-1\)

\(\Leftrightarrow x^2+5x+5=0\)

\(\Leftrightarrow4\left(x^2+5x+5\right)=0\)

\(\Leftrightarrow4x^2+20x+20=0\)

\(\Leftrightarrow\left(4x^2+20x+25\right)-5=0\)

\(\Leftrightarrow\left(2x+5\right)^2-5=0\)

\(\Leftrightarrow\left(2x+5-\sqrt{5}\right)\left(2x+5+\sqrt{5}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+5-\sqrt{5}=0\\2x+5+\sqrt{5}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-5+\sqrt{5}}{2}\\x=\frac{-5-\sqrt{5}}{2}\end{cases}}}\)

Vậy GTNN của C = 9 khi \(\orbr{\begin{cases}x=\frac{-5+\sqrt{5}}{2}\\x=\frac{-5-\sqrt{5}}{2}\end{cases}}\)