Question

Tìm giá trị nhỏ nhất

a)  A = x² - 4x + 5

b) B = 2x² - 4x -6 

c) C = 3x² + 9x +6

d) D = 5x² + 5x +1

Answer 1

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

\(\left(x-2\right)^2\ge0\forall x\Rightarrow\left(x-2\right)^2+1\ge1\)

Đẳng thức xảy ra <=> x - 2 = 0 => x = 2

Vậy A_Min = 1 khi x = 2

b) B = \(2x^2-4x-6=2\left(x^2-2x-3\right)=2\left(x^2-2x+1\right)-8=2\left(x-1\right)^2-8\)

\(\left(x-1\right)^2\ge0\forall x\Rightarrow2\left(x-1\right)^2\ge0\Rightarrow2\left(x-1\right)^2-8\ge-8\)

Đẳng thức xảy ra <=> x - 1 = 0 => x = 1

Vậy B_Min = -8 khi x = 1

c) C = \(3x^2+9x+6=3\left(x^2+3x+2\right)=3\left(x^2+3x+\frac{9}{4}\right)-\frac{3}{4}=3\left(x+\frac{3}{2}\right)^2-\frac{3}{4}\)

\(\left(x+\frac{3}{2}\right)^2\ge0\forall x\Rightarrow3\left(x+\frac{3}{2}\right)^2\ge0\Rightarrow3\left(x+\frac{3}{2}\right)^2-\frac{3}{4}\ge-\frac{3}{4}\forall x\)

Đẳng thức xảy ra <=> x + 3/2 = 0 => x = -3/2

Vậy C_Min = -3/4 khi x = -3/2

d) D = \(5x^2+5x+1=5\left(x^2+x+\frac{1}{5}\right)=5\left(x^2+x+\frac{1}{4}\right)-\frac{1}{4}=5\left(x+\frac{1}{2}\right)^2-\frac{1}{4}\)

\(\left(x+\frac{1}{2}\right)^2\ge0\forall x\Rightarrow5\left(x+\frac{1}{2}\right)^2\ge0\Rightarrow5\left(x+\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\forall x\)

Đẳng thức xảy ra <=> x + 1/2 = 0 => x = -1/2

Vậy D_Min = -1/4 khi x = -1/2