Question

[Nguyễn Trà Giang]

Tìm giá trị nhỏ nhất của x:

a, A=x^2-4.x+5

b, B=2.x^2+3.x+1

c, C=5.x-3+4.x^2

Answer 1

[FL.Hermit]

a)

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

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

=>    \(A\ge1\)

DẤU "=" XẢY RA <=>    \(x=2\)

b)

\(2B=4x^2+6x+2=\left(2x+\frac{3}{2}\right)^2-0,25\)

CÓ:     \(\left(2x+\frac{3}{2}\right)^2\ge0\forall x\Rightarrow\left(2x+\frac{3}{2}\right)^2-0,25\ge-0,25\)

DẤU "=" XẢY RA <=>      \(2x+\frac{3}{2}=0\Leftrightarrow x=-\frac{3}{4}\)

c)

\(C=\left(2x+\frac{5}{4}\right)^2-\frac{73}{16}\ge-\frac{73}{16}\)

DẤU "=" XẢY RA <=>     \(2x+\frac{5}{4}=0\Leftrightarrow x=-\frac{5}{8}\)

Answer 2

[Khánh Ngọc]

a. Ta có : 

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

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

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

b. \(B=2x^2+3x+1=2\left(x+\frac{3}{4}\right)^2-\frac{1}{8}\)

Vì \(\left(x+\frac{3}{4}\right)^2\ge0\forall x\)\(\Rightarrow2\left(x+\frac{3}{4}\right)^2-\frac{1}{8}\ge-\frac{1}{8}\)

Dấu "=" xảy ra \(\Leftrightarrow2\left(x+\frac{3}{4}\right)^2=0\Leftrightarrow x+\frac{3}{4}=0\Leftrightarrow x=-\frac{3}{4}\)

Vậy Bmin = - 1/8 <=> x = - 3/4

c. \(C=5x-3+4x^2=4\left(x+\frac{5}{8}\right)^2-\frac{73}{16}\)

Vì \(\left(x+\frac{5}{8}\right)^2\ge0\forall x\)\(\Rightarrow4\left(x+\frac{5}{8}\right)^2-\frac{73}{16}\ge-\frac{73}{16}\)

Dấu "=" xảy ra \(\Leftrightarrow4\left(x+\frac{5}{8}\right)^2=0\Leftrightarrow x+\frac{5}{8}=0\Leftrightarrow x=-\frac{5}{8}\)

Vậy Cmin = - 73/16 <=> x = - 5/8