Đáp án:
Giải thích các bước giải:
$\text{1)}$
$\text{a, (3x+1)²-(2x-1)(2x+1)}$
$\text{= 9x²+6x+1-4x²+1}$
$\text{= 5x²+6x+2}$
$\text{b, (2x+3)²-(x+2)(x-2)}$
$\text{= 4x²+12x+9-x²+4}$
$\text{= 3x²+12x+13}$
$\text{c, (2x+1)²-2(2x+1)(x-3)+(x-3)²}$
$\text{= (2x+1-x+3)²}$
$\text{= (x+4)²}$
$\text{= x²+8x+16}$
$\text{2)}$
$\text{b, B= 4x²+4x+7}$
$\text{B= 4x²+4x+1+6}$
$\text{B=(2x+1)²+6}$
$\text{Vì (2x+1)² ≥ 0 ∀ x}$
$\text{⇒ (2x+1)²+6 ≥ 6 ∀ x}$
$\text{Dấu = xảy ra ⇔ 2x+1=0 ⇔ x=$\frac{-1}{2}$ }$
$\text{Vậy $Min_{B}$ = 6 ⇔ x=$\frac{-1}{2}$}$
$\text{a, A= x²+3x-1}$
$\text{A= x²+3x+$\frac{9}{4}$ -$\frac{13}{4}$ }$
$\text{A=(x+$\frac{3}{2}$ )²-$\frac{13}{4}$ }$
$\text{Vì (x+$\frac{3}{2}$)² ≥ 0 ∀ x}$
$\text{⇒ (x+$\frac{3}{2}$ )²-$\frac{13}{4}$ ≥ -$\frac{13}{4}$ ∀ x}$
$\text{Dấu = xảy ra ⇔ x+$\frac{3}{2}$=0 ⇔ x=$\frac{-3}{2}$ }$
$\text{Vậy $Min_{A}$ = -$\frac{13}{4}$ ⇔ x=$\frac{-3}{2}$}$
$\text{c, C= x²-5x+2}$
$\text{C= x²-5x+$\frac{25}{4}$ -$\frac{17}{4}$ }$
$\text{C=(x-$\frac{5}{2}$ )²-$\frac{17}{4}$}$
$\text{Vì (x-$\frac{5}{2}$ )² ≥ 0 ∀ x}$
$\text{⇒ (x-$\frac{5}{2}$ )²-$\frac{17}{4}$ ≥ -$\frac{17}{4}$ ∀ x}$
$\text{Dấu = xảy ra ⇔ x-$\frac{5}{2}$ =0 ⇔ x=$\frac{5}{2}$ }$
$\text{Vậy $Min_{C}$ = -$\frac{17}{4}$ ⇔ x=$\frac{5}{2}$}$
$\text{d, D= x(x+2)}$
$\text{D= x²+2x+1-1}$
$\text{D=(x+1)²-1}$
$\text{Vì (x+1)² ≥ 0 ∀ x}$
$\text{⇒ (x+1)²-1 ≥ -1 ∀ x}$
$\text{Dấu = xảy ra ⇔ x+1=0 ⇔ x=-1 }$
$\text{Vậy $Min_{D}$ = -1 ⇔ x=-1 }$
$\text{e, E=2x²+4x+2}$
$\text{E=2(x²+2x+1)}$
$\text{E= 2(x+1)²}$
$\text{Vì (x+1)² ≥ 0 ∀ x}$
$\text{⇒ 2(x+1)² ≥ 0 ∀ x}$
$\text{Dấu = xảy ra ⇔ x+1=0 ⇔ x=-1 }$
$\text{Vậy $Min_{E}$ = 0 ⇔ x= -1}$
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