# bài 1: tìm GTNN a) A = (2x + 1/3 ) ^2 = 1 b) B = ( 3x + 2/5)^4 + 20 c ) C = ( y + 1)^2 + ( x – 2/3 )^4 – 2020

bài 1: tìm GTNN
a) A = (2x + 1/3 ) ^2 = 1
b) B = ( 3x + 2/5)^4 + 20
c ) C = ( y + 1)^2 + ( x – 2/3 )^4 – 2020

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1. A = ( 2x + 1/3 )² ± 1 ( không biết để dấu nào nên để ± vậy .-. )

Vì ( 2x + 1/3 )² ≥ 0 ∀ x => ( 2x + 1/3 )² ± 1 ≥ ± 1

Dấu “=” xảy ra khi 2x + 1/3 = 0 => x = -1/6

=> MinA = ±1 <=> x = -1/6

B = ( 3x + 2/5 )^4 + 20

Vì ( 3x + 2/5 )^4 ≥ 0 ∀ x => ( 3x + 2/5 )^4 + 20 ≥ 20

Dấu “=” xảy ra khi 3x + 2/5 = 0 => x = -2/15

=> MinB = 20 <=> x = -2/15

C = ( y + 1 )² + ( x – 2/3 )^4 – 2020

Vì ( y + 1 )² ≥ 0 ∀ y

( x – 2/3 )^4 ≥ 0 ∀ x

=> ( y + 1 )² + ( x – 2/3 )^4 – 2020 ≥ -2020 ∀ x, y

Dấu “=” xảy ra khi x = 2/3 ; y = -1

=> MinC = -2020 <=> x = 2/3 ; y= -1

2. a, A=(2x+1/3)^2-1

Vì (2x+1/3)^2≥0 ∀x

⇒A≥-1

Dấu $”=”$ xảy ra khi : 2x+1/3=0⇒x={-1}/6

b, B=(3x+2/5)^4+20

Vì (3x+2/5)^4≥0 ∀x

⇒B≥20

Dấu $”=”$ xảy ra khi : 3x+2/5=0⇒x={-2}/5

c, C=(y+1)^2+(x-2/3)^4-2020

Vì : (y+1)^2;(x-2/3)^4≥0

⇒(y+1)^2+(x-2/3)^4≥0

⇒C≥-2020

Dấu $”=”$ xảy ra khi : $\begin{cases}(y+1)^2=0\\(x-\dfrac23)^4=0\end{cases}$

$⇒\begin{cases}y=-1\\x=\dfrac23\end{cases}$