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

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

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

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