Share
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
Question
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Answers ( )
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
`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 : $\begin{cases}(y+1)^2=0\\(x-\dfrac23)^4=0\end{cases}$
$⇒\begin{cases}y=-1\\x=\dfrac23\end{cases}$