( 5x- 3x ) ² = ( 1- 2x) ² = 0 Tìm GTNN,GTLN 2x ² – 3x -9 D = 6x – x^2 + 5 F= x-x^2 – 2020 E= ( x -1) ² + ( x-3) ²

Question

( 5x- 3x ) ² = ( 1- 2x) ² = 0
Tìm GTNN,GTLN
2x ² – 3x -9
D = 6x – x^2 + 5
F= x-x^2 – 2020
E= ( x -1) ² + ( x-3) ²

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Camila 1 tháng 2021-07-27T01:57:42+00:00 2 Answers 4 views 0

Answers ( )

    0
    2021-07-27T01:58:45+00:00

    ( 5x- 3x ) ² = ( 1- 2x) ² = 0

           ???

    Đề sai ak?

    Tìm GTNN,GTLN:

    +, Ta có: 2x ² – 3x -9

    = 2(x² – $\frac{3}{2}$x – $\frac{9}{2}$)

    = 2[(x² – 2.x.$\frac{3}{4}$ + $\frac{9}{16}$) – $\frac{81}{16}$]

    = 2(x – $\frac{3}{4}$)² – $\frac{81}{8}$

    Vì (x – $\frac{3}{4}$)² ≥ 0 ⇒ 2x² – 3x – 9 ≥ -$\frac{81}{8}$

    Dấu “=” xảy ra khi (x – $\frac{3}{4}$)² = 0 ⇒ x = $\frac{3}{4}$

    Vậy Min của 2x² – 3x – 9 = -$\frac{81}{8}$ khi x = $\frac{3}{4}$

    +, Ta có: D = 6x – x^2 + 5

    ⇔ D = -(x² – 6x – 5)

    ⇔ D = -[(x² – 6x + 9) – 14]

    ⇔ D = -(x – 3)² + 14

    Vì (x – 3)² ≥ 0 ⇒ D ≤ 14

    Dấu “=” xảy ra khi (x – 3)² = 0 ⇒ x = 3

    Vậy Max D = 14 khi x = 3

    +, Ta có: F= x-x^2 – 2020

    ⇔ F = -(x² – x + 2020)

    ⇔ F = -[(x² – 2.x.$\frac{1}{2}$ + $\frac{1}{4}$) + $\frac{8079}{4}$]

    ⇔ F = -(x – $\frac{1}{2}$)² – $\frac{8079}{4}$

    Vì (x – $\frac{1}{2}$)² ≥ 0 ⇒ F ≤ -$\frac{8079}{4}$

    Dấu “=” xảy ra khi (x – $\frac{1}{2}$)² = 0 ⇒ x = $\frac{1}{2}$

    Vậy Max F = -$\frac{8079}{4}$

    +, Ta có: E= ( x -1) ² + ( x-3) ²

    ⇔ E = x² – 2x + 1 + x² – 6x + 9

    ⇔ E = 2x² – 8x + 10

    ⇔ E = 2(x² – 4x + 5)

    ⇔ E = 2[(x² – 4x + 4) + 1]

    ⇔ E = 2(x – 2)² + 2

    Vì (x – 2)² ≥ 0 ⇒ E ≥ 2

    Dấu “=” xảy ra khi (x – 2)² = 0 ⇒ x = 2

    Vậy Min E = 2 khi x = 2

    0
    2021-07-27T01:59:15+00:00

    Đáp án:

     $(5x-3x)^2 – (1-2x)^2 = 0$

    $⇔ (2x)^2 – (1-2x)^2 =0$

    $⇔(2x+1-2x)(2x-1+2x) =0$

    $⇔4x-1=0$

    $⇔4x=1$

    $⇔x = \dfrac{1}{4}$

    Vậy $x=\dfrac{1}{4}$

    $2x^2-3x-9$

    $=(\sqrt[]{2}x)^2 -2 . \sqrt[]{2}x . \dfrac{3\sqrt[]{2}}{4} + \dfrac{9}{8} -\dfrac{81}{8}$

    $= (\sqrt[]{2}x – \dfrac{3\sqrt[]{2}}{4})^2-\dfrac{81}{8}$

    Vì $(\sqrt[]{2}x -\dfrac{3\sqrt[]{2}}{4})^2 ≥ 0 $

    Nên $(\sqrt[]{2}x -\dfrac{3\sqrt[]{2}}{4})^2 -\dfrac{81}{8} ≥ – \dfrac{81}{8}$

    Dấu ”=” xảy ra khi $\sqrt[]{2}x – \dfrac{3\sqrt[]{2}}{4} =0⇔ x = \dfrac{3}{4}$

    Vậy  giá trị nhỏ nhất $=-\dfrac{81}{8}$ tại $x=\dfrac{3}{4}$

    $D =6x-x^2+5$

    $ = -(x^2-6x-5)$

    $= -(x^2 -2.3.x +9-14)$

    $ = -(x-3)^2+14$

    Vì $-(x-3)^2 ≤ 0$

    Nên $-(x-3)^2 +14 ≤ 14$

    Dấu ”=” xảy ra khi $x-3=0⇔x=3$

    Vậy Max D =14 tại $x=3$

    $F = x – x^2 -2020$

    $= -(x^2-x+2020)$

    $= -(x^2 -2. x . \dfrac{1}{2} +\dfrac{1}{4} +2019,75)$

    $ = -(x-\dfrac{1}{2})^2 – 2019,75$

    Vì $-(x-\dfrac{1}{2})^2 ≤ 0$

    Nên $-(x-\dfrac{1}{2})^2 – 2019,75 ≤ -2019,75$

    Dấu ”=” xảy ra khi $x-\dfrac{1}{2} =0⇔x=\dfrac{1}{2}$

    Vậy Max F $=-2019,75$ tại $x=\dfrac{1}{2}$

    $E =(x-1)^2 +(x-3)^2$

     $ = x^2 -2x+1 +x^2-6x+9$

    $ =2x^2-8x+10$

    $ = (\sqrt[]{2}x)^2 – 2 . \sqrt[]{2}x . 2√2 + 8+2$

    $  =(\sqrt[]{2}x – 2√2)^2 +2$

    Vì $(\sqrt[]{2}x -2√2)^2 ≥ 0$

    Nên $(\sqrt[]{2}x -2√2)^2 +2 ≥ 2$

    Dấu ”=” xảy ra khi $\sqrt[]{2}x-2√2 =0⇔x=2$

    Vậy Min E $=2$ tại $x=2$

     

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