( 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
( 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
( 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
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 $x=2$