Tìm Max a) P = 5 – 2x – x^2 Q= -3y^2 + 2y – 1 02/07/2021 Bởi Katherine Tìm Max a) P = 5 – 2x – x^2 Q= -3y^2 + 2y – 1
$a,P=5-2x-x^2 \\=-(x^2+2x-5) \\=-(x^2+2x+1-6) \\=-(x+1)^2+6 \\Vì\ -(x+1)^2≤0∀x \\⇒P=-(x+1)^2+6≤6∀x$ Dấu = xảy ra khi và chỉ khi $-(x+1)^2=0 \\⇔-(x+1)=0 \\⇔x+1=0 \\⇔x=-1 \\Vậy\ Max\ P=6⇔x=-1 \\b,Q=-3y^2+2y-1 \\=-(3y^2-2y+1) \\=-\bigg[(\sqrt{3}y)^2-2.\sqrt{3}y.\dfrac{1}{\sqrt{3}}+\bigg(\dfrac{1}{\sqrt{3}}\bigg)^2+\dfrac{2}{3}\bigg] \\=-\bigg[\bigg(\sqrt{3}y-\dfrac{1}{\sqrt{3}}\bigg)^2+\dfrac{2}{3}\bigg] \\=-\bigg(\sqrt{3}y-\dfrac{1}{\sqrt{3}}\bigg)^2-\dfrac{2}{3} \\Vì\ -\bigg(\sqrt{3}y-\dfrac{1}{\sqrt{3}}\bigg)^2-≤0∀x \\⇒Q=-\bigg(\sqrt{3}y-\dfrac{1}{\sqrt{3}}\bigg)^2-\dfrac{2}{3}≤-\dfrac{2}{3}∀x$ Dấu = xảy ra khi và chỉ khi $-\bigg(\sqrt{3}y-\dfrac{1}{\sqrt{3}}\bigg)^2=0 \\⇔\sqrt{3}y-\dfrac{1}{\sqrt{3}}=0 \\⇔\sqrt{3}y=\dfrac{1}{\sqrt{3}} \\⇔y=\dfrac{1}{3}$ Vậy $Max\ Q=\dfrac{-2}{3}⇔y=\dfrac{1}{3}$ Bình luận
`a)` `P = 5 – 2x – x^2` ` = – (x^2 + 2x – 5)` ` = – (x^2 + 2x + 1) + 6` ` = -( x^2 + 2 . x . 1 + 1^2) + 6` ` = – (x+1)^2 + 6` `\forall x ` ta có : `(x+1)^2 \ge 0` `=> -(x+1)^2 \le 0` `=> -(x+1)^2 + 6 \le 6` `=> P \le 6` Dấu `=` xảy ra `<=> x + 1 = 0` `<=> x = -1` Vậy `Max_P = 6 <=> x = -1` `b)` `Q = -3y^2 + 2y -1` `= -3 . (y^2 – 2/3y ) -1` ` = -3 . (y^2 – 2/3y + 1/9) – 2/3` ` = -3 . [y^2 – 2 . y . 1/3 + (1/3)^2 ] -2/3` ` = -3 . (y-1/3)^2 – 2/3` `\forall x` ta có : `(y-1/3)^2 \ge 0` `=> -3 . (y-1/3)^2 \le 0` `=> -3 . (y-1/3)^2 – 2/3 \le -2/3` `=> Q \le -2/3` Dấu `=` xảy ra `<=> y – 1/3 =0` `<=> y=1/3` Vậy `Max_Q = -2/3 <=> y = 1/3` Bình luận
$a,P=5-2x-x^2 \\=-(x^2+2x-5) \\=-(x^2+2x+1-6) \\=-(x+1)^2+6 \\Vì\ -(x+1)^2≤0∀x \\⇒P=-(x+1)^2+6≤6∀x$
Dấu = xảy ra khi và chỉ khi
$-(x+1)^2=0 \\⇔-(x+1)=0 \\⇔x+1=0 \\⇔x=-1 \\Vậy\ Max\ P=6⇔x=-1 \\b,Q=-3y^2+2y-1 \\=-(3y^2-2y+1) \\=-\bigg[(\sqrt{3}y)^2-2.\sqrt{3}y.\dfrac{1}{\sqrt{3}}+\bigg(\dfrac{1}{\sqrt{3}}\bigg)^2+\dfrac{2}{3}\bigg] \\=-\bigg[\bigg(\sqrt{3}y-\dfrac{1}{\sqrt{3}}\bigg)^2+\dfrac{2}{3}\bigg] \\=-\bigg(\sqrt{3}y-\dfrac{1}{\sqrt{3}}\bigg)^2-\dfrac{2}{3} \\Vì\ -\bigg(\sqrt{3}y-\dfrac{1}{\sqrt{3}}\bigg)^2-≤0∀x \\⇒Q=-\bigg(\sqrt{3}y-\dfrac{1}{\sqrt{3}}\bigg)^2-\dfrac{2}{3}≤-\dfrac{2}{3}∀x$
Dấu = xảy ra khi và chỉ khi
$-\bigg(\sqrt{3}y-\dfrac{1}{\sqrt{3}}\bigg)^2=0 \\⇔\sqrt{3}y-\dfrac{1}{\sqrt{3}}=0 \\⇔\sqrt{3}y=\dfrac{1}{\sqrt{3}} \\⇔y=\dfrac{1}{3}$
Vậy $Max\ Q=\dfrac{-2}{3}⇔y=\dfrac{1}{3}$
`a)`
`P = 5 – 2x – x^2`
` = – (x^2 + 2x – 5)`
` = – (x^2 + 2x + 1) + 6`
` = -( x^2 + 2 . x . 1 + 1^2) + 6`
` = – (x+1)^2 + 6`
`\forall x ` ta có :
`(x+1)^2 \ge 0`
`=> -(x+1)^2 \le 0`
`=> -(x+1)^2 + 6 \le 6`
`=> P \le 6`
Dấu `=` xảy ra `<=> x + 1 = 0`
`<=> x = -1`
Vậy `Max_P = 6 <=> x = -1`
`b)`
`Q = -3y^2 + 2y -1`
`= -3 . (y^2 – 2/3y ) -1`
` = -3 . (y^2 – 2/3y + 1/9) – 2/3`
` = -3 . [y^2 – 2 . y . 1/3 + (1/3)^2 ] -2/3`
` = -3 . (y-1/3)^2 – 2/3`
`\forall x` ta có :
`(y-1/3)^2 \ge 0`
`=> -3 . (y-1/3)^2 \le 0`
`=> -3 . (y-1/3)^2 – 2/3 \le -2/3`
`=> Q \le -2/3`
Dấu `=` xảy ra `<=> y – 1/3 =0`
`<=> y=1/3`
Vậy `Max_Q = -2/3 <=> y = 1/3`