Cho `x,y>0` tm `\sqrt{x}+\sqrt{y} ≥1` Tìm min `P=(2x+8\sqrt{x}+17)/(\sqrt{x}+2)+(3y+6\sqrt{y}+5)/(\sqrt{y}+1`

Question

Cho `x,y>0` tm `\sqrt{x}+\sqrt{y} ≥1`
Tìm min `P=(2x+8\sqrt{x}+17)/(\sqrt{x}+2)+(3y+6\sqrt{y}+5)/(\sqrt{y}+1`

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Piper 1 năm 2021-10-27T20:29:42+00:00 2 Answers 5 views 0

Answers ( )

    0
    2021-10-27T20:31:14+00:00

    Đáp án:

    $GTNN$ $của$ $P$ $là$ $14$ $tại$ $x = 1; y = 0$

    Giải thích các bước giải:

    $P=\frac{2x + 8√x + 17}{√x + 2} + \frac{3y + 6√y + 5}{√y + 1}$

    $P=\frac{(2x + 8√x + 8) + 9}{√x + 2} + \frac{(3y + 6√y + 3) + 2}{√y + 1}$

    $P= \frac{2(x + 4√x + 4)+9}{√x+2} + \frac{3(y + 2√y + 1) + 2}{√y +1}$

    $P = \frac{2(√x + 2)² + 9}{√x + 2} + \frac{3(√y + 1)² + 2}{√y +1}$

    $Đặt$ $a = √x + 2 ; b = √y + 1$

    $ĐK : a \geq 2 ; b \geq 1$ $thỏa$ $mãn$ $a+b \geq4$$(vì √x +√y \geq1)$

    $Khi$ $đó$ $ta$ $có:$

    $P=\frac{2a² + 9}{a} + \frac{3b² + 2}{b}$

    $P=\frac{2a²}{a} + \frac{9}{a} + \frac{3b²}{b} +\frac{2}{b}$

    $P = 2a + \frac{9}{a} + 3b + \frac{2}{b}$

    $P = (a + b) + (a + \frac{9}{a}) + (2b + \frac{2}{b})$

    $Theo$ $BĐT$ $Cauchy$ $ta$ $có:$

    $a + \frac{9}{a} \geq 2√(a.\frac{9}{a}) = 2.3=6$

    $2b + \frac{2}{b} \geq 2√(2b.\frac{2}{b}) = 2.2 =4$

    $→ P \geq 4 + 6 + 4 = 14$

    $Dấu$ $”=”$ $xảy$ $ra$ $⇒$ $\left \{ {a + b = 4} \atop {a = \frac{9}{a} \atop {2b = \frac{2}{b}}}\right.⇒\left \{ {{a = 3} \atop {b = 1}} \right. ⇒\left \{ {{x = 1} \atop {y = 0}} \right.$

    $Vậy$ $GTNN$ $của$ $P$ $là$ $14$ $tại$ $x = 1; y = 0$

    0
    2021-10-27T20:31:26+00:00

    Đáp án:

    `min_P=14<=>x=1,y=0`

    Giải thích các bước giải:

    `P=(2x+8\sqrt{x}+17)/(\sqrt{x}+2)+(3y+6\sqrt{y}+5)/(\sqrt{y}+1)`

    `=(2(\sqrt{x}+2)^2+9)/(\sqrt{x}+2)+(3(\sqrt{y}+1)^2+2)/(\sqrt{y}+1)`

    Đặt `a=\sqrt{x}+2,b=\sqrt{y}+1(a,b>0)`

    `=>P=(2a^2+9)/a+(3b^2+2)/b`

    `=2a+9/a+3b+2/b`

    `=a+b+a+9/a+2b+2/b`

    `=4+a+9/a+2b+2/b`

    BĐT AM-GM:

    `=>a+9/a>=6`

    `=>2b+2/b>=4`

    `=>P>=4+6+4=14`

    Dấu “=” xảy ra khi `a+b=4,a^2=9,b^2=1`

    `<=>a=3,b=1`

    `<=>x=1,y=0`

    Vậy `min_P=14<=>x=1,y=0`

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