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24 tháng 6 2021

a) đk: x\(\ge0\);

P = \(\left[\dfrac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}+1}\right].\dfrac{4\sqrt{x}}{3}\)

\(\dfrac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\dfrac{4\sqrt{x}}{3}\)

\(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}.\dfrac{4\sqrt{x}}{3}=\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)

b) Để P = \(\dfrac{8}{9}\)

<=> \(\dfrac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\dfrac{8}{9}\)

<=> \(\dfrac{\sqrt{x}}{x-\sqrt{x}+1}=\dfrac{2}{3}\)

<=> \(\dfrac{3\sqrt{x}-2x+2\sqrt{x}-2}{3\left(x-\sqrt{x}+1\right)}=0\)

<=> \(-2x+5\sqrt{x}-2=0\)

<=> \(\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)

<=> \(\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{1}{4}\left(tm\right)\end{matrix}\right.\)

c)

Đặt \(\sqrt{x}=a\) (\(a\ge0\))

P = \(\dfrac{4a}{3\left(a^2-a+1\right)}\)

Xét P + \(\dfrac{4}{9}\) = \(\dfrac{4a}{3a^2-3a+3}+\dfrac{4}{9}=\dfrac{12a+4a^2-4a+4}{9\left(a^2-a+1\right)}=\dfrac{4a^2+8a+4}{9\left(a^2-a+1\right)}=\dfrac{4\left(a+1\right)^2}{9\left(a^2-a+1\right)}\ge0\)

Dấu "=" <=> a = -1 (loại)

=> Không tìm được Min của P

Xét P - \(\dfrac{4}{3}\) = \(\dfrac{4a}{3\left(a^2-a+1\right)}-\dfrac{4}{3}=\dfrac{4a-4a^2+4a-4}{3\left(a^2-a+1\right)}=\dfrac{-4a^2+8a-4}{3\left(a^2-a+1\right)}=\dfrac{-4\left(a-1\right)^2}{3\left(a^2-a+1\right)}\le0\)

<=> \(P\le\dfrac{4}{3}\)

Dấu "=" <=> a = 1 <=> x = 1 (tm)

24 tháng 6 2021

Ai bảo cậu là không tìm được minP vậy?

9 tháng 8 2017

\(A=\sqrt{x-1}+\sqrt{8-x}-2\sqrt{-x^2+9x-8}\)

\(=\left(\sqrt{x-1}-\sqrt{8-x}\right)^2\ge0\)

Đẳng thức xảy ra \(\Leftrightarrow\sqrt{x-1}=\sqrt{8-x}\Leftrightarrow x=\dfrac{9}{2}\)

Vậy \(A_{min}=0\Leftrightarrow x=\dfrac{9}{2}\)

9 tháng 8 2017

ấy chết nhầm to

7 tháng 8 2016

diều kiện x >= 0

P=\(\left(\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right).\frac{4\sqrt{x}}{3}\)

\(\frac{x+2-x+\sqrt{x}-1}{x\sqrt{x}+1}.\frac{4\sqrt{x}}{3}\)

=\(\frac{\sqrt{x}+1}{x\sqrt{x}+1}.\frac{4\sqrt{x}}{3}\)=\(\frac{4\sqrt{x}}{3x-3\sqrt{x}+3}\)

P=8/9

<=> \(\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\frac{8}{9}\)

<=> \(3\sqrt{x}=2x-2\sqrt{x}+1\)

<=> \(2x-5\sqrt{x}+2=0\)

<=> \(\left[\begin{array}{nghiempt}x=4\\x=\frac{1}{4}\end{array}\right.\)

vậy x=4 hoặc x=1/4 thì p=8/9

 

 

7 tháng 8 2016

a) \(P=\left(\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\right)\cdot\frac{4\sqrt{x}}{3}\left(ĐK:x\ge0;x\ne-1\right)\)

\(=\left[\frac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}+1}\right]\cdot\frac{4\sqrt{x}}{3}\)

\(=\frac{x+2-x+\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\frac{4\sqrt{x}}{3}\)

\(=\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\cdot\frac{4\sqrt{x}}{3}\)

\(=\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}\)

b) Để P=8/9

\(\Leftrightarrow\)\(\frac{4\sqrt{x}}{3\left(x-\sqrt{x}+1\right)}=\frac{8}{9}\)

\(\Leftrightarrow24\left(x-\sqrt{x}+1\right)=36\sqrt{x}\)

\(\Leftrightarrow24x-24\sqrt{x}+24-36\sqrt{x}=0\)

\(\Leftrightarrow24x-60\sqrt{x}+24=0\)

\(\Leftrightarrow12\left(2x-5\sqrt{x}+2\right)=0\)

\(\Leftrightarrow\left(2x-\sqrt{x}\right)-\left(4\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\sqrt{x}\left(2\sqrt{x}-1\right)-2\left(2\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}2\sqrt{x}-1=0\\\sqrt{x}-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}\sqrt{x}=\frac{1}{2}\\\sqrt{x}=2\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{1}{4}\left(tm\right)\\x=4\left(tm\right)\end{array}\right.\)

15 tháng 8 2018

sử dụng bđt     \(\hept{\begin{cases}\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\\\sqrt{a}-\sqrt{b}\le\sqrt{a-b}\end{cases}}\)            

cái trên bđt xảy ra khi a=0 hoặc b=0

cái dưới xảy ra khi a=b hoặc b=0

\(B\ge\sqrt{x-5+13-x}\ge\sqrt{8}\)

dấu ''='' xảy ra khi \(\orbr{\begin{cases}x=5\\x=13\end{cases}}\)

\(C\le\sqrt{x-1-x+8}\le\sqrt{7}\)

dấu ''='' xảy ra khi 

\(x=8\)

D ,tương tự a

6 tháng 10 2020

Bạn nguyễn thị lan hương sai maxC rồi nhé, mình chỉ bổ sung phần còn lại 

  \(B\le\sqrt{\left(1^2+1^2\right)\left(x-5+13-x\right)}=4\)(Bunhiacopski)  Dấu bằng xảy ra khi  x=9

Tìm maxD cũng vậy  

Ta có: \(P=\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{9x-1}\right):\left(1-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\)

\(=\dfrac{3x+\sqrt{x}-3\sqrt{x}-1-3\sqrt{x}+1+8\sqrt{x}}{\left(3\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)}:\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}\)

\(=\dfrac{3x+3\sqrt{x}}{3\sqrt{x}-1}\cdot\dfrac{1}{3}\)

\(=\dfrac{x+\sqrt{x}}{3\sqrt{x}-1}\)

20 tháng 8 2021

\(\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}-1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{9x-1}\right):\left(1-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)=\dfrac{\left(\sqrt{x}-1\right)\left(3\sqrt{x}+1\right)-\left(3\sqrt{x}-1\right)+8\sqrt{x}}{9x-1}:\dfrac{3\sqrt{x}+1-3\sqrt{x}+2}{3\sqrt{x}+1}=\dfrac{3x+3\sqrt{x}-1}{9x-1}.\dfrac{3\sqrt{x}+1}{3}=\dfrac{3x+3\sqrt{x}-1}{9\sqrt{x}-3}\)

21 tháng 7 2017

*)Minimize : Áp dụng BĐT \(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\) ta có:

\(M=\sqrt{x-1}+\sqrt{y+3}\)

\(\ge\sqrt{x-1+y+3}=\sqrt{x+y+2}=\sqrt{10}\)

Xảy ra khi \(x=1;y=7\)

*)Maximize: Áp dụng BĐT Cauchy-Schwarz ta có:

\(M^2=\left(\sqrt{x-1}+\sqrt{y+3}\right)^2\)

\(\le\left(1+1\right)\left(x-1+y+3\right)\)

\(=2\left(x+y+2\right)=2\cdot\left(8+2\right)=20\)

\(\Rightarrow M^2\le20\Rightarrow M\le\sqrt{20}\)

Xảy ra khi \(x=6;y=2\)

ghi rõ phần min dc ko bạn

với lại bđt đó xảy ra dấu = khi nào thế

28 tháng 7 2021

\(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\left(x\ge0,x\ne1\right)\)

\(=\left(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\right).\dfrac{2}{\sqrt{x}-1}\)

\(=\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)

\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)

\(=\dfrac{2}{x+\sqrt{x}+1}\)

b) Ta có: \(x\ge0\Rightarrow x+\sqrt{x}+1\ge1\Rightarrow\dfrac{2}{x+\sqrt{x}+1}\le2\)

\(\Rightarrow max=2\) khi \(x=0\)

Ta có: \(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)

\(=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)

\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\cdot\dfrac{2}{x+\sqrt{x}+1}\)

\(=\dfrac{2}{x+\sqrt{x}+1}\)

11 tháng 7 2019

\(đkxđ\Leftrightarrow x\ge4\)

\(P=\frac{\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}}{\sqrt{\frac{16}{x^2}-\frac{8}{x}+1}}\)

\(=\frac{\sqrt{x-4+4\sqrt{x-4}+4}+\sqrt{x-4-4\sqrt{x-4}+4}}{\sqrt{\frac{4^2}{x^2}-2.\frac{4}{x}+1}}\)

\(=\frac{\sqrt{\left(x-4+2\right)^2}+\sqrt{\left(x-4-2\right)^2}}{\sqrt{\left(\frac{4}{x}-1\right)^2}}\)

\(=\frac{|x-2|+|x-6|}{|\frac{4}{x}-1|}=\frac{x-2+|x-6|}{|\frac{4}{x}-1|}\)

Dùng bảng xét dấu nha