Cho \(P=\frac{1}{\sqrt{x}+1}\)
Tính minA=\(p+\frac{\sqrt{x}+19}{9}\)
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11 tháng 4 2020
ĐK : \(x\ge0\)
Ta có :
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{1}{x+\sqrt{x}}\right).\left(\frac{1}{\sqrt{x}+1}+\frac{1}{x-1}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)\(.\left(\frac{1}{\sqrt{x}+1}+\frac{1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\frac{x-1}{\sqrt{x}\left(\sqrt{x}+1\right)}.\frac{\sqrt{x}-1+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(x-1\right)\sqrt{x}}{\sqrt{x}\left(x-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{1}{\sqrt{x}+1}\)
Vậy ta có
\(A=\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}+19}{9}\)
\(=\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}+1}{9}+\frac{18}{9}\)
\(=\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}+1}{9}+2\)
Áp dụng BĐT Cauchy ta có
\(\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}+1}{9}\ge2\sqrt{\frac{1}{\sqrt{x}+1}.\frac{\sqrt{x}+1}{9}}=\frac{2}{3}\)
\(\Leftrightarrow\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}+1}{9}+2\ge\frac{8}{3}\)
\(\Leftrightarrow A\ge\frac{8}{3}\)
Dấu "=" xảy ra khi
\(\frac{1}{\sqrt{x}+1}=\frac{\sqrt{x}+1}{9}\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)^2=9\)
\(\Leftrightarrow\sqrt{x}+1=3\)
\(\Leftrightarrow\sqrt{x}=2\)
\(\Leftrightarrow x=4\)
Vậy GTNN của A là \(\frac{8}{3}\) đạt được khi x = 4
15 tháng 7 2019
\(E=\left(\frac{\sqrt{\sqrt{x}-1}}{\sqrt{\sqrt{x}+1}}+\frac{\sqrt{\sqrt{x}+1}}{\sqrt{\sqrt{x}-1}}\right):\sqrt{\frac{1}{x-1}}\) \(ĐKXĐ:x>1\)
\(E=\left(\frac{\left(\sqrt{\sqrt{x}-1}\right)^2}{\left(\sqrt{\sqrt{x}+1}\right)\left(\sqrt{\sqrt{x}-1}\right)}+\frac{\left(\sqrt{\sqrt{x}+1}\right)^2}{\left(\sqrt{\sqrt{x}-1}\right)\left(\sqrt{\sqrt{x}+1}\right)}\right)\cdot\sqrt{\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{1}}\)
\(E=\left(\frac{\sqrt{x}-1}{\left(\sqrt{\sqrt{x}+1}\right)\left(\sqrt{\sqrt{x}-1}\right)}+\frac{\sqrt{x}+1}{\left(\sqrt{\sqrt{x}-1}\right)\left(\sqrt{\sqrt{x}+1}\right)}\right)\cdot\sqrt{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(E=\frac{\sqrt{x}-1+\sqrt{x}+1}{\left(\sqrt{\sqrt{x}+1}\right)\left(\sqrt{\sqrt{x}-1}\right)}\cdot\sqrt{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(E=\frac{2\sqrt{x}}{\sqrt{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}}\cdot\sqrt{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=2\sqrt{x}\)
Ta có:\(x=19-8\sqrt{3}=16-2.4\sqrt{3}+3=\left(4-\sqrt{3}\right)^2\)
\(\Rightarrow2\sqrt{x}=2.\sqrt{\left(4-\sqrt{3}\right)^2}=2.\left(4-\sqrt{3}\right)=8-2\sqrt{3}\)
21 tháng 2 2022
a: \(P=\left(\dfrac{2\sqrt{x}}{\left(x+1\right)\left(\sqrt{x}+1\right)}+\dfrac{1}{\sqrt{x}+1}\right):\dfrac{x+1+\sqrt{x}}{x+1}\)
\(=\dfrac{2\sqrt{x}+x+1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\cdot\dfrac{x+1}{x+\sqrt{x}+1}=\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
b: Thay \(x=9+2\sqrt{7}\) vào P, ta được:
\(P=\dfrac{\sqrt{9+2\sqrt{7}}+1}{9+2\sqrt{7}+\sqrt{9+2\sqrt{7}+1}}\simeq0,25\)
23 tháng 5 2021
Mình ghi nhầm. \(x=\frac{\sqrt{4+2\sqrt{3}}.\left(\sqrt{3}-1\right)}{\sqrt{6+2\sqrt{5}}-\sqrt{5}}\)nhé
9 tháng 9 2018
ta có:\(\frac{x-2\sqrt{x}+1}{x-\sqrt{x}+1}=\frac{1}{2}\)
\(\Rightarrow x-3\sqrt{x}+1=0\)
\(\Rightarrow\hept{\begin{cases}x+1=3\sqrt{x}\\x-3\sqrt{x}=-1\end{cases}}\)
lại có \(B=\frac{3x\sqrt{x}+10x+19}{x^2+7x+15}\)
\(=\frac{3x\sqrt{x}-9x+19x+19}{x^2-9x+16x+15}\)
\(=\frac{3\sqrt{x}\left(x-3\sqrt{x}\right)+19\left(x+1\right)}{\left(x+3\sqrt{x}\right)\left(x-3\sqrt{x}\right)+16x+15}\)
\(=\frac{-3\sqrt{x}+19\times3\sqrt{x}}{-1\times\left(x+3\sqrt{x}\right)+16x+15}\)
\(=\frac{57\sqrt{x}-3\sqrt{x}}{15x+15-3\sqrt{x}}\)
\(=\frac{54\sqrt{x}}{15\left(x+1\right)-3\sqrt{x}}\)
\(=\frac{54\sqrt{x}}{45\sqrt{x}-3\sqrt{x}}\)
\(=\frac{54\sqrt{x}}{42\sqrt{x}}=\frac{27}{21}\)
TB
27 tháng 7 2018
a, \(A=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}+\frac{1}{1-\sqrt{x}}\) (ĐKXĐ: \(x\ne1,x\ge0\))
\(=\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)
\(=\frac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\sqrt{x}}{x+\sqrt{x}+1}\)
b, \(A-\frac{1}{3}\Leftrightarrow\frac{\sqrt{x}}{x+\sqrt{x}+1}-\frac{1}{3}\)\(=\frac{3\sqrt{x}-x-\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}=\frac{-x+2\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}=-\frac{-\left(x-2\sqrt{x}+1\right)}{3\left(x+\sqrt{x}+1\right)}=-\frac{\left(\sqrt{x}+1\right)^2}{3\left(x+\sqrt{x}+1\right)}< 0\)
\(\Rightarrow A-\frac{1}{3}< 0\Leftrightarrow A< \frac{1}{3}\)
c, ĐKXĐ: \(x\ge0,x\ne1\)
Ta có: x = \(19-8\sqrt{3}\)(TMĐK) \(\Leftrightarrow\sqrt{x}=\sqrt{19-8\sqrt{3}}\Leftrightarrow\sqrt{x}=\sqrt{\left(4-\sqrt{3}\right)^2}\Leftrightarrow\sqrt{x}=4-\sqrt{3}\)
Thay \(\sqrt{x}=4-\sqrt{3}\)vào A ta có:
\(A=\frac{4-\sqrt{3}}{\left(4-\sqrt{3}\right)^2+4-\sqrt{3}+1}=\frac{4-\sqrt{3}}{19-8\sqrt{3}+4-\sqrt{3}+1}=\frac{4-\sqrt{3}}{24-9\sqrt{3}}\)
Vậy với \(x=19-8\sqrt{3}\)thì \(A=\frac{4-\sqrt{3}}{24-9\sqrt{3}}\)
bài này chỉ ngồi mò được điểm rơi là xong
Áp dụng bất đẳng thức AM-GM có ;
\(\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}+19}{9}=\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}+1}{9}+\frac{18}{9}\ge2\sqrt{\frac{\sqrt{x}+1}{\left(\sqrt{x}+1\right).9}}+2\)
\(=2.\sqrt{\frac{1}{9}}+2=2.\frac{1}{3}+2=\frac{2}{3}+2=\frac{8}{3}\)
Dấu "=" xảy ra khi và chỉ khi \(x=4\)
Vậy Min A = 8/3 khi x = 4
bài này mình không kiếm được điểm rơi nên mình đoán bừa nhé , nếu sai thì nhờ cao thủ nào đó đến cứu =))))))))