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LM
0
6 tháng 2 2022
a: \(P=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\dfrac{1}{\sqrt{x}+1}\right):\dfrac{x+1+\sqrt{x}}{x+1}\)
\(=\dfrac{x+2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\cdot\dfrac{x+1}{x+\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\)
NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 10 2019
ĐKXĐ: ...
\(P=\left(\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\frac{2x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{-\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{\left(5-\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{-\sqrt{x}}{\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(5-\sqrt{x}\right)}=\frac{x}{\sqrt{x}-5}\)
\(x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\Rightarrow x=\sqrt{5}-1\)
\(\Rightarrow P=\frac{6-2\sqrt{5}}{\sqrt{5}-6}=...\)
\(P=\sqrt{x}+5+\frac{25}{\sqrt{x}-5}=\sqrt{x}-5+\frac{25}{\sqrt{x}-5}+10\)
\(\Rightarrow P\ge2\sqrt{\frac{25\left(\sqrt{x}-5\right)}{\sqrt{x}-5}}+10=20\)
\(\Rightarrow P_{min}=20\) khi \(x=100\)
\(P< \sqrt{x}\Rightarrow\frac{x}{\sqrt{x}-5}< \sqrt{x}\Rightarrow\frac{\sqrt{x}}{\sqrt{x}-5}< 1\Rightarrow\frac{\sqrt{x}}{\sqrt{x}-5}-1< 0\)
\(\Rightarrow\frac{5}{\sqrt{x}-5}< 0\Rightarrow\sqrt{x}-5< 0\Rightarrow x< 25\)
Kết hợp ĐKXĐ \(\Rightarrow\left\{{}\begin{matrix}0< x< 25\\x\ne3\\\end{matrix}\right.\)
\(Q=\frac{\sqrt{x}}{\sqrt{x}-5}=1+\frac{5}{\sqrt{x}-5}\)
Để Q nguyên \(\Rightarrow\sqrt{x}-5=Ư\left(5\right)=\left\{-5;-1;1;5\right\}\)
\(\Rightarrow\sqrt{x}=\left\{0;4;6;10\right\}\Rightarrow x=\left\{16;36;100\right\}\)
HH
16 tháng 8 2016
a) Rút gọn : Q =\(\left(\frac{\sqrt{x}-3}{\sqrt{x}+3}+\frac{\sqrt{x}+3}{\sqrt{x}-3}-\frac{14}{9-x}\right).\frac{\sqrt{x}-3}{2}\left(x\ge0,x\ne9\right)\)
Q =\(\left(\frac{\sqrt{x}-3}{\sqrt{x}+3}+\frac{\sqrt{x}+3}{\sqrt{x}-3}+\frac{14}{x-9}\right).\frac{\sqrt{x}-3}{2}\)
Q =\(\left(\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{14}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right).\frac{\sqrt{x}-3}{2}\)
Q = \(\frac{x-6\sqrt{x}+9+x+6\sqrt{x}+9+14}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{2}\)
Q = \(\frac{2x+32}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{2}\)
Q = \(\frac{2\left(x+16\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{2}\)
Q = \(\frac{x+16}{\sqrt{x}+3}\)
thay \(x=7-4\sqrt{3}\) vào Q ta được
Q =\(\frac{7-4\sqrt{3}+16}{\sqrt{7-4\sqrt{3}}+3}\) =\(\frac{23-4\sqrt{3}}{\sqrt{\left(2-\sqrt{3}\right)^2+3}}\)
=\(\frac{23-4\sqrt{3}}{2-\sqrt{3}+3}\)
=\(\frac{23-4\sqrt{3}}{5-\sqrt{3}}\)
Với x > 9
\(C=\frac{x}{\sqrt{x}-3}=\frac{x-6\sqrt{x}+9+6\sqrt{x}-18-9+18}{\sqrt{x}-3}\)
\(=\frac{\left(\sqrt{x}-3\right)^2+6\left(\sqrt{x}-3\right)+9}{\sqrt{x}-3}\)
\(=\sqrt{x}-3+\frac{9}{\sqrt{x}-3}+6\ge2\sqrt{9}+6=12\) ( AM - GM cho 2 số không âm )
Dấu "=" xảy ra <=> \(\sqrt{x}-3=\frac{9}{\sqrt{x}-3}\Leftrightarrow x=36\)thỏa mãn
Vậy min c = 9 đạt tại x = 36.