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a) \(A=\left(\frac{1}{\sqrt{x}+3}-\frac{4}{9-x}\right).\frac{2\sqrt{x}-6}{\sqrt{x}+1}\)
\(A=\left[\frac{\sqrt{x}-3}{x-9}+\frac{4}{x-9}\right].\frac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}\)
\(A=\frac{\sqrt{x}-3+4}{x-9}.\frac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}\)
\(A=\frac{\sqrt{x}+1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}\)
\(A=\frac{2}{\sqrt{x}+3}\)
vậy \(A=\frac{2}{\sqrt{x}+3}\)
a, Ta có : \(x=\sqrt{3+2\sqrt{2}}+\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}=4\)
Thay x = 4 => \(\sqrt{x}=2\) vào B ta được :
\(B=\frac{2+5}{2-3}=-7\)
b, Ta có : Với \(x\ge0;x\ne9\)
\(A=\frac{4}{\sqrt{x}+3}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\frac{4\left(\sqrt{x}-3\right)+2x-\sqrt{x}-13-\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}\)
\(=\frac{4\sqrt{x}-12+2x-\sqrt{x}-13-x-3\sqrt{x}}{x-9}=\frac{x-25}{x-9}\)
Lại có \(P=\frac{A}{B}\Rightarrow P=\frac{\frac{x-25}{x-9}}{\frac{\sqrt{x}+5}{\sqrt{x}-3}}=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)
\(a,B=\frac{10\sqrt{x}+12+\sqrt{x}\left(\sqrt{x}-3\right)-\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{x+6\sqrt{x}+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}+3}{\sqrt{x}-3}\)
\(b,C=\frac{x-1}{\sqrt{x}-3}:\frac{\sqrt{x}+3}{\sqrt{x}-3}=\frac{x-1}{\sqrt{x}+3}\)
Vì\(\hept{\begin{cases}x\ge0\\\sqrt{x}+3>0\end{cases}\Rightarrow}x-1\ge-1\)
\(\Rightarrow C_{min}=-1\Leftrightarrow x=0\)
Vậy................
Với x = 0 thì C = -1/3 chứ có phải là -1 đâu .
b)
Ta có: \(C=\frac{x-1}{\sqrt{x}+3}=\sqrt{x}-3+\frac{8}{\sqrt{x}+3}=\left(\sqrt{x}+3+\frac{9}{\sqrt{x}+3}\right)-6-\frac{1}{\sqrt{x}+3}\)
\(\ge2\sqrt{\left(\sqrt{x}+3\right).\frac{9}{\sqrt{x}+3}}-6-\frac{1}{3}=-\frac{1}{3}\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}\sqrt{x}+3=\frac{9}{\sqrt{x}+3}\\x=0\end{cases}}\Leftrightarrow x=0\)
Vậy min C = -1/3 tại x =0
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 =\(\frac{7-4\sqrt{3}+16}{\sqrt{7-4\sqrt{3}}+3}\) =\(\frac{23-4\sqrt{3}}{\sqrt{\left(2-\sqrt{3}\right)^2+3}}\)
=\(\frac{23-4\sqrt{3}}{2-\sqrt{3}+3}\)
=\(\frac{23-4\sqrt{3}}{5-\sqrt{3}}\)
a, Q = \(\left(\frac{\sqrt{x}-3}{\sqrt{x}+3}+\frac{\sqrt{x}+3}{\sqrt{x}-3}-\frac{14}{9-x}\right)\times\frac{\sqrt{x}-3}{2}\)
= \(\left[\frac{\left(\sqrt{x}-3\right)^2+\left(\sqrt{x}+3\right)^2+14}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]\times\frac{\sqrt{x}-3}{2}\)
= \(\left[\frac{x-6\sqrt{x}+9+x+6\sqrt{x}+9+14}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]\times\frac{\sqrt{x}-3}{2}\)
= \(\frac{2x+32}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\times\frac{\sqrt{x}-3}{2}\)
= \(\frac{2\left(x+16\right)\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
= \(\frac{x+16}{\sqrt{x}+3}\)
Thay \(x=7-4\sqrt{3}\) vào Q ta được:
Q= \(\frac{7-4\sqrt{3}+16}{\sqrt{7-4\sqrt{3}}+3}\) = \(\frac{23-4\sqrt{3}}{\sqrt{\left(2-\sqrt{3}\right)^2}+3}\)=\(\frac{23-4\sqrt{3}}{2+3-\sqrt{3}}=\frac{23-4\sqrt{3}}{5-\sqrt{3}}=\frac{\left(23-4\sqrt{3}\right)\left(5+\sqrt{3}\right)}{\left(5+\sqrt{3}\right)\left(5-\sqrt{3}\right)}\) =\(\frac{103+3\sqrt{3}}{22}\)
b,
\(Q=\frac{x+16}{\sqrt{x}+3}=\frac{x+9+7}{\sqrt{x}+3}=2+\frac{7}{\sqrt{x}+3}\)
Ta có \(2+\frac{7}{\sqrt{x}+3}\) nhỏ nhất khi \(\sqrt{x}+3\) nhỏ nhất
Mà với điều kiện \(x\ge0\) nên GTNNQ=\(2+\frac{7}{3}=\frac{13}{3}\)
a: \(P=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\cdot\dfrac{\sqrt{x}-7+\sqrt{x}+1}{\sqrt{x}+1}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\cdot\dfrac{2\sqrt{x}-6}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}-3}{\sqrt{x}+1}\cdot\dfrac{2}{\sqrt{x}+3}=-\dfrac{6}{\sqrt{x}+3}\)
b: P>=-1/2
=>P+1/2>=0
=>\(\dfrac{-6}{\sqrt{x}+3}+\dfrac{1}{2}>=0\)
=>\(\dfrac{-12+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}>=0\)
=>căn x-9>=0
=>x>=81
c: căn x+3>=3
=>6/căn x+3<=6/3=2
=>-6/căn x+3>=-2
Dấu = xảy ra khi x=0
Bổ sung điều kiện là x>0
a/\(P=\frac{x}{\sqrt{x}+3}+\frac{\left(\sqrt{x}-3\right)\left(5\sqrt{x}-6\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{x+5\sqrt{x}-6}{\sqrt{x}+3}\)
b/\(x=9-4\sqrt{5}=\left(\sqrt{5}\right)^2-2.2.\sqrt{5}+2^2=\left(\sqrt{5}-2\right)^2\Rightarrow\sqrt{x}=\sqrt{5}-2\)
Thay vào ta được \(P=\frac{9-4\sqrt{5}+5\sqrt{5}-10-6}{\sqrt{5}-2+3}=\frac{\sqrt{5}-7}{\sqrt{5}+1}\)