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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=\(2+\frac{7}{3}=\frac{13}{3}\)
\(A=\frac{\left(x-9\right)+25}{\sqrt{x}+3}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+25}{\sqrt{x}+3}=\sqrt{x}-3+\frac{25}{\sqrt{x}+3}\)\(=\left(\sqrt{x}+3\right)+\frac{25}{\sqrt{x}+3}-6\ge2\sqrt{\left(\sqrt{x}+3\right).\frac{25}{\sqrt{x}+3}}-6=2.5-4=6\)
Dấu'=' xảy ra khi và chỉ khi \(\sqrt{x}+3=\frac{25}{\sqrt{x}+3}\)
\(\Rightarrow\left(\sqrt{x}+3\right)^2=25\Rightarrow\sqrt{x}+3=5\left(do\sqrt{x}+3>0\right)\Rightarrow\sqrt{x}=2\Rightarrow x=4\)
Vậy MinA=4 khi và chỉ khi x=4
\( a)A = \dfrac{{a - \sqrt a - 6}}{{4 - a}} - \dfrac{1}{{\sqrt a - 2}}\\ A = \dfrac{{a + 2\sqrt a - 3\sqrt a - 6}}{{\left( {2 - \sqrt a } \right)\left( {2 + \sqrt a } \right)}} - \dfrac{1}{{\sqrt a - 2}}\\ A = \dfrac{{\left( {\sqrt a + 2} \right)\left( {\sqrt a - 3} \right)}}{{\left( {2 - \sqrt a } \right)\left( {2 + \sqrt a } \right)}} - \dfrac{1}{{\sqrt a - 2}}\\ A = - \dfrac{{\sqrt a - 3}}{{\sqrt a - 2}} - \dfrac{1}{{\sqrt a - 2}}\\ A = - \dfrac{{\sqrt a - 2}}{{\sqrt a - 2}} = - 1 \)
\( b)B = \dfrac{1}{{\sqrt x - 1}} + \dfrac{1}{{\sqrt x + 1}} - \dfrac{2}{{x - 1}}\\ B = \dfrac{1}{{\sqrt x - 1}} + \dfrac{1}{{\sqrt x + 1}} - \dfrac{2}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\ B = \dfrac{{\sqrt x + 1 + \sqrt x - 1 - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\ B = \dfrac{{2\sqrt x - 2}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\ B = \dfrac{{2\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}} = \dfrac{2}{{\sqrt x + 1}} \)
a) Thay x = 25 vào biểu thức A , ta có
\(A=\frac{5-2}{5-1}=\frac{3}{4}\)
b) \(B=\frac{x-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{4\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B =\frac{x+1+2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B =\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
a, Ta có : \(x=25\Rightarrow\sqrt{x}=5\)
Thay vào biểu thức A ta được :
\(A=\frac{5-2}{5-1}=\frac{3}{4}\)
Vậy với x = 25 thì A = 3/4
b, Với \(x\ge0;x\ne1\)
\(B=\frac{x-5}{x-1}-\frac{2}{\sqrt{x}+1}+\frac{4}{\sqrt{x}-1}\)
\(=\frac{x-5-2\left(\sqrt{x}-1\right)+4\left(\sqrt{x}+1\right)}{x-1}=\frac{x-5-2\sqrt{x}+2+4\sqrt{x}+4}{x-1}\)
\(=\frac{x+1+2\sqrt{x}}{x-1}=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}\pm1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
c, Ta có P = A/B hay \(P=\frac{\sqrt{x}-2}{\sqrt{x}-1}.\frac{\sqrt{x}-1}{\sqrt{x}+1}=\frac{\sqrt{x}-2}{\sqrt{x}+1}\)
\(\sqrt{P}< \frac{1}{2}\)hay \(\sqrt{\frac{\sqrt{x}-2}{\sqrt{x}+1}}< \frac{1}{2}\Rightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}< \frac{1}{4}\)
\(\Leftrightarrow\frac{\sqrt{x}-2}{\sqrt{x}+1}-\frac{1}{4}< 0\Leftrightarrow\frac{4\sqrt{x}-8-\sqrt{x}-1}{4\left(\sqrt{x}+1\right)}< 0\)
\(\Rightarrow3\sqrt{x}-9>0\)do \(4\left(\sqrt{x}+1\right)>0\)
\(\Leftrightarrow3\sqrt{x}>9\Leftrightarrow\sqrt{x}>3\Leftrightarrow x>9\)
\(\left[\frac{\left(1-\sqrt{x}\right)\left(x+\sqrt{x}+1\right)}{1-\sqrt{x}}\right]\left[\frac{1-\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\right]^2=\left(x+\sqrt{x}+1\right)\frac{1}{\left(1+\sqrt{x}\right)^2}=\frac{x+\sqrt{x}+1}{x+2\sqrt{x}+1}\)
Đề bài sai
\(\sqrt{2012}-\sqrt{2011}=\frac{1}{\sqrt{2012}+\sqrt{2011}}\)
\(\sqrt{2011}-\sqrt{2010}=\frac{1}{\sqrt{2011}+\sqrt{2010}}\)
Do \(\sqrt{2012}>\sqrt{2010}\) \(\Rightarrow\sqrt{2012}+\sqrt{2011}>\sqrt{2011}+\sqrt{2010}>0\)
\(\Rightarrow\frac{1}{\sqrt{2012}+\sqrt{2011}}< \frac{1}{\sqrt{2011}+\sqrt{2010}}\Rightarrow\sqrt{2012}-\sqrt{2011}< \sqrt{2011}-\sqrt{2010}\)
\(A=\frac{x+2\sqrt{xy}+y-4\sqrt{xy}}{\sqrt{x}-\sqrt{y}}+\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}=\frac{\left(\sqrt{x}-\sqrt{y}\right)^2}{\sqrt{x}-\sqrt{y}}+\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}\)
\(=\sqrt{x}-\sqrt{y}+\sqrt{x}-\sqrt{y}=2\sqrt{x}-2\sqrt{y}\)
\(M^2=\left(\sqrt{x-1}+\sqrt{9-x}\right)^2\le2\left(x-1+9-x\right)=16\)
\(\Rightarrow M\le4\Rightarrow M_{max}=4\) khi \(x-1=9-x\Leftrightarrow x=5\)
\(A=\frac{\sqrt{x}+1+x}{\sqrt{x}}=\frac{1}{\sqrt{x}}+1+\sqrt{x}\ge1+2\sqrt{\frac{1}{\sqrt{x}}.\sqrt{x}}=1+2=3\)
dấu "=" xảy ra khi và chỉ khi \(x=1\)
\(MIN:A=3\)
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