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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) 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=\(2+\frac{7}{3}=\frac{13}{3}\)
Mới đc câu a ak, thog cảm nha, trih độ mih thấp lắm:
\(\frac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\frac{2b}{a-b}\)
=\(\frac{a+\sqrt{ab}-\sqrt{ab}+b}{a-b}-\frac{2b}{a-b}\)
=\(\frac{a+b-2b}{a-b}=\frac{a-b}{a-b}=1\)
a) Ta có: \(\frac{a-b}{\sqrt{a}-\sqrt{b}}-\frac{\sqrt{a^3}-\sqrt{b^3}}{a-b}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)
\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\sqrt{a}+\sqrt{b}}-\frac{a+\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{a+2\sqrt{ab}+b-a-\sqrt{ab}-b}{\sqrt{a}+\sqrt{b}}\)
\(=\frac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\)
b)Sửa đề: \(\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
Ta có: \(\frac{\left(\sqrt{a}+\sqrt{b}\right)^2-4\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)}-\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}\)
\(=\sqrt{a}-\sqrt{b}-\sqrt{a}-\sqrt{b}\)
\(=-2\sqrt{b}\)
c) Ta có: \(\left(\frac{1}{\sqrt{a}-1}-\frac{1}{\sqrt{a}}\right):\left(\frac{\sqrt{a}+1}{\sqrt{a}-2}-\frac{\sqrt{a}+2}{\sqrt{a}-1}\right)\)
\(=\left(\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\sqrt{a}-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\frac{a-1-a+4}{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-2\right)\left(\sqrt{a}-1\right)}{3}\)
\(=\frac{\sqrt{a}-2}{3\sqrt{a}}\)
d) Ta có: \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)
\(=\left(\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}{\left(\sqrt{a}+\sqrt{b}\right)}-\sqrt{ab}\right)\left(\frac{\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\right)^2\)
\(=\left(a-\sqrt{ab}+b-\sqrt{ab}\right)\cdot\left(\frac{1}{\sqrt{a}-\sqrt{b}}\right)^2\)
\(=\left(a-2\sqrt{ab}+b\right)\cdot\frac{1}{\left(\sqrt{a}-\sqrt{b}\right)^2}\)
\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}=1\)
e) Ta có: \(\left(\frac{\sqrt{x}}{3+\sqrt{x}}+\frac{x+9}{9-x}\right):\left(\frac{3\sqrt{x}+1}{x-3\sqrt{x}}-\frac{1}{\sqrt{x}}\right)\)
\(=\left(\frac{\sqrt{x}\left(3-\sqrt{x}\right)}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}+\frac{x+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}\right):\left(\frac{3\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\frac{3\sqrt{x}+9}{\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)}:\frac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\frac{3\left(\sqrt{x}+3\right)}{-\left(\sqrt{x}-3\right)\cdot\left(\sqrt{x}+3\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\left(\sqrt{x}+2\right)}\)
\(=\frac{-3\sqrt{x}}{2\sqrt{x}+4}\)
\(P=\left(\frac{3x+3}{x-9}-\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}}{3-\sqrt{x}}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right).ĐKXĐ:x\ge0,x\ne9\)
\(=\left(\frac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\frac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\left(\frac{3x+3-2x+6\sqrt{x}-x-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right):\left(\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
\(=\frac{3\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\frac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{3}{\sqrt{x}+3}\)
\(b,x=20-6\sqrt{11}=11-2.3\sqrt{11}+9\)
\(=\left(\sqrt{11}-3\right)^2\)
\(P=\frac{3}{\sqrt{x}+3}=\frac{3}{\sqrt{\left(\sqrt{11}-3\right)^2}+3}=\frac{3}{\sqrt{11}-3+3}=\frac{3\sqrt{11}}{11}\)
\(c,P>\frac{1}{2}\Rightarrow\frac{3}{\sqrt{x}+3}>\frac{1}{2}\)
\(\Leftrightarrow\frac{3}{\sqrt{x}+3}-\frac{1}{2}>0\)
\(\Leftrightarrow\frac{6-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>0\)
\(\Leftrightarrow\frac{6-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>0\)\(\Leftrightarrow\frac{3-\sqrt{x}}{2\left(\sqrt{x}+3\right)}>0\)
vì \(2\left(\sqrt{x}+3\right)>0\) (nếu x=0 =>pt vô nghiệm)
\(\Rightarrow3-\sqrt{x}>0\Rightarrow\sqrt{x}< 3\Rightarrow x< 9\)
Kết hợp ĐKXĐ: \(0< x< 9\)
Với x >= 0 ; x khác 9
\(B=\frac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}=\frac{-3\sqrt{x}-3}{x-9}=\frac{-3\left(\sqrt{x}+1\right)}{x-9}\)
\(\frac{B}{A}=\frac{-3\left(\sqrt{x}+1\right)}{x-9}:\frac{\sqrt{x}+1}{\sqrt{x}-3}=\frac{-3}{\sqrt{x}+3}+\frac{1}{2}< 0\)
\(\Leftrightarrow\frac{-6+\sqrt{x}+3}{2\left(\sqrt{x}+3\right)}< 0\Rightarrow\sqrt{x}-3< 0\Leftrightarrow x< 9\)
Kết hợp đk vậy 0 =< x < 9
Tui nhầm đề xíu, cái A kia phải là: A=\(\sqrt{\left(1-\sqrt{5}\right)^2}-\frac{5-2\sqrt{5}}{\sqrt{5}}\)
thảo nào rút gọn mãi nó chả mất căn :))
\(A=\sqrt{\left(1-\sqrt{5}\right)^2}-\frac{5-2\sqrt{5}}{\sqrt{5}}\)
\(=\sqrt{5}-1-\frac{5\sqrt{5}-10}{5}=\frac{5\sqrt{5}-5-5\sqrt{5}+10}{5}=\frac{5}{5}=1\)
Với \(x\ge0;x\ne4;9\)
\(P=\left(\frac{3\sqrt{x}+6}{x-4}+\frac{\sqrt{x}}{\sqrt{x}-2}\right):\frac{x-9}{\sqrt{x}-3}\)
\(=\left(\frac{3\sqrt{x}+6+\sqrt{x}\left(\sqrt{x}+2\right)}{x-4}\right):\left(\sqrt{x}+3\right)\)
\(=\left(\frac{x+5\sqrt{x}+6}{x-4}\right):\left(\sqrt{x}+3\right)=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+3\right)}{\left(x-4\right)\left(\sqrt{x}+3\right)}=\frac{1}{\sqrt{x}-2}\)
b, \(2P-A< 0\Rightarrow\frac{2}{\sqrt{x}-2}-1< 0\)
\(\Leftrightarrow\frac{4-\sqrt{x}}{\sqrt{x}-2}< 0\Leftrightarrow\frac{\sqrt{x}-4}{\sqrt{x}-2}>0\)
TH1 : \(\hept{\begin{cases}\sqrt{x}-4>0\\\sqrt{x}-2>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>16\\x>4\end{cases}\Leftrightarrow x>16}\)
TH2 : \(\hept{\begin{cases}\sqrt{x}-4< 0\\\sqrt{x}-2< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 16\\x< 4\end{cases}}\Leftrightarrow x< 4}\)
Kết hợp với đk vậy \(0\le x< 4;x>16\)
ĐKXĐ: ...
\(A=\left(\frac{\sqrt{x}\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}-1\right):\left(\frac{25-x+\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}-\frac{\sqrt{x}+3}{\sqrt{x}+5}\right)\)
\(=\left(\frac{\sqrt{x}}{\sqrt{x}+5}-\frac{\sqrt{x}+5}{\sqrt{x}+5}\right):\left(\frac{25-x+x-25}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+5\right)}-\frac{\sqrt{x}+3}{\sqrt{x}+5}\right)\)
\(=\frac{-5}{\left(\sqrt{x}+5\right)}.\frac{\left(\sqrt{x}+5\right)}{-\left(\sqrt{x}+3\right)}=\frac{5}{\sqrt{x}+3}\)
b/ \(B=\frac{x+16}{\sqrt{x}+3}=\sqrt{x}-3+\frac{25}{\sqrt{x}+3}=\sqrt{x}+3+\frac{25}{\sqrt{x}+3}-6\)
\(\Rightarrow B\ge2\sqrt{\frac{\left(\sqrt{x}+3\right).25}{\sqrt{x}+3}}-6=4\)
\(B_{min}=4\) khi \(\left(\sqrt{x}+3\right)^2=25\Rightarrow x=4\)
a) \(A=\frac{4}{\sqrt{x}+3}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\frac{4\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{4\sqrt{x}-12}{x-9}+\frac{2x-\sqrt{x}-13}{x-9}-\frac{x+3\sqrt{x}}{x-9}\)
\(=\frac{4\sqrt{x}-12+2x-\sqrt{x}-13-x-3\sqrt{x}}{x-9}\)
\(=\frac{x-25}{x-9}\)
b) \(P=\frac{A}{B}=\frac{\frac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}}{\frac{\sqrt{x}+5}{\sqrt{x}-3}}\)
\(=\frac{\sqrt{x}-5}{\sqrt{x}+3}\)
\(\sqrt{P}< \frac{1}{3}\Rightarrow\sqrt{\frac{\sqrt{x}-5}{\sqrt{x}+3}}< \frac{1}{3}\)
\(\Rightarrow\frac{\sqrt{x}-5}{\sqrt{x}+3}< \frac{1}{9}\Leftrightarrow9\sqrt{x}-45< \sqrt{x}+3\)
\(\Leftrightarrow8\sqrt{x}< 48\Leftrightarrow\sqrt{x}< 6\Rightarrow0\le x< 36\)
\(a,\)\(A=\frac{4}{\sqrt{x}+3}+\frac{2x-\sqrt{x}-13}{x-9}=\frac{4\left(\sqrt{x}-3\right)+2x-\sqrt{x}-13}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{4\sqrt{x}-12+2x-\sqrt{x}-13}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)\(=\frac{2x+3\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(b,P=\frac{A}{B}=\frac{2x+3\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\frac{\sqrt{x}+5}{\sqrt{x}-3}\)
\(=\frac{2x+3\sqrt{x}-1}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\frac{\left(\sqrt{x}-3\right)}{\sqrt{x}+5}=\frac{2x+3\sqrt{x}-1}{\sqrt{x}+5}\)
Để \(\sqrt{P}< \frac{1}{3}\Rightarrow\frac{2x+3\sqrt{x}-1}{\sqrt{x}+5}< \frac{1}{3}\)
\(\Rightarrow\frac{2x+3\sqrt{x}-1}{\sqrt{x}+5}-\frac{1}{3}< 0\)
\(\Rightarrow\frac{3\left(2x+3\sqrt{x}-1\right)-\sqrt{x}-5}{3\left(\sqrt{x}+5\right)}< 0\)
\(\Rightarrow6x+9\sqrt{x}-3-\sqrt{x}-5< 0\)( do \(3\left(\sqrt{x}+5\right)>0\))
\(\Rightarrow6x-8\sqrt{x}-8< 0\Rightarrow3x-4\sqrt{x}-4< 0\)
\(\Rightarrow3x-6\sqrt{x}+2\sqrt{x}-4< 0\)
\(\Rightarrow3\sqrt{x}\left(\sqrt{x}-2\right)+2\left(\sqrt{x}-2\right)< 0\)
\(\Rightarrow\left(\sqrt{x}-2\right)\left(3\sqrt{x}+2\right)< 0\)
Vì \(3\sqrt{x}+2>0\Rightarrow\sqrt{x}-2< 0\)
\(\Rightarrow\sqrt{x}< 2\Rightarrow x< 4\)
Vậy để \(\sqrt{P}< \frac{1}{3}\)thì \(0\le x< 4\)