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1, đk: \(x>0\) và \(x\ne4\)
Ta có: A=\(\dfrac{1}{2\sqrt{x}-x}=\dfrac{1}{-\left(x-2\sqrt{x}+1\right)+1}=\dfrac{1}{-\left(\sqrt{x}-1\right)^2+1}\)
Ta luôn có: \(-\left(\sqrt{x}-1\right)^2\le0\) với \(x>0\) và \(x\ne4\)
\(\Rightarrow-\left(\sqrt{x}-1\right)^2+1\le1\)
\(\Rightarrow A\ge1\). Dấu "=" xảy ra <=> x=1 (t/m)
Vậy MinA=1 khi x=1
2, đk: \(x\ge0;x\ne1;x\ne9\)
Ta có: B=\(\dfrac{1}{x-4\sqrt{x}+3}=\dfrac{1}{\left(x-4\sqrt{x}+4\right)-1}=\dfrac{1}{\left(\sqrt{x}-2\right)^2-1}\)
Ta luôn có: \(\left(\sqrt{x}-2\right)^2\ge0\) với \(x\ge0;x\ne1;x\ne9\)
\(\Rightarrow\left(\sqrt{x}-2\right)^2-1\ge-1\)
\(\Rightarrow B\le-1\). Dấu "=" xảy ra <=> x=4 (t/m)
Vậy MaxB=-1 khi x=4
3, đk: \(x\ge0;x\ne15+4\sqrt{11}\)
Ta có: C=\(\dfrac{1}{4\sqrt{x}-x+7}=\dfrac{1}{-\left(x-4\sqrt{x}+4\right)+11}=\dfrac{1}{-\left(\sqrt{x}-2\right)^2+11}\)
Ta luôn có: \(-\left(\sqrt{x}-2\right)^2\le0\) với \(x\ge0;x\ne15+4\sqrt{11}\)
\(\Rightarrow-\left(\sqrt{x}-2\right)^2+11\le11\)
\(\Rightarrow C\ge\dfrac{1}{11}\). Dấu "=" xảy ra <=> x=4 (t/m)
Vậy MinC=\(\dfrac{1}{11}\) khi x=4
mình đang cần gấp lắm~!!!
a) \(\sqrt{\left|x\right|-1}\) biểu thức sau có nghĩa \(\Leftrightarrow\) \(\left|x\right|-1\ge0\)
\(\Leftrightarrow\left|x\right|\ge1\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\hoac\\x\le-1\end{matrix}\right.\)
b) \(\sqrt{\left|x-1\right|-3}\) biểu thức sau có nghĩa \(\Leftrightarrow\left|x-1\right|-3\ge0\)
\(\Leftrightarrow\left|x-1\right|\ge3\) \(\left\{{}\begin{matrix}x-1\ge3\\hoac\\x-1\le-3\end{matrix}\right.\)
c) \(\sqrt{4-\left|x\right|}\) biểu thức sau có nghĩa \(\Leftrightarrow4-\left|x\right|\ge0\)
\(\Leftrightarrow4\ge\left|x\right|\) \(\Leftrightarrow-4\le x\le4\)
ĐKXĐ: x>0
\(\left(\dfrac{x}{x+3\sqrt{x}}+\dfrac{1}{\sqrt{x}+3}\right):\left(1-\dfrac{2}{\sqrt{x}}+\dfrac{6}{x+3\sqrt{x}}\right)\)
= \(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}:\dfrac{x+3\sqrt{x}-2\left(\sqrt{x}+3\right)+6}{\sqrt{x}\left(\sqrt{x}+3\right)}\)
= \(\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}:\dfrac{x+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=1\)
1a) \(\sqrt{4+\sqrt{8}}.\sqrt{2+\sqrt{2+\sqrt{2}}}.\sqrt{2-\sqrt{2+\sqrt{2}}}\)
\(=\sqrt{4+\sqrt{8}}.\sqrt{\left(2+\sqrt{2+\sqrt{2}}\right)\left(\sqrt{2-\sqrt{2+\sqrt{2}}}\right)}\)
\(=\sqrt{4+\sqrt{8}}.\sqrt{4-2-\sqrt{2}}\)
\(=\sqrt{4+\sqrt{8}}.\sqrt{2-\sqrt{2}}=\sqrt{\left(4+\sqrt{8}\right)\left(2-\sqrt{2}\right)}\)
\(=\sqrt{8-4\sqrt{2}-\sqrt{16}+2\sqrt{8}}\)
\(=\sqrt{8-4\sqrt{2}-4+4\sqrt{2}}\)
\(=\sqrt{4}=2\)
1b) \(\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{4+4\sqrt{3}+3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{25-10\sqrt{3}+3}}\)
\(=\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}\)
\(=\sqrt{5\sqrt{3}+25-5\sqrt{3}}\)
\(=\sqrt{25}=5\)
a) điều kiện : \(x>0;x\ne4\)
\(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}}\right)\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{x-4}\right)\)
\(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)
\(P=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}-2+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(x=4+2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}+1\right)^2}\Leftrightarrow\sqrt{x}=\sqrt{3}+1\) \(\left(x>0\right)\)
thay vào P ta có \(P=\dfrac{\sqrt{3}+1+2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}+1-2\right)}=\dfrac{\sqrt{3}+3}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}+3}{2}\)
\(P>0\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}>0\)
ta có : \(\sqrt{x}+2>0\) và \(\sqrt{x}>0\) \(\left(x>0\right)\)
\(\Rightarrow p>0\) thì \(\sqrt{x}-2>0\Leftrightarrow\sqrt{x}>2\Leftrightarrow x>4\)
vậy \(x>4\) thì P > 0
câu : a ; b ; c đầy đủ rồi nha quênh gi câu : a ; b ; c 
1) a) \(\sqrt{27}\) + \(\sqrt{75}\) - \(\sqrt{\dfrac{1}{3}}\) = \(3\sqrt{3}\) + \(5\sqrt{3}\) - \(\dfrac{\sqrt{3}}{3}\) = \(8\sqrt{3}\) - \(\dfrac{\sqrt{3}}{3}\)
= \(\dfrac{23\sqrt{3}}{3}\)
b) \(\sqrt{4+2\sqrt{3}}\) \(-\sqrt{4-2\sqrt{3}}\)
= \(\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1^2}\) \(-\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2}\)
= \(\sqrt{\left(\sqrt{3}+1\right)^2}\) \(-\sqrt{\left(\sqrt{3}-1\right)^2}\)
= \(\left(\sqrt{3}+1\right)\) \(-\left(\sqrt{3}-1\right)\)
= \(\sqrt{3}+1-\sqrt{3}+1\)
= 2
2) \(\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right)\) : \(\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{a-1}\right)\)
= \(\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right)\) : \(\left(\dfrac{1}{\sqrt{a}+1}+\dfrac{2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
= \(\left(\dfrac{a-1}{\left(\sqrt{a}-1\right)\sqrt{a}}\right)\) : \(\left(\dfrac{\left(\sqrt{a}-1\right)+2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
= \(\left(\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\sqrt{a}}\right)\) : \(\left(\dfrac{\left(\sqrt{a}-1\right)+2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
= \(\dfrac{\sqrt{a}+1}{\sqrt{a}}\) : \(\dfrac{2}{\sqrt{a}+1}\) = \(\dfrac{\sqrt{a}+1}{\sqrt{a}}\) . \(\dfrac{\sqrt{a}+1}{2}\) = \(\dfrac{\left(\sqrt{a}+1\right)^2}{2\sqrt{a}}\)