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A = \(\frac{8}{\sqrt{5}-1}\) - (\(2\sqrt{5}-1\) ) ( chúng ta cần trục căn thức lên để khử mẫu )
= \(\frac{8\left(\sqrt{5}+1\right)}{5-1}\)- \(\left(2\sqrt{5}-1\right)\)
= \(2\sqrt{5}\)+ 2 - \(2\sqrt{5}\)+1
= 3
B = \(\frac{\left(1-\sqrt{x}\right)^2+4\sqrt{x}}{1+\sqrt{x}}\)( x \(\ge\)0 )
= \(\frac{1-2\sqrt{x}+x+4\sqrt{x}}{1+\sqrt{x}}\)
= \(\frac{1+2\sqrt{x}+x}{1+\sqrt{x}}\)
= \(\frac{\left(1+\sqrt{x}\right)^2}{1+\sqrt{x}}\)
= 1 +\(\sqrt{x}\)
#mã mã#
a: \(=4+\sqrt{11}+\dfrac{3}{2}-\dfrac{1}{2}\sqrt{7}-4-2\sqrt{7}-\dfrac{1}{2}\sqrt{7}+\dfrac{5}{2}\)
\(=4+\sqrt{11}-3\sqrt{7}\)
b: \(VT=\dfrac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y+2x+2y}{2\left(x-y\right)}\)
\(=\dfrac{2x+4\sqrt{xy}+2y}{2\left(x-y\right)}=\dfrac{x+2\sqrt{xy}+y}{x-y}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
1 , ĐKXĐ : \(x\ge0,x\ne1\)
Với điều kiện xác định trên phương trình đã cho thánh :
\(\dfrac{1}{\sqrt{x}+1}-\dfrac{2}{\sqrt{x}-1}+\dfrac{x+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-1-2\left(\sqrt{x}+1\right)+x+3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-1}\)
Câu 1:
a: \(P=\dfrac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: Để \(2P=2\sqrt{5}+5\) thì \(P=\dfrac{2\sqrt{5}+5}{2}\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+5\right)=2\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+3\right)=2\)
hay \(x=\dfrac{4}{29+12\sqrt{5}}=\dfrac{4\left(29-12\sqrt{5}\right)}{121}\)
Câu 1:
a: \(P=\dfrac{x+\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b: Để \(2P=2\sqrt{5}+5\) thì \(P=\dfrac{2\sqrt{5}+5}{2}\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+5\right)=2\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow\sqrt{x}\left(2\sqrt{5}+3\right)=2\)
hay \(x=\dfrac{4}{29+12\sqrt{5}}=\dfrac{4\left(29-12\sqrt{5}\right)}{121}\)
Bài 2: a) Ta có: Q=\(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\) -\(\left(\dfrac{x+2}{\left(\sqrt{x}\right)^3-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}\right)\) =\(\dfrac{1}{\sqrt{x}-1}\) -\(\left(\dfrac{x+2+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\) =\(\dfrac{1}{\sqrt{x}-1}-\left(\dfrac{x+2+x-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\) =\(\dfrac{1}{\sqrt{x}-1}-\dfrac{2x}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\) =
a. \(Z=\left(\sqrt{x}-\dfrac{x+2}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}-4}{x-1}\right)\)\(\left(x\ge0,x\ne1\right)\)
\(Z=\dfrac{x\sqrt{x}-\sqrt{x}-x\sqrt{x}+x-2\sqrt{x}+2}{x-1}:\dfrac{x-\sqrt{x}+\sqrt{x}-4}{x-1}\)
\(Z=\dfrac{-3\sqrt{x}+x+2}{x-1}:\dfrac{x-4}{x-1}=\dfrac{x-3\sqrt{x}+2}{x-1}.\dfrac{x-1}{x-4}\)
\(Z=\dfrac{x-3\sqrt{x}+2}{x-4}\)
b. \(Z=\dfrac{x-3\sqrt{x}+2}{x-4}< \dfrac{1}{2}\)
\(\Leftrightarrow\dfrac{x-3\sqrt{x}+2}{x-4}-\dfrac{1}{2}< 0\)
\(\Leftrightarrow\dfrac{2x-6\sqrt{x}+4-x+4}{2x-8}< 0\)
\(\Leftrightarrow\dfrac{x-6\sqrt{x}+8}{2x-8}< 0\)\(\Leftrightarrow\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}-4\right)}{2\left(x-4\right)}< 0\)
*\(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x}-2\right)\left(\sqrt{x}-4\right)< 0\\2\left(x-4\right)>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\sqrt{x}-2< 0\&\sqrt{x}-4>0\\\sqrt{x}-2>0\&\sqrt{x}-4< 0\end{matrix}\right.\\2\left(x-4\right)>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x< 4\&x>16\left(l\right)\\16>x>4\end{matrix}\right.\\x>4\end{matrix}\right.\)
* \(\left\{{}\begin{matrix}\left(\sqrt{x}-2\right)\left(\sqrt{x}-4\right)>0\\2\left(x-4\right)< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\sqrt{x}-2>0\&\sqrt{x}-4>0\\\sqrt{x}-2< 0\&\sqrt{x}-4< 0\end{matrix}\right.\\2\left(x-4\right)< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>16\\x< 4\end{matrix}\right.\\x< 4\end{matrix}\right.\) \(\Leftrightarrow16>x>4\)
Vậy: \(Z< \dfrac{1}{2}\Leftrightarrow16>x>4\)
Lời giải:
ĐK: \(x>0; x\neq 4\)
Có: \(K=\left(\frac{4\sqrt{x}(2-\sqrt{x})}{(2+\sqrt{x})(2-\sqrt{x})}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}(\sqrt{x}-2)}-\frac{2(\sqrt{x}-2)}{\sqrt{x}(\sqrt{x}-2)}\right)\)
\(=\frac{8\sqrt{x}-4x+8x}{(2+\sqrt{x})(2-\sqrt{x})}: \frac{\sqrt{x}-1-2(\sqrt{x}-2)}{\sqrt{x}(\sqrt{x}-2)}\)
\(=\frac{8\sqrt{x}+4x}{(2+\sqrt{x})(2-\sqrt{x})}.\frac{\sqrt{x}(\sqrt{x}-2)}{-\sqrt{x}+3}\)
\(=\frac{4\sqrt{x}(2+\sqrt{x})}{2+\sqrt{x}}. \frac{-\sqrt{x}}{3-\sqrt{x}}=\frac{-4\sqrt{x}.\sqrt{x}}{3-\sqrt{x}}=\frac{4x}{\sqrt{x}-3}\)
b)
\(K=-1\Leftrightarrow \frac{4x}{\sqrt{x}-3}=-1\Rightarrow 4x=-(\sqrt{x}-3)\)
\(\Leftrightarrow 4x+\sqrt{x}-3=0\)
\(\Leftrightarrow (4\sqrt{x}-3)(\sqrt{x}+1)=0\)
Vì \(\sqrt{x}+1>0\Rightarrow 4\sqrt{x}-3=0\Rightarrow x=\frac{9}{16}\)
c) \(m(\sqrt{x}-3)K>x+1\)
\(\Leftrightarrow m. (\sqrt{x}-3).\frac{4x}{\sqrt{x}-3}>x+1\)
\(\Leftrightarrow m> \frac{x+1}{4x}\)
\(\Leftrightarrow m> max(\frac{4x}{x+1}), \forall x< 9\)
Với đk đã cho thì ta thấy \(\frac{4x}{x+1}\) có min thôi.
a. \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}}\right)\cdot\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{x-4}\right)\)
<=> \(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
<=> \(P=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
<=> \(P=\dfrac{\sqrt{x}+2}{x-2\sqrt{x}}\)
b. Khi \(x=7+4\sqrt{3}=\left(2+\sqrt{3}\right)^2\) => \(\sqrt{x}=2+\sqrt{3}\)
=> \(P=\dfrac{2+\sqrt{3}+2}{7+4\sqrt{3}-2\left(2+\sqrt{3}\right)}=\dfrac{4+\sqrt{3}}{7+4\sqrt{3}-4-2\sqrt{3}}=\dfrac{4+\sqrt{3}}{3+2\sqrt{3}}=\dfrac{5\sqrt{3}-6}{3}\)
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