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a: \(A=\dfrac{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}+3}-\dfrac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)
\(=-\sqrt{x}+3-\sqrt{x}+3-6=-2\sqrt{x}\)
b: \(\left(\dfrac{2\sqrt{x}}{x\sqrt{x}+x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\right):\left(\dfrac{2\sqrt{x}}{\sqrt{x}+1}-1\right)\)
\(=\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x+1\right)}-\dfrac{1}{\sqrt{x}+1}\right):\dfrac{2\sqrt{x}-\sqrt{x}-1}{\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}-x-1}{\left(\sqrt{x}+1\right)\left(x+1\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{1}{x+1}\)
g: \(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\left(\dfrac{x-1}{\sqrt{x}+1}-2\right)\)
\(=\dfrac{\sqrt{x}+1+\sqrt{x}-1}{x-1}\cdot\left(\sqrt{x}-1-2\right)\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{x-1}\)
\(=\left(\sqrt{x}+\sqrt{x-1}-\sqrt{x-1}+\sqrt{2}\right)\cdot\left(\dfrac{2\sqrt{x}-\sqrt{x}-\sqrt{2}}{\sqrt{x}\left(2-\sqrt{x}\right)}\right)\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{2}\right)}{-\sqrt{x}}\)
a: Sửa đề; \(P=\left(\dfrac{3x+3\sqrt{x}-3}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}+\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\right)\cdot\left(\dfrac{1}{1-\sqrt{x}}-1\right)\)
\(=\dfrac{3x+3\sqrt{x}-3-x+1+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{1-1+\sqrt{x}}{1-\sqrt{x}}\)
\(=\dfrac{3x+3\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\sqrt{x}}{1-\sqrt{x}}=\dfrac{3\sqrt{x}}{1-\sqrt{x}}\)
b: Để \(P=\sqrt{x}\) thì \(3\sqrt{x}=\sqrt{x}-x\)
\(\Leftrightarrow x+2\sqrt{x}=0\)
hay x=0
a/ ĐKXĐ: \(x\ge0,x\ne1\)
\(P=\left(\dfrac{3}{\sqrt{x}-1}+\dfrac{\sqrt{x}-3}{x-1}\right):\left(\dfrac{x+2}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+2}\right)\)
= \(\dfrac{3\left(\sqrt{x}+1\right)+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
= \(\dfrac{4\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}:\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)
= \(\dfrac{4\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
= \(\dfrac{4\sqrt{x}}{\sqrt{x}+1}\)
b/ Với \(x\ge0,x\ne1\)
Để \(P=\sqrt{x}-1\Leftrightarrow\dfrac{4\sqrt{x}}{\sqrt{x}+1}=\sqrt{x}-1\)
\(\Leftrightarrow4\sqrt{x}=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(\Leftrightarrow x-4\sqrt{x}-1=0\)
\(\Leftrightarrow\left(\sqrt{x}-2+\sqrt{5}\right)\left(\sqrt{x}-2-\sqrt{5}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-2+\sqrt{5}=0\\\sqrt{x}-2-\sqrt{5}=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=2-\sqrt{5}\left(ktm\right)\\\sqrt{x}=2+\sqrt{5}\left(tm\right)\end{matrix}\right.\)
\(\Leftrightarrow x=9+4\sqrt{5}\)
Vậy để \(P=\sqrt{x}-1\) thì \(x=9+4\sqrt{5}\)
Cho \(5\sqrt{x}7\) mk viet nham
Sua lai thanh \(5\sqrt{x}-7\)
a: \(A=\left(\dfrac{2}{\sqrt{x}-2}+\dfrac{3}{2\sqrt{x}+1}-\dfrac{5\sqrt{x}-7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\right)\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\dfrac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\dfrac{2\sqrt{x}+3}{\left(2\sqrt{x}+1\right)}\cdot\dfrac{5\sqrt{x}}{2\sqrt{x}+3}=\dfrac{5\sqrt{x}}{2\sqrt{x}+1}\)
b: Để A là số nguyên thì \(5\sqrt{x}⋮2\sqrt{x}+1\)
=>10 căn x+5-5 chia hết cho 2 căn x+1
=>\(2\sqrt{x}+1\in\left\{1;5\right\}\)
hay \(x\in\varnothing\)
a: \(A=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}+\dfrac{-3+2\sqrt{x}+2}{x\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}+\dfrac{2\sqrt{x}-1}{x\sqrt{x}+1}\)
\(=\dfrac{2x^2+2\sqrt{x}-9x\sqrt{x}-9+2x\sqrt{x}-10x+12\sqrt{x}-x+5\sqrt{x}-6}{\left(x\sqrt{x}+1\right)\left(x-5\sqrt{x}+6\right)}\)
\(=\dfrac{2x^2+19\sqrt{x}-7x\sqrt{x}-11x-15}{\left(x\sqrt{x}+1\right)\left(x-5\sqrt{x}+6\right)}\)
b: \(=\dfrac{x-\sqrt{x}+1-3+2\sqrt{x}+2}{x\sqrt{x}+1}\)
\(=\dfrac{x+\sqrt{x}}{x\sqrt{x}+1}=\dfrac{\sqrt{x}}{x-\sqrt{x}+1}\)
a) ĐK:\(x\ge0,x\ne4\)
\(P=\dfrac{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}+2\right)-2-5\sqrt{x}}{x-4}\)
\(=\dfrac{x\sqrt{x}+4x}{x-4}\)
b) ĐK: \(x\ge0,x\ne1\)
\(A=\dfrac{\sqrt{x}\left(x-1\right)+3\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)+4-6\sqrt{x}}{\left(\sqrt{x}+3\right)\left(x-1\right)}\)
\(=\dfrac{x\sqrt{x}+3x-\sqrt{x}-5}{\left(\sqrt{x}+3\right)\left(x-1\right)}\)
\(A=\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-....-\frac{1}{\sqrt{24}-\sqrt{25}}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{(\sqrt{1}-\sqrt{2})(\sqrt{1}+\sqrt{2})}-\frac{\sqrt{2}+\sqrt{3}}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})}+\frac{\sqrt{3}+\sqrt{4}}{(\sqrt{3}-\sqrt{4})(\sqrt{3}+\sqrt{4})}-...-\frac{\sqrt{24}+\sqrt{25}}{(\sqrt{24}-\sqrt{25})(\sqrt{24}+\sqrt{25})}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{-1}-\frac{\sqrt{2}+\sqrt{3}}{-1}+\frac{\sqrt{3}+\sqrt{4}}{-1}-...-\frac{\sqrt{24}+\sqrt{25}}{-1}\)
\(=\frac{(1+\sqrt{2})-(\sqrt{2}+\sqrt{3})+(\sqrt{3}+\sqrt{4})-...-(\sqrt{24}+\sqrt{25})}{-1}\)
\(=\frac{1-\sqrt{25}}{-1}=4\)
\(B=\frac{5}{4+\sqrt{11}}+\frac{11-3\sqrt{11}}{\sqrt{11}-3}-\frac{4}{\sqrt{5}-1}+\sqrt{(\sqrt{5}-2)^2}\)
\(=\frac{5(4-\sqrt{11})}{(4+\sqrt{11})(4-\sqrt{11})}+\frac{\sqrt{11}(\sqrt{11}-3)}{\sqrt{11}-3}-\frac{4(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}+\sqrt{5}-2\)
\(=\frac{5(4-\sqrt{11})}{5}+\sqrt{11}-\frac{4(\sqrt{5}+1)}{4}+\sqrt{5}-2\)
\(=4-\sqrt{11}+\sqrt{11}-(\sqrt{5}+1)+\sqrt{5}-2\)
\(=1\)
a: \(\dfrac{5}{4-\sqrt{11}}+\dfrac{1}{3+\sqrt{7}}-\dfrac{6}{\sqrt{7}-2}-\dfrac{\sqrt{7}-5}{2}\)
\(=4+\sqrt{11}+\dfrac{3}{2}-\dfrac{\sqrt{7}}{2}-4-2\sqrt{7}-\dfrac{1}{2}\sqrt{7}+\dfrac{5}{2}\)
\(=4+\sqrt{11}-3\sqrt{7}\)
b: \(\dfrac{\sqrt{x}+\sqrt{y}}{2\left(\sqrt{x}-\sqrt{y}\right)}-\dfrac{\sqrt{x}-\sqrt{y}}{2\left(\sqrt{x}+\sqrt{y}\right)}-\dfrac{y+x}{y-x}\)
\(=\dfrac{x+2\sqrt{xy}+y-x+2\sqrt{xy}-y+2x+2y}{2\left(x-y\right)}\)
\(=\dfrac{2\left(x+2\sqrt{xy}+y\right)}{2\left(x-y\right)}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(\dfrac{2\sqrt{x}}{x-5\sqrt{x}+6}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{\sqrt{x}+3}{2-\sqrt{x}}\left(ĐK:x\ge0;x\ne4;x\ne9\right)\)
\(=\dfrac{2\sqrt{x}}{x-2\sqrt{x}-3\sqrt{x}+6}+\dfrac{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)}+\dfrac{2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}+\dfrac{2x-3\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{x-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{2\sqrt{x}+2x-3\sqrt{x}-2-x+9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-\sqrt{x}+7}{x-5\sqrt{x}+6}\)
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