Rút gọn
A=\(\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-...-\dfrac{1}{\sqrt{24}-\sqrt{25}}\)
B=\(\dfrac{5}{4+\sqrt{11}}+\dfrac{11-3\sqrt{11}}{\sqrt{11}-3}-\dfrac{4}{\sqrt{5}-1}+\sqrt{\left(\sqrt{5}-2\right)^2}\)
C=\(\dfrac{\sqrt{x}+1}{x\sqrt[]{x}+x+\sqrt{x}}:\dfrac{1}{x^2-\sqrt{x}}\) (với x>0; x#1)
D=\(\dfrac{\sqrt{x^2-10x+25}}{x-5}\)
\(A=\frac{1}{\sqrt{1}-\sqrt{2}}-\frac{1}{\sqrt{2}-\sqrt{3}}+\frac{1}{\sqrt{3}-\sqrt{4}}-....-\frac{1}{\sqrt{24}-\sqrt{25}}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{(\sqrt{1}-\sqrt{2})(\sqrt{1}+\sqrt{2})}-\frac{\sqrt{2}+\sqrt{3}}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})}+\frac{\sqrt{3}+\sqrt{4}}{(\sqrt{3}-\sqrt{4})(\sqrt{3}+\sqrt{4})}-...-\frac{\sqrt{24}+\sqrt{25}}{(\sqrt{24}-\sqrt{25})(\sqrt{24}+\sqrt{25})}\)
\(=\frac{\sqrt{1}+\sqrt{2}}{-1}-\frac{\sqrt{2}+\sqrt{3}}{-1}+\frac{\sqrt{3}+\sqrt{4}}{-1}-...-\frac{\sqrt{24}+\sqrt{25}}{-1}\)
\(=\frac{(1+\sqrt{2})-(\sqrt{2}+\sqrt{3})+(\sqrt{3}+\sqrt{4})-...-(\sqrt{24}+\sqrt{25})}{-1}\)
\(=\frac{1-\sqrt{25}}{-1}=4\)
\(B=\frac{5}{4+\sqrt{11}}+\frac{11-3\sqrt{11}}{\sqrt{11}-3}-\frac{4}{\sqrt{5}-1}+\sqrt{(\sqrt{5}-2)^2}\)
\(=\frac{5(4-\sqrt{11})}{(4+\sqrt{11})(4-\sqrt{11})}+\frac{\sqrt{11}(\sqrt{11}-3)}{\sqrt{11}-3}-\frac{4(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}+\sqrt{5}-2\)
\(=\frac{5(4-\sqrt{11})}{5}+\sqrt{11}-\frac{4(\sqrt{5}+1)}{4}+\sqrt{5}-2\)
\(=4-\sqrt{11}+\sqrt{11}-(\sqrt{5}+1)+\sqrt{5}-2\)
\(=1\)
\(C=\frac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}}: \frac{1}{x^2-\sqrt{x}}=\frac{\sqrt{x}+1}{\sqrt{x}(x+\sqrt{x}+1)}.(x^2-\sqrt{x})\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}(x+\sqrt{x}+1)}.\sqrt{x}(\sqrt{x^3}-1)\)
\(=\frac{(\sqrt{x}+1)(\sqrt{x^3}-1)}{x+\sqrt{x}+1}\)
\(=\frac{(\sqrt{x}+1)(\sqrt{x}-1)(x+\sqrt{x}+1)}{x+\sqrt{x}+1}\)
\(=(\sqrt{x}+1)(\sqrt{x}-1)=x-1\)
\(D=\frac{\sqrt{x^2-10x+25}}{x-5}=\frac{\sqrt{x^2-2.5x+5^2}}{x-5}=\frac{\sqrt{(x-5)^2}}{x-5}=\frac{|x-5|}{x-5}\)
Nếu $x\geq 5$ thì:
\(D=\frac{|x-5|}{x-5}=\frac{x-5}{x-5}=1\)
Nếu \(x< 5\) thì:
\(D=\frac{|x-5|}{x-5}=\frac{5-x}{x-5}=-1\)