\(\dfrac{3}{1.4}\)+\(\dfrac{3}{4.7}\)+\(\dfrac{3}{7.10}\)+...+\(\dfrac{3}{94.97}\)+\(\dfrac{3}{97.100}\)
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ĐK: \(x\ge0,x\ne4\)
\(\dfrac{1}{\sqrt{x}-2}+\dfrac{5\sqrt{x}-4}{2\sqrt{x}-x}=\dfrac{-1}{2-\sqrt{x}}+\dfrac{5\sqrt{x}-4}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{-\sqrt{x}+5\sqrt{x}-4}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(2-\sqrt{x}\right)}\)
\(\dfrac{2+\sqrt{x}}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}-2}=\dfrac{2+\sqrt{x}}{\sqrt{x}}+\dfrac{\sqrt{x}}{2-\sqrt{x}}=\dfrac{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)+\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{4-x+x}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{4}{\sqrt{x}\left(2-\sqrt{x}\right)}\)
\(\left(\dfrac{2}{\sqrt{x}-2}+\dfrac{5\sqrt{x}-4}{2\sqrt{x}-x}\right):\left(\dfrac{2+\sqrt{x}}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)=\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(2-\sqrt{x}\right)}:\dfrac{4}{\sqrt{x}\left(2-\sqrt{x}\right)}=\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(2-\sqrt{x}\right)}.\dfrac{\sqrt{x}\left(2-\sqrt[]{x}\right)}{4}=\sqrt{x}-1\)
\(\dfrac{5}{8}+\dfrac{5}{12}=\dfrac{15}{24}+\dfrac{10}{24}=\dfrac{25}{24}\)
=1 - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{10}\) + ....... + \(\dfrac{1}{94}\) - \(\dfrac{1}{97}\) + \(\dfrac{1}{97}\) - \(\dfrac{1}{100}\)
= 1 - \(\dfrac{1}{100}\)
=\(\dfrac{99}{100}\)
\(\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{94.97}+\dfrac{3}{97.100}\)
3A = 3.\(\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+...+\dfrac{1}{94.97}+\dfrac{1}{97.100}\right)\)
3A = 3.\(\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{94}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
3A = 3.\(\left(1-\dfrac{1}{100}\right)\)
3A = 3.\(\dfrac{99}{100}\)
3A = \(\dfrac{297}{100}\)
A = \(\dfrac{297}{100}:3\)
A = \(\dfrac{297}{100}.\dfrac{1}{3}\)
A = \(\dfrac{99}{100}\)