Toán Lớp 8: 1.A=(x/căn x -1 -căn x):(căn x+1/căn x-1/1-căn x+2-x/x-căn x)
2.A=(1-4/căn x+1+1/x-1):x-2căn x/x-1
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Toán Lớp 8: 1.A=(x/căn x -1 -căn x):(căn x+1/căn x-1/1-căn x+2-x/x-căn x)
2.A=(1-4/căn x+1+1/x-1):x-2căn x/x-1
Comments ( 1 )
1)\\
A = \left( {\dfrac{x}{{\sqrt x – 1}} – \sqrt x } \right):\left( {\dfrac{{\sqrt x + 1}}{{\sqrt x }} – \dfrac{1}{{1 – \sqrt x }} + \dfrac{{2 – x}}{{x – \sqrt x }}} \right)\\
= \dfrac{{x – \sqrt x \left( {\sqrt x – 1} \right)}}{{\sqrt x – 1}}:\dfrac{{\left( {\sqrt x + 1} \right).\left( {\sqrt x – 1} \right) + \sqrt x + 2 – x}}{{\sqrt x \left( {\sqrt x – 1} \right)}}\\
= \dfrac{{x – x + \sqrt x }}{{\sqrt x – 1}}.\dfrac{{\sqrt x \left( {\sqrt x – 1} \right)}}{{x – 1 + \sqrt x + 2 – x}}\\
= \dfrac{{\sqrt x }}{1}.\dfrac{{\sqrt x }}{{\sqrt x + 1}}\\
= \dfrac{x}{{\sqrt x + 1}}\\
2)\\
A = \left( {1 – \dfrac{4}{{\sqrt x + 1}} + \dfrac{1}{{x – 1}}} \right):\dfrac{{x – 2\sqrt x }}{{x – 1}}\\
= \dfrac{{x – 1 – 4\left( {\sqrt x – 1} \right) + 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x – 1} \right)}}.\dfrac{{x – 1}}{{\sqrt x \left( {\sqrt x – 2} \right)}}\\
= \dfrac{{x – 1 – 4\sqrt x + 4 + 1}}{{\sqrt x \left( {\sqrt x – 2} \right)}}\\
= \dfrac{{x – 4\sqrt x + 4}}{{\sqrt x \left( {\sqrt x – 2} \right)}}\\
= \dfrac{{{{\left( {\sqrt x – 2} \right)}^2}}}{{\sqrt x \left( {\sqrt x – 2} \right)}}\\
= \dfrac{{\sqrt x – 2}}{{\sqrt x }}
\end{array}$