Toán Lớp 9: gấp ạ
A=(1/x+2 √x – 1/ √x+1): 1- √x/x+4 √x+4 (với x>0;x khác 1)
a) rút gọn biểu thức A
b) tìm x để A=5/3
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Toán Lớp 9: gấp ạ
A=(1/x+2 √x – 1/ √x+1): 1- √x/x+4 √x+4 (với x>0;x khác 1)
a) rút gọn biểu thức A
b) tìm x để A=5/3
Comments ( 1 )
a)A = \dfrac{{ – x\sqrt x – 3x – \sqrt x + 2}}{{ – x\sqrt x + \sqrt x }}\\
b)\left[ \begin{array}{l}
x = \dfrac{1}{4}\\
x = 14 + 4\sqrt {10}
\end{array} \right.
\end{array}\)
a)A = \left( {\dfrac{1}{{x + 2\sqrt x }} – \dfrac{1}{{\sqrt x + 1}}} \right):\dfrac{{1 – \sqrt x }}{{x + 4\sqrt x + 4}}\\
= \dfrac{{\sqrt x + 1 – x – 2\sqrt x }}{{\sqrt x \left( {\sqrt x + 2} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{{{\left( {\sqrt x + 2} \right)}^2}}}{{1 – \sqrt x }}\\
= \dfrac{{ – x – \sqrt x + 1}}{{\sqrt x \left( {\sqrt x + 2} \right)\left( {\sqrt x + 1} \right)}}.\dfrac{{{{\left( {\sqrt x + 2} \right)}^2}}}{{1 – \sqrt x }}\\
= \dfrac{{\left( { – x – \sqrt x + 1} \right)\left( {\sqrt x + 2} \right)}}{{\sqrt x \left( {1 – x} \right)}}\\
= \dfrac{{ – x\sqrt x – 2x – x – 2\sqrt x + \sqrt x + 2}}{{ – x\sqrt x + \sqrt x }}\\
= \dfrac{{ – x\sqrt x – 3x – \sqrt x + 2}}{{ – x\sqrt x + \sqrt x }}\\
b)A = \dfrac{5}{3}\\
\to \dfrac{{ – x\sqrt x – 3x – \sqrt x + 2}}{{ – x\sqrt x + \sqrt x }} = \dfrac{5}{3}\\
\to – 3x\sqrt x – 9x – 3\sqrt x + 6 = – 5x\sqrt x + 5\sqrt x \\
\to 2x\sqrt x – 9x – 8\sqrt x + 6 = 0\\
\to 2x\sqrt x – x – 8x + 4\sqrt x – 12\sqrt x + 6 = 0\\
\to x\left( {2\sqrt x – 1} \right) – 4\sqrt x \left( {2\sqrt x – 1} \right) – 6\left( {2\sqrt x – 1} \right) = 0\\
\to \left( {2\sqrt x – 1} \right)\left( {x – 4\sqrt x – 6} \right) = 0\\
\to \left[ \begin{array}{l}
2\sqrt x – 1 = 0\\
x – 4\sqrt x – 6 = 0
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt x = \dfrac{1}{2}\\
\sqrt x = 2 + \sqrt {10} \\
\sqrt x = 2 – \sqrt {10} \left( l \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = \dfrac{1}{4}\\
x = 14 + 4\sqrt {10}
\end{array} \right.
\end{array}\)