Toán Lớp 9: A = ( x-7 / x-3 √x – 3+ √x / √x ) ÷ 1 / 2 √x-6
a,Tính A khi x = 2/10-3√11
Leave a reply
Register Now
Login
Lost Password
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Toán Lớp 9: A = ( x-7 / x-3 √x – 3+ √x / √x ) ÷ 1 / 2 √x-6
a,Tính A khi x = 2/10-3√11
Comments ( 2 )
a)DK:x > 0;x \ne 9\\
A = \left( {\dfrac{{x – 7}}{{x – 3\sqrt x }} – \dfrac{{3 + \sqrt x }}{{\sqrt x }}} \right):\dfrac{1}{{2\left( {\sqrt x – 3} \right)}}\\
= \dfrac{{x – 7 – \left( {3 + \sqrt x } \right)\left( {\sqrt x – 3} \right)}}{{\sqrt x \left( {\sqrt x – 3} \right)}}.2\left( {\sqrt x – 3} \right)\\
= \dfrac{{x – 7 – x + 9}}{{\sqrt x \left( {\sqrt x – 3} \right)}}.2\left( {\sqrt x – 3} \right)\\
= \dfrac{2}{{\sqrt x \left( {\sqrt x – 3} \right)}}.2\left( {\sqrt x – 3} \right)\\
= \dfrac{4}{{\sqrt x }}\\
Thay:x = \dfrac{2}{{10 – 3\sqrt {11} }}\\
= \dfrac{{2\left( {10 + 3\sqrt {11} } \right)}}{{{{10}^2} – {{\left( {3\sqrt {11} } \right)}^2}}}\\
= \dfrac{{2\left( {10 + 3\sqrt {11} } \right)}}{{100 – 9.11}}\\
= \dfrac{{20 + 6\sqrt {11} }}{1}\\
= 20 + 6\sqrt {11} = 11 + 2.3.\sqrt {11} + 9\\
= {\left( {\sqrt {11} + 3} \right)^2}\\
\to A = \dfrac{4}{{\sqrt {{{\left( {\sqrt {11} + 3} \right)}^2}} }}\\
= \dfrac{4}{{\sqrt {11} + 3}} = \dfrac{{4\left( {\sqrt {11} – 3} \right)}}{{11 – 9}}\\
= 2\left( {\sqrt {11} – 3} \right)\\
= 2\sqrt {11} – 6
\end{array}\)