Toán Lớp 9: A=(1/căn x-1/căn x-1):(căn x+2/căn x-1-căn x+1/căn x-2)
a) rút gọn A
b) tính giá trị của A khi x=7+4 căn 3
Mọi người giúp e vs ạ
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Toán Lớp 9: A=(1/căn x-1/căn x-1):(căn x+2/căn x-1-căn x+1/căn x-2)
a) rút gọn A
b) tính giá trị của A khi x=7+4 căn 3
Mọi người giúp e vs ạ
Comments ( 1 )
a)A = \dfrac{{\sqrt x – 2}}{{3\sqrt x }}\\
b)\dfrac{{ – 3 + 2\sqrt 3 }}{3}
\end{array}\)
a)DK:x > 0;x \ne \left\{ {1;4} \right\}\\
A = \left( {\dfrac{1}{{\sqrt x }} – \dfrac{1}{{\sqrt x – 1}}} \right):\left( {\dfrac{{\sqrt x + 2}}{{\sqrt x – 1}} – \dfrac{{\sqrt x + 1}}{{\sqrt x – 2}}} \right)\\
= \dfrac{{\sqrt x – 1 – \sqrt x }}{{\sqrt x \left( {\sqrt x – 1} \right)}}:\dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 2} \right) – \left( {\sqrt x + 1} \right)\left( {\sqrt x – 1} \right)}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x – 2} \right)}}\\
= \dfrac{{ – 1}}{{\sqrt x \left( {\sqrt x – 1} \right)}}.\dfrac{{\left( {\sqrt x – 1} \right)\left( {\sqrt x – 2} \right)}}{{x – 4 – x + 1}}\\
= \dfrac{{ – 1}}{{\sqrt x \left( {\sqrt x – 1} \right)}}.\dfrac{{\left( {\sqrt x – 1} \right)\left( {\sqrt x – 2} \right)}}{{ – 3}}\\
= \dfrac{{\sqrt x – 2}}{{3\sqrt x }}\\
b)Thay:x = 7 + 4\sqrt 3 = 4 + 2.2.\sqrt 3 + 3\\
= {\left( {2 + \sqrt 3 } \right)^2}\\
\to A = \dfrac{{\sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} – 2}}{{3\sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} }}\\
= \dfrac{{2 + \sqrt 3 – 2}}{{3\left( {2 + \sqrt 3 } \right)}}\\
= \dfrac{{\sqrt 3 }}{{6 + 3\sqrt 3 }} = \dfrac{{ – 3 + 2\sqrt 3 }}{3}
\end{array}\)