Toán Lớp 9: Cho biểu thức: P=(4 căn x/ căn x + 2 + 8x/4 – x): ( căn x – 1/x – 2 căn x – 2/căn x)
a) Tìm điều kiện của x để P có nghĩa
b) Rút gọn P
c) Tính P khi x = 4 – 2 căn 3
d) Tìm x thuộc Z để P thuộc Z
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Toán Lớp 9: Cho biểu thức: P=(4 căn x/ căn x + 2 + 8x/4 – x): ( căn x – 1/x – 2 căn x – 2/căn x)
a) Tìm điều kiện của x để P có nghĩa
b) Rút gọn P
c) Tính P khi x = 4 – 2 căn 3
d) Tìm x thuộc Z để P thuộc Z
Comments ( 1 )
a)x > 0;x \ne 4\\
b)P = \dfrac{{4x}}{{\sqrt x – 3}}\\
c)\dfrac{{16 – 8\sqrt 3 }}{{\sqrt 3 – 4}}\\
d)\left[ \begin{array}{l}
x = 1521\\
x = 441\\
x = 225\\
x = 144\\
x = 49\\
x = 36\\
x = 25\\
x = 16
\end{array} \right.
\end{array}\)
a)DK:x > 0;x \ne 4\\
b)P = \left( {\dfrac{{4\sqrt x }}{{\sqrt x + 2}} + \dfrac{{8x}}{{4 – x}}} \right):\left( {\dfrac{{\sqrt x – 1}}{{x – 2\sqrt x }} – \dfrac{2}{{\sqrt x }}} \right)\\
= \dfrac{{4\sqrt x \left( {\sqrt x – 2} \right) – 8x}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 2} \right)}}:\dfrac{{\sqrt x – 1 – 2\left( {\sqrt x – 2} \right)}}{{\sqrt x \left( {\sqrt x – 2} \right)}}\\
= \dfrac{{4x – 8\sqrt x – 8x}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x – 2} \right)}}{{ – \sqrt x + 3}}\\
= \dfrac{{ – 4x – 8\sqrt x }}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x – 2} \right)}}{{ – \sqrt x + 3}}\\
= \dfrac{{4\sqrt x \left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x – 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x – 2} \right)}}{{\sqrt x – 3}}\\
= \dfrac{{4x}}{{\sqrt x – 3}}\\
c)Thay:x = 4 – 2\sqrt 3 \\
= 3 – 2\sqrt 3 .1 + 1\\
= {\left( {\sqrt 3 – 1} \right)^2}\\
\to P = \dfrac{{4\left( {4 – 2\sqrt 3 } \right)}}{{\sqrt {{{\left( {\sqrt 3 – 1} \right)}^2}} – 3}} = \dfrac{{16 – 8\sqrt 3 }}{{\sqrt 3 – 1 – 3}}\\
= \dfrac{{16 – 8\sqrt 3 }}{{\sqrt 3 – 4}}\\
d)P = \dfrac{{4x}}{{\sqrt x – 3}} = \dfrac{{4\sqrt x \left( {\sqrt x – 3} \right) + 12\sqrt x }}{{\sqrt x – 3}}\\
= 4\sqrt x + \dfrac{{12\left( {\sqrt x – 3} \right) + 36}}{{\sqrt x – 3}}\\
= 4\sqrt x + 12 + \dfrac{{36}}{{\sqrt x – 3}}\\
P \in Z \to \dfrac{{36}}{{\sqrt x – 3}} \in Z\\
\to \sqrt x – 3 \in U\left( {36} \right)\\
\to \left[ \begin{array}{l}
\sqrt x – 3 = 36\\
\sqrt x – 3 = 18\\
\sqrt x – 3 = 12\\
\sqrt x – 3 = 9\\
\sqrt x – 3 = 4\\
\sqrt x – 3 = 3\\
\sqrt x – 3 = – 3\\
\sqrt x – 3 = 2\\
\sqrt x – 3 = – 2\\
\sqrt x – 3 = 1\\
\sqrt x – 3 = – 1
\end{array} \right. \to \left[ \begin{array}{l}
\sqrt x = 39\\
\sqrt x = 21\\
\sqrt x = 15\\
\sqrt x = 12\\
\sqrt x = 7\\
\sqrt x = 6\\
\sqrt x = 0\left( l \right)\\
\sqrt x = 5\\
\sqrt x = 4\\
\sqrt x = 2\left( l \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 1521\\
x = 441\\
x = 225\\
x = 144\\
x = 49\\
x = 36\\
x = 25\\
x = 16
\end{array} \right.
\end{array}\)