Toán Lớp 9: Cho A = √2-2 / √x
a, tính A khi x= 6 – 4 √2
b, tìm x khi biết A ≥ 2
c, tìm x khi biết A = √x -2
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Toán Lớp 9: Cho A = √2-2 / √x
a, tính A khi x= 6 – 4 √2
b, tìm x khi biết A ≥ 2
c, tìm x khi biết A = √x -2
Comments ( 1 )
a)Dkxd:x > 0\\
A = \dfrac{{\sqrt 2 – 2}}{{\sqrt x }}\\
x = 6 – 4\sqrt 2 \left( {tm} \right)\\
= 4 – 2.2.\sqrt 2 + 2\\
= {\left( {2 – \sqrt 2 } \right)^2}\\
\Leftrightarrow \sqrt x = \sqrt {{{\left( {2 – \sqrt 2 } \right)}^2}} = 2 – \sqrt 2 \\
\Leftrightarrow A = \dfrac{{\sqrt 2 – 2}}{{2 – \sqrt 2 }} = – 1\\
b)A \ge 2\\
\Leftrightarrow \dfrac{{\sqrt 2 – 2}}{{\sqrt x }} \ge 2\\
\Leftrightarrow \dfrac{{\sqrt 2 – 2}}{{\sqrt x }} – 2 \ge 0\\
\Leftrightarrow \dfrac{{\sqrt 2 – 2 – 2\sqrt x }}{{\sqrt x }} \ge 0\\
\Leftrightarrow \sqrt 2 – 2 – 2\sqrt x \ge 0\\
\Leftrightarrow 2\sqrt x \le \sqrt 2 – 2\\
\Leftrightarrow \sqrt x \le \dfrac{{\sqrt 2 – 2}}{2}\\
Do:\sqrt 2 – 2 < 0\\
\Leftrightarrow \sqrt x \le \dfrac{{\sqrt 2 – 2}}{2} < 0\left( {ktm} \right)
\end{array}$
c)A = \sqrt x – 2\\
\Leftrightarrow \dfrac{{\sqrt 2 – 2}}{{\sqrt x }} = \sqrt x – 2\\
\Leftrightarrow \sqrt 2 – 2 = \sqrt x \left( {\sqrt x – 2} \right)\\
\Leftrightarrow x – 2\sqrt x + 2 – \sqrt 2 = 0\\
\Leftrightarrow \left[ \begin{array}{l}
\sqrt x = 1,64\\
\sqrt x = 0,356
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 2,689\\
x = 0,1267
\end{array} \right.\left( {tmdk} \right)\\
Vậy\,x = 2,689;x = 0,1267
\end{array}$