Toán Lớp 9: 1.Cho bt A= √x +1/ √x -1; B= √x/ √x +1- √x -1/x -1 vs x ≥ 0, x ≠ 1.
a) Tìm x để A= 2.
b) Cm: A,B không phụ thuộc x.
c) Tìm x để A ≥ B.
2. Cho bt A= 8 -x/2 + ∛x : (2+ ∛x²/2 + ∛x) +( ∛x + 2∛x/ ∛x -2) . ∛x² -4/ ∛x² +2∛x .Vs x ≠ ±8, x ≠0.
Cm: A không phụ thuộc x
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Comments ( 1 )
1)a)A = 2\\
\Leftrightarrow \dfrac{{\sqrt x + 1}}{{\sqrt x – 1}} = 2\\
\Leftrightarrow \sqrt x + 1 = 2\sqrt x – 2\\
\Leftrightarrow \sqrt x = 3\\
\Leftrightarrow x = 9\left( {tmdk} \right)\\
Vậy\,x = 9\\
b)A.B\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x – 1}}.\left( {\dfrac{{\sqrt x }}{{\sqrt x + 1}} – \dfrac{{\sqrt x – 1}}{{x – 1}}} \right)\\
= \dfrac{{\sqrt x + 1}}{{\sqrt x – 1}}.\dfrac{{\sqrt x \left( {\sqrt x – 1} \right) – \sqrt x + 1}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x – 1} \right)}}\\
= \dfrac{1}{{\sqrt x – 1}}.\dfrac{{x – 2\sqrt x + 1}}{{\sqrt x – 1}}\\
= \dfrac{{{{\left( {\sqrt x – 1} \right)}^2}}}{{{{\left( {\sqrt x – 1} \right)}^2}}}\\
= 1
\end{array}$
c)A \ge B\\
\Leftrightarrow \dfrac{{\sqrt x + 1}}{{\sqrt x – 1}} \ge \dfrac{{{{\left( {\sqrt x – 1} \right)}^2}}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}}\\
\Leftrightarrow \dfrac{{\sqrt x + 1}}{{\sqrt x – 1}} – \dfrac{{{{\left( {\sqrt x – 1} \right)}^2}}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 1} \right)}} \ge 0\\
\Leftrightarrow \dfrac{{\sqrt x + 1}}{{\sqrt x – 1}} – \dfrac{{\sqrt x – 1}}{{\sqrt x + 1}} \ge 0\\
\Leftrightarrow \dfrac{{{{\left( {\sqrt x + 1} \right)}^2} – {{\left( {\sqrt x – 1} \right)}^2}}}{{x – 1}} \ge 0\\
\Leftrightarrow \dfrac{{4\sqrt x }}{{x – 1}} \ge 0\\
\Leftrightarrow x – 1 > 0\\
\Leftrightarrow x > 1\\
Vậy\,x > 1
\end{array}$