Toán Lớp 9: P= √x-3/√x-2 + √x-4/√x+3 + 5/x+√x-6
a, Tìm điều kiện xác định
b, Tìm x để x=1 và x=19+6 √2
c, Tìm x để P=1/3
d, Tìm x để |P| ≥ P
e, Tìm GTNN của P
f, Tìm x để P có giá trị nguyên
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Toán Lớp 9: P= √x-3/√x-2 + √x-4/√x+3 + 5/x+√x-6
a, Tìm điều kiện xác định
b, Tìm x để x=1 và x=19+6 √2
c, Tìm x để P=1/3
d, Tìm x để |P| ≥ P
e, Tìm GTNN của P
f, Tìm x để P có giá trị nguyên
Comments ( 1 )
a)x \ge 0;x \ne 4\\
b)18 – 12\sqrt 2 \\
c)x = \dfrac{{81}}{{25}}\\
d)0 \le x \le 1\\
e)Min = – \dfrac{2}{3}\\
f)\left[ \begin{array}{l}
x = 25\\
x = 1
\end{array} \right.
\end{array}\)
a)DK:x \ge 0;x \ne 4\\
b)P = \dfrac{{\sqrt x – 3}}{{\sqrt x – 2}} + \dfrac{{\sqrt x – 4}}{{\sqrt x + 3}} + \dfrac{5}{{x + \sqrt x – 6}}\\
= \dfrac{{\left( {\sqrt x – 3} \right)\left( {\sqrt x + 3} \right) + \left( {\sqrt x – 4} \right)\left( {\sqrt x – 2} \right) + 5}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{x – 9 + x – 6\sqrt x + 8 + 5}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2x – 6\sqrt x + 4}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2\left( {\sqrt x – 2} \right)\left( {\sqrt x – 1} \right)}}{{\left( {\sqrt x – 2} \right)\left( {\sqrt x + 3} \right)}}\\
= \dfrac{{2\left( {\sqrt x – 1} \right)}}{{\sqrt x + 3}}\\
Thay:x = 1\\
\to P = \dfrac{{2\left( {\sqrt 1 – 1} \right)}}{{\sqrt 1 + 3}} = 0\\
Thay:x = 19 + 6\sqrt 2 \\
= 18 + 2.3\sqrt 2 .1 + 1\\
= {\left( {3\sqrt 2 + 1} \right)^2}\\
\to P = \dfrac{{2\left( {\sqrt {{{\left( {3\sqrt 2 + 1} \right)}^2}} – 1} \right)}}{{\sqrt {{{\left( {3\sqrt 2 + 1} \right)}^2}} + 3}}\\
= \dfrac{{2\left( {3\sqrt 2 + 1 – 1} \right)}}{{3\sqrt 2 + 1 + 3}}\\
= \dfrac{{6\sqrt 2 }}{{3\sqrt 2 + 4}}\\
= 18 – 12\sqrt 2 \\
c)P = \dfrac{1}{3}\\
\to \dfrac{{2\left( {\sqrt x – 1} \right)}}{{\sqrt x + 3}} = \dfrac{1}{3}\\
\to 6\left( {\sqrt x – 1} \right) = \sqrt x + 3\\
\to 5\sqrt x = 9\\
\to \sqrt x = \dfrac{9}{5}\\
\to x = \dfrac{{81}}{{25}}\\
d)\left| P \right| \ge P\\
\to \left[ \begin{array}{l}
P \ge P\left( {ld} \right)\\
P \le – P
\end{array} \right.\\
\to 2P \le 0\\
\to P \le 0\\
\to \dfrac{{2\left( {\sqrt x – 1} \right)}}{{\sqrt x + 3}} \le 0\\
\to \sqrt x – 1 \le 0\left( {do:\sqrt x + 3 > 0\forall x \ge 0} \right)\\
\to 0 \le x \le 1\\
e)P = \dfrac{{2\left( {\sqrt x – 1} \right)}}{{\sqrt x + 3}} = \dfrac{{2\left( {\sqrt x + 3} \right) – 8}}{{\sqrt x + 3}}\\
= 2 – \dfrac{8}{{\sqrt x + 3}}\\
Do:\sqrt x \ge 0\forall x \ge 0\\
\to \sqrt x + 3 \ge 3\\
\to \dfrac{8}{{\sqrt x + 3}} \le \dfrac{8}{3}\\
\to – \dfrac{8}{{\sqrt x + 3}} \ge – \dfrac{8}{3}\\
\to 2 – \dfrac{8}{{\sqrt x + 3}} \ge – \dfrac{2}{3}\\
\to Min = – \dfrac{2}{3}\\
\Leftrightarrow x = 0\\
f)P \in Z \to \dfrac{8}{{\sqrt x + 3}} \in Z\\
\to \sqrt x + 3 \in U\left( 8 \right)\\
\to \left[ \begin{array}{l}
\sqrt x + 3 = 8\\
\sqrt x + 3 = 4
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt x = 5\\
\sqrt x = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
x = 25\\
x = 1
\end{array} \right.
\end{array}\)