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1.
\(A=-3x+2\sqrt{x}+6\\ =-3\left(x-\frac{2}{3}\sqrt{x}-2\right)\\ =-3\left[\left(\sqrt{x}\right)^2-2\cdot\sqrt{x}\cdot\frac{1}{3}+\left(\frac{1}{3}\right)^2-\frac{19}{9}\right]\\ =-3\left[\left(\sqrt{x}-\frac{1}{3}\right)^2\right]+\frac{19}{3}\le\frac{19}{3}\forall x\ge0\)
Vậy Max A = \(\frac{19}{3}\Leftrightarrow x=\frac{1}{9}\)
a: \(=-xy\cdot\dfrac{\sqrt{xy}}{x}=-y\sqrt{yx}\)
b: \(=\sqrt{\dfrac{-105x^3}{35^2}}=\sqrt{-105x}\cdot\dfrac{x}{35}\)
c: \(=\sqrt{\dfrac{5a^3b}{49b^2}}=\sqrt{5ab}\cdot\dfrac{a}{7b}\)
d: \(=-7xy\cdot\dfrac{\sqrt{3}}{\sqrt{xy}}=-7\sqrt{3}\cdot\sqrt{xy}\)
a. \(\sqrt{\dfrac{2}{3}}=\sqrt{\dfrac{2.3}{3^2}}=\dfrac{1}{3}.\sqrt{6}\)
b. \(\sqrt{\dfrac{x^2}{5}}=\sqrt{\dfrac{5x^2}{5^2}}=\dfrac{x}{5}.\sqrt{5}\) (vì x \(\ge\) 0)
c. \(\sqrt{\dfrac{3}{x}}=\sqrt{\dfrac{3.x}{x^2}}=\dfrac{1}{x}.\sqrt{3x}\) (vì x > 0)
d. \(\sqrt{x^2-\dfrac{x^2}{7}}=\sqrt{\dfrac{6x^2}{7}}=\sqrt{\dfrac{6x^2.7}{7.7}}=\sqrt{\dfrac{42.x^2}{7^2}}=-\dfrac{x}{7}.\sqrt{42}\) (vì x < 0)
Đặt \(\sqrt{x}+4=t\left(t\ge4\right)\)
\(\Rightarrow P=\dfrac{7-3\left(t-4\right)}{t}\)
\(\Leftrightarrow P=\dfrac{7+12-3t}{t}=\dfrac{19-3t}{t}\)
\(\Leftrightarrow P=\dfrac{19}{t}-3\)
Mà \(t\ge4\)
\(\Rightarrow P\le\dfrac{19}{4}-3\)
\(\Leftrightarrow P\le\dfrac{7}{4}\)
Dấu "=" xảy ra khi x = 0 (thoả mãn)
Vậy GTLN của P là \(\dfrac{7}{4}\) khi x = 0 .
c: \(=\sqrt{\dfrac{4}{16-6\sqrt{7}}}+\sqrt{7}\)
\(=\dfrac{2}{3-\sqrt{7}}+\sqrt{7}\)
\(=3+2\sqrt{7}\)
d: \(=\dfrac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{x-4}\)
\(=\dfrac{3x-6\sqrt{x}}{x-4}=\dfrac{3\sqrt{x}}{\sqrt{x}+2}\)
@Mysterious Person
\(D=\dfrac{3\sqrt{x}+7}{\sqrt{x}+2}=\dfrac{3\sqrt{x}+6+1}{\sqrt{x}+2}=3+\dfrac{1}{\sqrt{x}+2}\le3+\dfrac{1}{2}=\dfrac{7}{2}\)
ta có : \(D=\dfrac{3\sqrt{x}+7}{\sqrt{x}+2}=\dfrac{3\sqrt{x}+6+1}{\sqrt{x}+2}=3+\dfrac{1}{\sqrt{x}+2}\ge3+\dfrac{1}{2}=\dfrac{7}{2}\)
dấu "=" xảy ra \(\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\) vậy .......................................................