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ĐKXĐ: \(x\ge0,x\ne9\)
a) \(P=\frac{3\sqrt{x}+2}{\sqrt{x}+1}+\frac{2\sqrt{x}+3}{\sqrt{x}-3}-\frac{3\left(3\sqrt{x}-5\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x-3}\right)}\)
\(=\frac{\left(3\sqrt{x}+2\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)+3\left(3\sqrt{x}-5\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{3x-9\sqrt{x}+2\sqrt{x}-6+2x+2\sqrt{x}-3\sqrt{x}-3-9\sqrt{x}+15}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{5x-17\sqrt{x}+6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{5x-15\sqrt{x}-2\sqrt{x}+6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\left(5\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\frac{5\sqrt{x}-2}{\sqrt{x}+1}\)
b) Ta có: \(x=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\Rightarrow\sqrt{x}=\sqrt{3}+1\)
Do đó: \(P=\frac{5\left(\sqrt{3}+1\right)-2}{\left(\sqrt{3}+1\right)+1}=\frac{5\sqrt{3}+3}{\sqrt{3}+2}=\frac{\left(5\sqrt{3}+3\right)\left(2-\sqrt{3}\right)}{\left(\sqrt{3}+2\right)\left(2-\sqrt{3}\right)}=7\sqrt{3}-9\)
c) Ta có \(P=\frac{5\sqrt{x}-2}{\sqrt{x}+1}=\frac{5\sqrt{x}+5-7}{\sqrt{x}+1}\)
\(P=5-\frac{7}{\sqrt{x}+1}\)
Vì \(\frac{7}{\sqrt{x}+1}>0\)nên \(P\)có giá trị nhỏ nhất khi và chỉ khi \(\frac{7}{\sqrt{x}+1}\)lớn nhất
\(\Leftrightarrow\sqrt{x}+1\)nhỏ nhất \(\Leftrightarrow x=0\)
Khi đó minP=5-7=-2
ĐK \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)
a, \(R=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\frac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\frac{3x-6\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}=\frac{3\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\frac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}\)
b. \(R< -1\Rightarrow R+1< 0\Rightarrow\frac{3\sqrt{x}-9+\sqrt{x}+3}{\sqrt{x}+3}< 0\Rightarrow\frac{4\sqrt{x}-6}{\sqrt{x}+3}< 0\)
\(\Rightarrow0\le x< \frac{9}{4}\)
c. \(R=\frac{3\left(\sqrt{x}-3\right)}{\sqrt{x}+3}=3+\frac{-18}{\sqrt{x}+3}\)
Ta thấy \(\sqrt{x}+3\ge3\Rightarrow\frac{-18}{\sqrt{x}+3}\ge-6\Rightarrow3+\frac{-18}{\sqrt{x}+3}\ge-3\Rightarrow R\ge-3\)
Vậy \(MinR=-3\Leftrightarrow x=0\)
a) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\)
\(=\left[-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\cdot\left(\sqrt{2}-\sqrt{5}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\)
\(=-\left(2-5\right)\)
\(=-\left(-3\right)\)
\(=3\)
b) Ta có:
\(x^2-x\sqrt{3}+1\)
\(=x^2-2\cdot\dfrac{\sqrt{3}}{2}\cdot x+\left(\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
\(=\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\)
Mà: \(\left(x-\dfrac{\sqrt{3}}{2}\right)^2\ge0\forall x\) nên
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}\ge\dfrac{1}{4}\forall x\)
Dấu "=" xảy ra:
\(\left(x-\dfrac{\sqrt{3}}{2}\right)^2+\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow x=\dfrac{\sqrt{3}}{2}\)
Vậy: GTNN của biểu thức là \(\dfrac{1}{4}\) tại \(x=\dfrac{\sqrt{3}}{2}\)
a)
\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{\sqrt{5}-5}{1-\sqrt{5}}\right):\dfrac{1}{\sqrt{2}-\sqrt{5}}\\ =\left(-\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =\left(-\sqrt{2}-\sqrt{5}\right).\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}+\sqrt{5}\right)\left(\sqrt{2}-\sqrt{5}\right)\\ =-\left(\sqrt{2}^2-\sqrt{5}^2\right)\\ =-\left(2-5\right)\\ =-\left(-3\right)\\ =3\)
\(P=\frac{x-9+25}{\sqrt{x}+3}=\sqrt{x}-3+\frac{25}{\sqrt{x}+3}=\sqrt{x}+3+\frac{25}{\sqrt{x}+3}-6\)
\(P\ge2\sqrt{\frac{25\left(\sqrt{x}+3\right)}{\sqrt{x}+3}}-6=4\)
\(P_{min}=4\) khi \(\sqrt{x}+3=5\Rightarrow x=4\)
\(a^3=4+3a.\sqrt[3]{2^2-3}=4+3a\)
\(\Rightarrow a^3-3a=4\)
\(\Leftrightarrow a\left(a^2-3\right)=4\Rightarrow a^2-3=\frac{4}{a}\)
\(\Rightarrow\frac{64}{\left(a^2-3\right)^2}-3a=\frac{64}{\left(\frac{4}{a}\right)^3}-3a=a^3-3a=4\)