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\(A=\frac{2\sqrt{x}+x-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}×\frac{x+\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\frac{1}{\sqrt{x}+2}\)
A đạt GTLN khi \(2+\sqrt{x}\)đạt GTNN hay x là nhỏ nhất. Vậy A đạt GTLN là \(\frac{1}{2}\)khi x = 0
a) điều kiện xác định : \(x\ge2;x\ne5\)
b) \(P=\dfrac{x-5}{\sqrt{x-2}-\sqrt{3}}=\dfrac{\left(\sqrt{x-2}-\sqrt{3}\right)\left(\sqrt{x-2}+\sqrt{3}\right)}{\sqrt{x-2}-\sqrt{3}}\)
\(\Leftrightarrow P=\sqrt{x-2}+\sqrt{3}\)
c) ta có : \(P=\sqrt{x-2}+\sqrt{3}\ge\sqrt{3}\) \(\Rightarrow\) GTNN của \(P\) là \(\sqrt{3}\)
dấu "=" xảy ra khi \(x=2\)
Bài 1:
\(A=\sqrt{8}-2\sqrt{2}+\sqrt{20}-2\sqrt{5}-2=2\sqrt{2}-2\sqrt{2}+2\sqrt{5}-2\sqrt{5}-2=-2\)\(B=\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}}\)
\(a.P=\dfrac{1}{\sqrt{x}+2}-\dfrac{5}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}-2}{3-\sqrt{x}}=\dfrac{\sqrt{x}-3-5+\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}-8+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+4}{\sqrt{x}+2}\) ( x ≥ 0 ; x # 9 )
\(b.\) \(P=\dfrac{\sqrt{x}+4}{\sqrt{x}+2}=\dfrac{2\left(\sqrt{x}+2\right)-\sqrt{x}}{\sqrt{x}+2}=2-\dfrac{\sqrt{x}}{\sqrt{x}+2}\text{≤}2\)
⇒ \(P_{Max}=2."="\) ⇔ \(x=0\)
đkxđ: x≥0; x≠4
\(A=\dfrac{1}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{4-x}\)
\(=\dfrac{2-\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}+\dfrac{2+\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}-\dfrac{2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}\)
\(=\dfrac{4-2\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}=\dfrac{2\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}=\dfrac{2}{2+\sqrt{x}}\)
+) A = 1/4 <=> \(\dfrac{2}{2+\sqrt{x}}=\dfrac{1}{4}\Leftrightarrow2+\sqrt{x}=8\Leftrightarrow\sqrt{x}=6\Leftrightarrow x=36\)(tm)
Vậy x = 36
đkxđ \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)
\(A=\dfrac{2+\sqrt{x}+2-\sqrt{x}-2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)}\)
\(A=\dfrac{4-2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(2-\sqrt{x}\right)}\)
\(A=\dfrac{2}{\sqrt{x}+2}\)
để \(A=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{1}{4}\)
\(\Leftrightarrow\sqrt{x}+2=8\)
\(\Leftrightarrow x=36\left(tm\right)\)
vậy tại x=36 thì A=1/4
ĐKXĐ: ...
\(P=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{2}{x}-\frac{2-x}{x\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\frac{2\left(\sqrt{x}+1\right)-2+x}{x\left(\sqrt{x}+1\right)}\right)\)
\(=\frac{\left(x+2\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\frac{x\left(\sqrt{x}+1\right)}{\left(x+2\sqrt{x}\right)}=\frac{x}{\sqrt{x}-1}\)
\(x=\frac{2}{2-\sqrt{3}}=\frac{4}{4-2\sqrt{3}}=\left(\frac{2}{\sqrt{3}-1}\right)^2\)
\(\Rightarrow P=\frac{\frac{2}{2-\sqrt{3}}}{\frac{2}{\sqrt{3}-1}-1}=\frac{\frac{2}{2-\sqrt{3}}}{\frac{3-\sqrt{3}}{\sqrt{3}-1}}=\frac{2}{2\sqrt{3}-3}\)
\(\sqrt{P}\) xác định khi \(x>1\)
Khi đó: \(\sqrt{P}=\sqrt{\frac{x}{\sqrt{x}-1}}=\sqrt{\frac{x}{\sqrt{x}-1}-4+4}=\sqrt{\frac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-1}+4}\ge2\)
\(\sqrt{P}_{min}=2\) khi \(x=4\)
\(=\left(\frac{1}{\sqrt{x}}+\frac{\sqrt{x}}{\sqrt{x}+1}\right):\frac{\sqrt{x}}{x+\sqrt{x}}\)
\(=\left(\frac{\sqrt{x}+1+x}{\sqrt{x}(\sqrt{x}+1)}\right):\frac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}+1+x}{\sqrt{x}(\sqrt{x}+1)}\cdot\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)
\(B=\left(\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\frac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}\right).\left(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}\right)\)
\(=\frac{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)
\(x=4\Rightarrow B=\frac{4+2+1}{2}=\frac{7}{2}\)
\(B=\sqrt{x}+\frac{1}{\sqrt{x}}+1\ge2\sqrt{\frac{\sqrt{x}}{\sqrt{x}}}+1=3\)
\(B_{min}=3\) khi \(x=1\)
a: Ta có: \(A=\dfrac{1}{\sqrt{x}+1}-\dfrac{x+2}{x\sqrt{x}+1}\)
\(=\dfrac{x-\sqrt{x}+1-x-2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)
\(=\dfrac{-1}{x-\sqrt{x}+1}\)