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a.\(\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right)=\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}.\left(\sqrt{a}-\sqrt{b}\right)=a-b\)
b.\(DK:x\ge0\)
\(\frac{x+4\sqrt{x}+4}{2+\sqrt{x}}=\frac{\left(\sqrt{x}+2\right)^2}{\sqrt{x}+2}=\sqrt{x}+2\)
1. ĐK \(\hept{\begin{cases}x\ge0\\x\ne4\end{cases}}\)
a. Ta có \(R=\left(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right).\left(\frac{1}{\sqrt{x}+2}+\frac{4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)
\(=\frac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}.\frac{\sqrt{x}-2+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\frac{\sqrt{x}+2}{\sqrt{x}}.\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
b. Với \(x=4+2\sqrt{3}\Rightarrow R=\frac{\sqrt{4+2\sqrt{3}}+2}{\sqrt{4+2\sqrt{3}}\left(\sqrt{4+2\sqrt{3}}-2\right)}=\frac{\sqrt{\left(\sqrt{3}+1\right)^2}+2}{\sqrt{\left(\sqrt{3}+1\right)^2}\left(\sqrt{\left(\sqrt{3}+1\right)^2}-2\right)}\)
\(=\frac{\sqrt{3}+1+2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{\sqrt{3}+3}{3-1}=\frac{\sqrt{3}+3}{2}\)
c. Để \(R>0\Rightarrow\frac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}>0\Rightarrow\sqrt{x}-2>0\Rightarrow x>4\)
Vậy \(x>4\)thì \(R>0\)
2. Ta có \(A=6+2\sqrt{2}=6+\sqrt{8};B=9=6+3=6+\sqrt{9}\)
Vì \(\sqrt{8}< \sqrt{9}\Rightarrow A< B\)
3. a. \(VT=\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\left(\sqrt{a}+\sqrt{b}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right).\left(\sqrt{a}+\sqrt{b}\right)=a-b=VP\left(đpcm\right)\)
b. Ta có \(VT=\left(2+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right).\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(2+\sqrt{a}\right)\left(2-\sqrt{a}\right)=4-a=VP\left(đpcm\right)\)
Lời giải:
Áp dụng BĐT Bunhiacopxky:
\(\frac{a+b}{2}=\frac{(a+b)(1+1)}{4}\geq \frac{(\sqrt{a}+\sqrt{b})^2}{4}\)
Mà: \(\frac{(\sqrt{a}+\sqrt{b})^2}{4}=\frac{[(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})]^2}{4(\sqrt{a}-\sqrt{b})^2}=\frac{(a-b)^2}{4(\sqrt{a}-\sqrt{b})^2}\)
Do đó: \(\frac{a+b}{2}\geq \frac{(a-b)^2}{4(\sqrt{a}-\sqrt{b})^2}\)
Dấu "=" xảy ra khi \(\frac{\sqrt{a}}{1}=\frac{\sqrt{b}}{1}\Leftrightarrow a=b\) (sai vì $a\neq b$). Do đó dấu "=" không xảy ra, hay \(\frac{a+b}{2}> \frac{(a-b)^2}{4(\sqrt{a}-\sqrt{b})^2}\)
Mặt khác:
\(\frac{(\sqrt{a}+\sqrt{b})^2}{4}=\frac{a+b+2\sqrt{ab}}{4}\geq \frac{2\sqrt{ab}+2\sqrt{ab}}{4}=\sqrt{ab}\)
\(\Leftrightarrow \frac{(a-b)^2}{4(\sqrt{a}-\sqrt{b})^2}\geq \sqrt{ab}\)
Dấu "=" xảy ra khi \(a=b\) (sai do $a\neq b$). Do đó dấu "=" không xảy ra, hay \( \frac{(a-b)^2}{4(\sqrt{a}-\sqrt{b})^2}> \sqrt{ab}\)
Ta có đpcm.