Tìm giá trị của A= \(\frac{2b\sqrt{x^2-1}}{x-\sqrt{x^2-1}}\) với \(x=\frac{1}{2}\left(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\right)\)và a>b>0
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27 tháng 4 2019
\(a,A=\sqrt{27}+\frac{2}{\sqrt{3}-2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)
\(=3\sqrt{3}+\frac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\left(\sqrt{3}-1\right)\)
\(=3\sqrt{3}+\frac{2\sqrt{3}+4}{3-4}-\sqrt{3}+1\)
\(=3\sqrt{3}-2\sqrt{3}-4-\sqrt{3}+1\)
\(=-3\)
\(B=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}}\)
b, Ta có \(B< A\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}< -3\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}+3< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-1+3\sqrt{x}}{\sqrt{x}}< 0\)
\(\Leftrightarrow\frac{4\sqrt{x}-1}{\sqrt{x}}< 0\)
\(\Leftrightarrow4\sqrt{x}-1< 0\left(Do\sqrt{x}>0\right)\)
\(\Leftrightarrow\sqrt{x}< \frac{1}{4}\)
\(\Leftrightarrow0< x< \frac{1}{2}\)(Kết hợp ĐKXĐ)
Vậy ...
3 tháng 8 2017
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 \(VT=\left(2+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right).\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(2+\sqrt{a}\right)\left(2-\sqrt{a}\right)=4-a=VP\left(đpcm\right)\)
14 tháng 5 2021
1) Khi x = 49 thì:
\(A=\frac{4\sqrt{49}}{\sqrt{49}-1}=\frac{4\cdot7}{7-1}=\frac{28}{6}=\frac{14}{3}\)
2) Ta có:
\(B=\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-1}\)
\(B=\frac{\sqrt{x}-1+x+\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
c) \(P=A\div B=\frac{4\sqrt{x}}{\sqrt{x}-1}\div\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{4\sqrt{x}}{\sqrt{x}+1}\)
Ta có: \(P\left(\sqrt{x}+1\right)=x+4+\sqrt{x-4}\)
\(\Leftrightarrow\frac{4\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}=x+4+\sqrt{x-4}\)
\(\Leftrightarrow4\sqrt{x}=x+4+\sqrt{x-4}\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\sqrt{x-4}=0\)
Mà \(VT\ge0\left(\forall x\ge0,x\ne1\right)\)
\(\Rightarrow\hept{\begin{cases}\left(\sqrt{x}-2\right)^2=0\\\sqrt{x-4}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}=2\\x-4=0\end{cases}}\Rightarrow x=4\)
Vậy x = 4
Giao luu:
\(x-\sqrt{x^2-1}\ne0\Rightarrow A.xacdinh.moi.x\)
\(0\le\left(\sqrt[4]{\frac{a}{b}}-\sqrt[4]{\frac{b}{a}}\right)^2=\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}-2\Rightarrow\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\ge2\)
\(\Rightarrow x\ge1\Rightarrow\hept{\begin{cases}x-1\ge0\\x^2-1\ge0\end{cases}\left(1\right)}\)
\(A=\frac{2b\sqrt{x^2-1}}{x-\sqrt{x^2-1}}=\frac{2b\sqrt{x^2-1}\left(x+\sqrt{x^2-1}\right)}{x^2-\left(x^2-1\right)}=2b\sqrt{x^2-1}\left(x+\sqrt{x^2-1}\right)\)
\(\frac{A}{b}=2x\sqrt{x^2-1}+2\sqrt{\left(x^2-1\right)^2}=\left(x^2+2x\sqrt{x^2-1}+\sqrt{\left(x^2-1\right)^2}\right)-1\)
\(\frac{A}{b}+1=\left(x+\sqrt{x^2-1}\right)^2=\frac{1}{2}\left(x+1+2\sqrt{\left(x-1\right)\left(x+1\right)}+x-1\right)\)
\(\frac{A}{2b}+1=\left(\sqrt{x-1}+\sqrt{x+1}\right)^2=\left(\frac{\sqrt{2x-2}+\sqrt{2x+2}}{\sqrt{2}}\right)^2\)
\(2\left(\frac{A}{2b}+1\right)=\left[\sqrt{\left(\sqrt[4]{\frac{a}{b}}-\sqrt[4]{\frac{b}{a}}\right)^2}+\sqrt{\left(\sqrt[4]{\frac{a}{b}}+\sqrt[4]{\frac{b}{a}}\right)^2}\right]^{^2}\)
\(2\left(\frac{A}{2b}+1\right)=\left[\left(\sqrt[4]{\frac{a}{b}}-\sqrt[4]{\frac{b}{a}}\right)+\left(\sqrt[4]{\frac{a}{b}}+\sqrt[4]{\frac{b}{a}}\right)\right]^{^2}=4\sqrt{\frac{a}{b}}\)
\(\frac{A}{2b}+1=2\sqrt{\frac{a}{b}}\)
\(A=4b\sqrt{\frac{a}{b}}-2b=4\sqrt{ab}-2b\)(hoa hết mắt có khi (+-,*/,) nhầm vì số liệu chưa đẹp...hihi)
Ta có:
\(x=\frac{1}{2}\left(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\right)=\frac{a+b}{2\sqrt{ab}}\)
\(x^2-1=\left(\frac{a+b}{2\sqrt{ab}}\right)^2-1=\frac{a^2+2ab+b^2-4ab}{4ab}\)
\(=\frac{\left(a-b\right)^2}{4ab}\)
Từ đây ta có
\(A=\frac{2b\sqrt{x^2-1}}{x-\sqrt{x^2-1}}=\frac{2b\sqrt{\frac{\left(a-b\right)^2}{4ab}}}{\frac{a+b}{2\sqrt{ab}}-\sqrt{\frac{\left(a-b\right)^2}{4ab}}}\)
\(=\frac{2b.\frac{a-b}{2\sqrt{ab}}}{\frac{a+b}{2\sqrt{ab}}-\frac{a-b}{2\sqrt{ab}}}=\frac{2ab-2b^2}{2b}=a-b\)