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=(\(\frac{\sqrt{a-b}\left(\sqrt{a+b}-\sqrt{a-b}\right)}{\left(\sqrt{a+b}+\sqrt{a-b}\right)\left(\sqrt{a+b}-\sqrt{a-b}\right)}\)+\(\frac{a-b}{\sqrt{a-b}\left(\sqrt{a+b}-\sqrt{a-b}\right)}\)):\(\frac{\sqrt{a^2-b^2}}{a^2+b^2}\)
=(\(\frac{\sqrt{a^2-b^2}-\left(a-b\right)}{a+b-a+b}+\frac{\sqrt{a^2-b^2}+a-b}{a+b-a+b}\)):\(\frac{\sqrt{a^2-b^2}}{a^2+b^2}\)
=\(\frac{2\sqrt{a^2-b^2}}{2b}\):\(\frac{\sqrt{a^2-b^2}}{a^2+b^2}\)
=\(\frac{\sqrt{a^2-b^2}}{b}\)*\(\frac{a^2+b^2}{\sqrt{a^2-b^2}}\)
=\(\frac{a^2+b^2}{b}\)
b/ Thế \(b=a-1\)thì ta có
\(P=\frac{a^2+\left(a-1\right)^2}{a-1}=\frac{2a^2-2a+1}{a-1}\)
\(\Leftrightarrow2a^2-\left(2+P\right)a+1+P=0\)
\(\Rightarrow\Delta_a=\left(2+P\right)^2-4.2.\left(1+P\right)\ge0\)
\(\Leftrightarrow P\ge2+2\sqrt{2}\)
\(a,Q=\frac{a}{\sqrt{a^2-b^2}}-\left(1+\frac{a}{\sqrt{a^2-b^2}}\right):\left(\frac{b}{a-\sqrt{a^2-b^2}}\right)\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\left(\frac{\sqrt{a^2-b^2}+a}{\sqrt{a^2-b^2}}\right):\frac{b}{a-\sqrt{a^2+b^2}}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{a+\sqrt{a^2-b^2}}{\sqrt{a^2-b^2}}.\frac{a-\sqrt{a^2-b^2}}{b}\)\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{a^2-\left(a^2-b^2\right)}{b\sqrt{a^2-b^2}}\)
\(=\frac{ab-a^2+a^2-b^2}{b\sqrt{a^2-b^2}}\)
\(=\frac{b\left(a-b\right)}{b\sqrt{a^2-b^2}}=\frac{\left(a-b\right)}{\sqrt{\left(a-b\right)\left(a+b\right)}}=\frac{\sqrt{a-b}}{\sqrt{a+b}}\)
\(b.\frac{\sqrt{3b-b}}{\sqrt{3b+b}}=\frac{\sqrt{2b}}{\sqrt{4b}}=\frac{\sqrt{2}.\sqrt{b}}{2\sqrt{b}}=\frac{\sqrt{2}}{2}\)
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 \(VT=\left(2+\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right).\left(2-\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\)
\(=\left(2+\sqrt{a}\right)\left(2-\sqrt{a}\right)=4-a=VP\left(đpcm\right)\)
\(\left(\frac{\sqrt{a}}{2}-\frac{1}{2\sqrt{a}}\right)^2\left(\frac{\sqrt{a}-1}{\sqrt{a}+1}-\frac{\sqrt{a}+1}{\sqrt{a}-1}\right)\)
\(=\left(\frac{a-1}{2\sqrt{a}}\right)^2\left(\frac{\left(\sqrt{a}-1\right)^2-\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\frac{\left(a-1\right)^2}{4a}\left(\frac{\left(\sqrt{a}-1-\sqrt{a}-1\right)\left(\sqrt{a}-1+\sqrt{a}+1\right)}{a-1}\right)\)
\(=\frac{\left(a-1\right)\left(-2\right)2\sqrt{a}}{4a}=-\frac{\left(a-1\right)}{\sqrt{a}}\)
:") Làm bừa nhezzz
a) \(Q=\frac{a}{\sqrt{a^2-b^2}}-\left(1+\frac{a}{\sqrt{a^2-b^2}}\right):\frac{b}{a-\sqrt{a^2}-b^2}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{a+\sqrt{a^2-b^2}}{\sqrt{a^2-b^2}}.\frac{a-\sqrt{a^2-b^2}}{b}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{a^2-\left(\sqrt{a^2-b^2}\right)^2}{b.\left(\sqrt{a^2-b^2}\right)}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\left(\frac{a^2-\left(a^2-b^2\right)}{b.\left(\sqrt{a^2-b^2}\right)}\right)\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{b^2}{b\sqrt{a^2-b^2}}\)
\(=\frac{a}{\sqrt{a^2-b^2}}-\frac{b}{\sqrt{a^2-b^2}}\)
\(=\frac{a-b}{\sqrt{a^2-b^2}}=\frac{a-b}{\sqrt{\left(a-b\right)\left(a+b\right)}}=\frac{\sqrt{a-b}}{\sqrt{a+b}}\)
b) Thay a = 3b vào , ta được :
\(Q=\frac{\sqrt{3b-b}}{\sqrt{3b+b}}=\frac{\sqrt{2b}}{\sqrt{4b}}=\sqrt{\frac{2b}{4b}}=\sqrt{\frac{1}{2}}=\frac{\sqrt{2}}{2}\)