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![](https://rs.olm.vn/images/avt/0.png?1311)
Xét biểu thức phụ : \(\frac{1}{\left(k+1\right)\sqrt{k}+k\left(\sqrt{k+1}\right)}=\frac{1}{\sqrt{k\left(k+1\right)}\left(\sqrt{k}+\sqrt{k+1}\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}\left(k+1-k\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\)
Áp dụng : \(\frac{1}{2.\sqrt{1}+1.\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+\frac{1}{5\sqrt{4}+4\sqrt{5}}+...+\frac{1}{2012\sqrt{2011}+2011\sqrt{2012}}=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2011}}-\frac{1}{\sqrt{2012}}=1-\frac{1}{\sqrt{2012}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a). \(\frac{1}{\sqrt{5-\sqrt{7}}}+\frac{\sqrt{5}}{\sqrt{5+\sqrt{7}}})-1\)
\(\Leftrightarrow\frac{1}{\sqrt{25-\sqrt{49}}}-1\)
\(\Leftrightarrow\frac{1}{\sqrt{25-7}}-1\)
\(\Leftrightarrow\frac{1}{\sqrt{18}}-1\)
\(\Leftrightarrow\frac{1}{3\sqrt{2}}-1\)
ĐẾN ĐÂY BN QUY ĐỒNG LÀ ĐC
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^2-9\ge0\Leftrightarrow x^2\ge9\Leftrightarrow\orbr{\begin{cases}x\ge3\\x\ge-3\end{cases}}\)
b) \(-x-2\ge0\Leftrightarrow-x\ge2\Leftrightarrow x\ge-2\)
c) \(x^2+2x+1=\left(x+1\right)^2\)
\(\Rightarrow\left(x+1\right)^2\ge0\Leftrightarrow x+1\ge0\Leftrightarrow x\ge-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=\sqrt{4-2\sqrt{3}}-\sqrt{7+4\sqrt{3}}\)
\(\Leftrightarrow C=\sqrt{3-2\sqrt{3}+1}-\sqrt{4+4\sqrt{3}+3}\)
\(\Leftrightarrow C=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(\Leftrightarrow C=\left|\sqrt{3}-1\right|-\left|2+\sqrt{3}\right|\)
\(\Leftrightarrow C=\sqrt{3}-1-2-\sqrt{3}\)
\(\Leftrightarrow C=-3\)
\(\frac{B}{\sqrt{2}}=\frac{\frac{2+\sqrt{3}}{2}}{\sqrt{2}+\sqrt{\frac{4+2\sqrt{3}}{2}}}+\frac{\frac{2-\sqrt{3}}{2}}{\sqrt{2}-\sqrt{\frac{4-2\sqrt{3}}{2}}}\)
\(=\frac{\frac{2+\sqrt{3}}{2}}{\frac{2}{\sqrt{2}}+\sqrt{\frac{\left(\sqrt{3}+1\right)^2}{2}}}+\frac{\frac{2-\sqrt{3}}{2}}{\frac{2}{\sqrt{2}}-\sqrt{\frac{\left(\sqrt{3}-1\right)^2}{2}}}\)
\(=\frac{\frac{2+\sqrt{3}}{2}}{\frac{2}{\sqrt{2}}+\frac{\sqrt{3}+1}{\sqrt{2}}}+\frac{\frac{2-\sqrt{3}}{2}}{\frac{2}{\sqrt{2}}-\frac{\sqrt{3}-1}{\sqrt{2}}}=\frac{\frac{2+\sqrt{3}}{2}}{\frac{\sqrt{3}+3}{\sqrt{2}}}+\frac{\frac{2-\sqrt{3}}{2}}{\frac{3-\sqrt{3}}{\sqrt{2}}}\)
\(=\frac{\left(2+\sqrt{3}\right).\sqrt{2}}{2\cdot\left(3+\sqrt{3}\right)}+\frac{\left(2-\sqrt{3}\right).\sqrt{2}}{2.\left(3-\sqrt{3}\right)}\)
=> \(B=\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}}=\frac{\left(2+\sqrt{3}\right)\left(3-\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}+\frac{\left(2-\sqrt{3}\right)\left(3+\sqrt{3}\right)}{\left(3-\sqrt{3}\right)\left(3+\sqrt{3}\right)}\)
\(B=\frac{3+\sqrt{3}}{6}+\frac{3-\sqrt{3}}{6}=1\)
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Vài chỗ mình làm vắn tắt không hiểu cứ hỏi nhé, còn kết quả mình ấn máy tính ra chính xác rùi :)