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Bạn tham khảo câu số 9:
mọi người giúp em mấy bài này với ạ =((( - Hoc24
\(\dfrac{1}{\sqrt{k}+\sqrt{k+1}}=\dfrac{\sqrt{k}-\sqrt{k+1}}{k-k-1}=\sqrt{k+1}-\sqrt{k}\\ \Leftrightarrow\text{Đặt}\text{ }A=\dfrac{1}{3\left(\sqrt{2}+\sqrt{1}\right)}+\dfrac{1}{5\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{1}{4021\left(\sqrt{2011}+\sqrt{2010}\right)}< \dfrac{1}{2\left(\sqrt{2}+\sqrt{1}\right)}+\dfrac{1}{2\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{1}{2\left(\sqrt{2011}+\sqrt{2010}\right)}\\ \Leftrightarrow A< \dfrac{1}{2}\left(\dfrac{1}{\sqrt{2}+\sqrt{1}}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+...+\dfrac{1}{\sqrt{2011}+\sqrt{2010}}\right)\)
\(\Leftrightarrow A< \dfrac{1}{2}\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2011}-\sqrt{2010}\right)\\ \Leftrightarrow A< \dfrac{1}{2}\left(\sqrt{2011}-1\right)< \dfrac{1}{2}\cdot\dfrac{\sqrt{2011}-1}{\sqrt{2011}}=\dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{2011}}\right)\)
Ta có:
VT: \(\left(\dfrac{2}{\sqrt{3}-1}-\dfrac{2}{\sqrt{3}+1}\right):\dfrac{1}{\sqrt{2}}\)
\(=\left[\dfrac{2\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\right]:\dfrac{1}{\sqrt{2}}\)
\(=\left[\dfrac{2\sqrt{3}+2-2\sqrt{3}+2}{\left(\sqrt{3}\right)^2-1^2}\right]:\dfrac{1}{\sqrt{2}}\)
\(=\dfrac{4}{2}:\dfrac{1}{\sqrt{2}}\)
\(=2:\dfrac{1}{\sqrt{2}}\)
\(=2\sqrt{2}\left(dpcm\right)\)
\(VT=\left(\sqrt{3}+1-\sqrt{3}+1\right)\cdot\sqrt{2}=2\cdot\sqrt{2}=VP\)
Chứng minh
\(\sqrt[3]{\left(n+1\right)^2}-\sqrt[3]{n^2}< \frac{2}{3\sqrt[3]{n}}\)
\(\Leftrightarrow3\sqrt[3]{n\left(n+1\right)^2}< 2+3n\)
Lập phương 2 vế rồi rút gọn được
\(\Leftrightarrow9n+8>0\)
Đúng với mọi n dương. Ta có ĐPCM.
Cái còn lại tương tự
\(\left(4-\sqrt{7}\right)^2=4^2-2\cdot4\cdot\sqrt{7}+7\)
\(=16-8\sqrt{7}+7=23-8\sqrt{7}\)
\(\sqrt{9-4\sqrt{5}}-\sqrt{5}\)
\(=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}-\sqrt{5}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}\)
\(=\left|\sqrt{5}-2\right|-\sqrt{5}\)
\(=\sqrt{5}-2-\sqrt{5}=-2\)
\(\dfrac{\sqrt{4-2\sqrt{3}}}{1+\sqrt{2}}:\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)
\(=\dfrac{\sqrt{3-2\cdot\sqrt{3}\cdot1+1}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}+1}\cdot\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\)
\(=\dfrac{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}=\dfrac{3-1}{2-1}=2\)
\(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\dfrac{6\sqrt{6}}{3}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\left(\dfrac{1}{2}\sqrt{6}-2\sqrt{6}\right)\cdot\dfrac{1}{\sqrt{6}}\)
\(=\dfrac{1}{2}-2=-\dfrac{3}{2}=-1,5\)
Biến đổi vế trái ta được:
\(VT=2+\sqrt{2}+\sqrt{\left(\sqrt{2}-1\right)^2}-3\sqrt{2}=2+\sqrt{2}+\left|\sqrt{2}-1\right|-3\sqrt{2}\)
\(=2+\sqrt{2}+\sqrt{2}-1-3\sqrt{2}\)
\(=1-\sqrt{2}=VP\)
=>(đpcm)
Chúc bạn học tốt !!!