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\(2\left(\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+...+\dfrac{1}{\sqrt{97}+\sqrt{99}}\right)\)
\(>\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+...+\dfrac{1}{\sqrt{97}+\sqrt{99}}+\dfrac{1}{\sqrt{99}+\sqrt{101}}\)
\(=\dfrac{1}{2}\left(\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+...+\sqrt{101}-\sqrt{99}\right)\)
\(=\dfrac{1}{2}\left(\sqrt{101}-\sqrt{1}\right)>\dfrac{9}{2}\)
\(\Rightarrow\dfrac{1}{\sqrt{1}+\sqrt{3}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+...+\dfrac{1}{\sqrt{97}+\sqrt{99}}>\dfrac{9}{4}\)
Nhận xét 1: từng hạng tử của A có dạng:
\(\dfrac{1}{\sqrt{x}+\sqrt{x+2}}\left(x\ge3\right)\)
Nhận xét 2:
\(\left(\sqrt{x+2}-\sqrt{x}\right)\left(\sqrt{x}+\sqrt{x+2}\right)=\left(x+2\right)-x=2\)
\(\Rightarrow\dfrac{2}{\sqrt{x}+\sqrt[]{x+2}}=-\sqrt{x}+\sqrt{x+2}\)
Áp dụng vào A:
\(2A=\dfrac{2}{\sqrt{3}+\sqrt{5}}+\dfrac{2}{\sqrt{5}+\sqrt{7}}+...+\dfrac{2}{\sqrt{97}+\sqrt{99}}\)
\(=\left(-\sqrt{3}+\sqrt{5}\right)+\left(-\sqrt{5}+\sqrt{7}\right)+...+\left(-\sqrt{97}+\sqrt{99}\right)\)
\(=-\sqrt{3}+\sqrt{99}\Leftrightarrow A=-2\sqrt{3}+2\sqrt{99}\)
A = \(\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+\dfrac{1}{\sqrt{7}+\sqrt{9}}+...+\dfrac{1}{\sqrt{97}+\sqrt{99}}\)
=
\(\dfrac{\sqrt{5}-\sqrt{3}}{\left(\sqrt{3}+\sqrt{5}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)}+\dfrac{\sqrt{7}-\sqrt{5}}{\left(\sqrt{5}+\sqrt{7}\right)\cdot\left(\sqrt{7}-\sqrt{5}\right)}+\dfrac{\sqrt{9}-\sqrt{7}}{\left(\sqrt{7}+\sqrt{9}\right)\cdot\left(\sqrt{9}-\sqrt{7}\right)}+...+\dfrac{\sqrt{99}-\sqrt{97}}{\left(\sqrt{97}+\sqrt{99}\right)\cdot\left(\sqrt{99}-\sqrt{97}\right)}\)
= \(\dfrac{\sqrt{5}-\sqrt{3}}{5-3}+\dfrac{\sqrt{7}-\sqrt{5}}{7-5}+\dfrac{\sqrt{9}-\sqrt{7}}{9-7}+...+\dfrac{\sqrt{99}-\sqrt{97}}{99-97}\)
=\(\dfrac{\sqrt{5}-\sqrt{3}}{2}+\dfrac{\sqrt{7}-\sqrt{5}}{2}+\dfrac{\sqrt{9}-\sqrt{7}}{2}+...+\dfrac{\sqrt{99}-\sqrt{97}}{2}\)
=\(\dfrac{1}{2}\cdot\left(\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+\sqrt{9}-\sqrt{7}+...+\sqrt{99}-\sqrt{97}\right)\)
= \(\dfrac{1}{2}\cdot\left(-\sqrt{3}+\sqrt{99}\right)\)
= \(\dfrac{3\sqrt{11}-\sqrt{3}}{2}\)
\(b,\) Ta có:
\(\dfrac{1}{n\sqrt{n-1}+\left(n-1\right)\sqrt{n}}\\ =\dfrac{1}{\sqrt{n}.\sqrt{n-1}\left(\sqrt{n}+\sqrt{n-1}\right)}\\ =\dfrac{\sqrt{n}}{\sqrt{n}.\sqrt{n-1}}-\dfrac{\sqrt{n-1}}{\sqrt{n}.\sqrt{n-1}}\\ =\dfrac{1}{\sqrt{n-1}}-\dfrac{1}{\sqrt{n}}\)
Thay:
\(n=2\) \(\Leftrightarrow\dfrac{1}{2\sqrt{1}+1\sqrt{2}}=\dfrac{1}{1}-\dfrac{1}{\sqrt{2}}\)
\(n=3\Leftrightarrow\dfrac{1}{3\sqrt{2}+2\sqrt{3}}=\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}\)
\(...\)
\(n=2007\Leftrightarrow\dfrac{1}{2007\sqrt{2006}+2006\sqrt{2007}}=\dfrac{1}{\sqrt{2006}}-\dfrac{1}{\sqrt{2007}}\\ \)
A = \(\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+\dfrac{1}{\sqrt{7}+\sqrt{9}}+...+\dfrac{1}{\sqrt{97}+\sqrt{99}}\)
= \(\dfrac{1}{2}\left(\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+\sqrt{9}-\sqrt{7}+...+\sqrt{99}-\sqrt{97}\right)\)
= \(\dfrac{1}{2}\left(\sqrt{99}-\sqrt{3}\right)\)
B = 35 + 335 + 3335 + ... + 3333...(99 số 3)35
= 33 + 2 + 333 + 2 + 3333 + 2 + ... + 333...33 + 2
= 2 . 99 + (33 + 333 + 3333 + ... + 333...3)
= 198 + \(\dfrac{1}{3}\)(99 + 999 + 9999 + ... + 999...99)
= 198 + \(\dfrac{1}{3}\)(102 - 1 + 103 - 1 + 104 - 1 + ... + 10100 - 1)
= \(\left(\dfrac{10^{101}-10^2}{27}\right)+165\)
\(A=-\sqrt{2}-\sqrt{1}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+....-\sqrt{7}-\sqrt{8}+\sqrt{8}+\sqrt{9}\)
\(A=\sqrt{9}-\sqrt{1}=3-1=2\)
\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)
\(ĐK:x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)
\(\Leftrightarrow4x^2-9=4x+12\)
\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)
\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(ĐK:x\ge5\)
\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)
ĐK:x>=1
\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)
\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)
\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)
\(ĐK:x\ge3\)
\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)
\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)
\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}=0\) (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)
\(1.\text{ }\dfrac{1}{\sqrt{k}-\sqrt{k+1}}=\dfrac{\left(\sqrt{k}+\sqrt{k+1}\right)}{\left(\sqrt{k}+\sqrt{k+1}\right)\left(\sqrt{k}-\sqrt{k+1}\right)}\\ =-\left(\sqrt{k}+\sqrt{k+1}\right)\\ \Rightarrow\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-...-\dfrac{1}{\sqrt{8}-\sqrt{9}}\\ =-\left(\sqrt{1}+\sqrt{2}\right)+\left(\sqrt{2}+\sqrt{3}\right)-\left(\sqrt{3}+\sqrt{4}\right)+...+\left(\sqrt{8}+\sqrt{9}\right)\\ =-\sqrt{1}-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+...+\sqrt{8}+\sqrt{9}\\ \\ =\sqrt{9}-\sqrt{1}=2\)
\(2.\text{ }\dfrac{1}{\left(k+1\right)\sqrt{k}+\sqrt{k+1}k}=\dfrac{1}{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}+\sqrt{k}\right)}\\ =\dfrac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}+\sqrt{k}\right)\left(\sqrt{k+1}-\sqrt{k}\right)}\\ =\dfrac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}\left(k+1-k\right)}=\dfrac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}}\\ =\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\\ \Rightarrow\text{ }\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{7\sqrt{6}+6\sqrt{7}}\\ =\text{ }\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{6}}-\dfrac{1}{\sqrt{7}}\\ =\text{ }\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{7}}\\ \text{ }1-\dfrac{1}{\sqrt{7}}\)
1.\(\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-\dfrac{1}{\sqrt{4}-\sqrt{5}}+\dfrac{1}{\sqrt{5}-\sqrt{6}}-\dfrac{1}{\sqrt{6}-\sqrt{7}}+\dfrac{1}{\sqrt{7}-\sqrt{8}}-\dfrac{1}{\sqrt{8}-\sqrt{9}}=\dfrac{1+\sqrt{2}}{1-2}-\dfrac{\sqrt{2}+\sqrt{3}}{2-3}+\dfrac{\sqrt{3}+\sqrt{4}}{3-4}-\dfrac{\sqrt{4}+\sqrt{5}}{4-5}+\dfrac{\sqrt{5}+\sqrt{6}}{5-6}-\dfrac{\sqrt{6}+\sqrt{7}}{6-7}+\dfrac{\sqrt{7}+\sqrt{8}}{7-8}-\dfrac{\sqrt{8}+\sqrt{9}}{8-9}=-1-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+\sqrt{4}+\sqrt{5}-\sqrt{5}-\sqrt{6}+\sqrt{6}+\sqrt{7}-\sqrt{7}-\sqrt{8}+\sqrt{8}+\sqrt{9}=\sqrt{9}-1=3-1=2\)