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22 tháng 6 2019

\(\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}\right)^2-2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}-1\right)^2}=|\sqrt{2}-1|=\sqrt{2}-1\)

Tương tự  \(\sqrt{4-2\sqrt{3}}=\sqrt{3}-1\);   \(\sqrt{7-4\sqrt{3}}=2-\sqrt{3}\)

\(\Rightarrow BTT=\sqrt{2}-1+\sqrt{3}-1+2-\sqrt{3}=\sqrt{2}\)

23 tháng 6 2019

\(\sqrt{3-2\sqrt{2}}+\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)

\(=\sqrt{2-2\sqrt{2}+1}+\sqrt{3-2\sqrt{3}+1}-\sqrt{4-4\sqrt{3}+3}\)

\(=\sqrt{\left(\sqrt{2}-1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)

\(=\sqrt{2}-1+\sqrt{3}-1-2+\sqrt{3}\)

\(=2\sqrt{3}+\sqrt{2}-4\)

26 tháng 7 2018

\(1.\text{ }\dfrac{1}{\sqrt{k}-\sqrt{k+1}}=\dfrac{\left(\sqrt{k}+\sqrt{k+1}\right)}{\left(\sqrt{k}+\sqrt{k+1}\right)\left(\sqrt{k}-\sqrt{k+1}\right)}\\ =-\left(\sqrt{k}+\sqrt{k+1}\right)\\ \Rightarrow\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-...-\dfrac{1}{\sqrt{8}-\sqrt{9}}\\ =-\left(\sqrt{1}+\sqrt{2}\right)+\left(\sqrt{2}+\sqrt{3}\right)-\left(\sqrt{3}+\sqrt{4}\right)+...+\left(\sqrt{8}+\sqrt{9}\right)\\ =-\sqrt{1}-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+...+\sqrt{8}+\sqrt{9}\\ \\ =\sqrt{9}-\sqrt{1}=2\)

\(2.\text{ }\dfrac{1}{\left(k+1\right)\sqrt{k}+\sqrt{k+1}k}=\dfrac{1}{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}+\sqrt{k}\right)}\\ =\dfrac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}+\sqrt{k}\right)\left(\sqrt{k+1}-\sqrt{k}\right)}\\ =\dfrac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}\left(k+1-k\right)}=\dfrac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}}\\ =\dfrac{1}{\sqrt{k}}-\dfrac{1}{\sqrt{k+1}}\\ \Rightarrow\text{ }\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+...+\dfrac{1}{7\sqrt{6}+6\sqrt{7}}\\ =\text{ }\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{6}}-\dfrac{1}{\sqrt{7}}\\ =\text{ }\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{7}}\\ \text{ }1-\dfrac{1}{\sqrt{7}}\)

26 tháng 7 2018

1.\(\dfrac{1}{\sqrt{1}-\sqrt{2}}-\dfrac{1}{\sqrt{2}-\sqrt{3}}+\dfrac{1}{\sqrt{3}-\sqrt{4}}-\dfrac{1}{\sqrt{4}-\sqrt{5}}+\dfrac{1}{\sqrt{5}-\sqrt{6}}-\dfrac{1}{\sqrt{6}-\sqrt{7}}+\dfrac{1}{\sqrt{7}-\sqrt{8}}-\dfrac{1}{\sqrt{8}-\sqrt{9}}=\dfrac{1+\sqrt{2}}{1-2}-\dfrac{\sqrt{2}+\sqrt{3}}{2-3}+\dfrac{\sqrt{3}+\sqrt{4}}{3-4}-\dfrac{\sqrt{4}+\sqrt{5}}{4-5}+\dfrac{\sqrt{5}+\sqrt{6}}{5-6}-\dfrac{\sqrt{6}+\sqrt{7}}{6-7}+\dfrac{\sqrt{7}+\sqrt{8}}{7-8}-\dfrac{\sqrt{8}+\sqrt{9}}{8-9}=-1-\sqrt{2}+\sqrt{2}+\sqrt{3}-\sqrt{3}-\sqrt{4}+\sqrt{4}+\sqrt{5}-\sqrt{5}-\sqrt{6}+\sqrt{6}+\sqrt{7}-\sqrt{7}-\sqrt{8}+\sqrt{8}+\sqrt{9}=\sqrt{9}-1=3-1=2\)

2 tháng 7 2019

Lời giải :

a) \(\sqrt{\left(0,1-\sqrt{0,1}\right)^2}\)

\(=0,1-\sqrt{0,1}\)

b) \(\sqrt{4-2\sqrt{3}}=\sqrt{3-2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}-1\)

c) \(\sqrt{3+2\sqrt{2}}=\sqrt{2+2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)

d) \(\sqrt{9-4\sqrt{5}}=\sqrt{5-4\sqrt{5}+4}=\sqrt{\left(\sqrt{5}-2\right)^2}=\sqrt{5}-2\)

e) \(\sqrt{16-6\sqrt{7}}=\sqrt{9-2\cdot3\cdot\sqrt{7}+7}=\sqrt{\left(3-\sqrt{7}\right)^2}=3-\sqrt{7}\)

29 tháng 9 2020

1) \(\left(\sqrt{6}-\sqrt{8}\right)\left(\sqrt{6}+\sqrt{8}\right)\)

\(=\left(\sqrt{6}\right)^2-\left(\sqrt{8}\right)^2\)

\(=6-8=-2\)

2) \(\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\)

\(=3^2-\left(\sqrt{5}\right)^2\)

\(=9-5=4\)

29 tháng 9 2020

3) \(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}\)

\(=\sqrt{4-4\sqrt{3}+3}+\sqrt{4+4\sqrt{3}+3}\)

\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(2+\sqrt{3}\right)^2}\)

\(=2-\sqrt{3}+2+\sqrt{3}=4\)

4) Xét ta thấy: \(2\sqrt{3}=\sqrt{12}< \sqrt{16}=4\)

=> \(2\sqrt{3}-4< 0\) => vô lý không tm đk căn

3 tháng 8 2018

Câu 1 =3/10

3 tháng 8 2018

\(1,\sqrt{\left(-0,3\right)^2}=\sqrt{0,09}=0,3\)

\(2,-\frac{1}{2}\sqrt{\left(0,3\right)^2}=-\frac{1}{2}.0,3=-0,15\)

\(3,\sqrt{a^{10}}=\sqrt{\left(a^5\right)^2}=a^5\left(a\ge0\right)\)

\(4,\sqrt{\left(2-x\right)^2}=\left|2-x\right|=2-x\left(x\le2\right)\)

\(5,\sqrt{x^2+2x+1}=\sqrt{\left(x+1\right)^2}=\left|x+1\right|\)

\(6,\sqrt{\left(1-\sqrt{2}\right)^2}=\left|1-\sqrt{2}\right|=\sqrt{2}-1\)(Vì \(1< \sqrt{2}\))

\(7,\sqrt{11+6\sqrt{2}}=\sqrt{9+6\sqrt{2}+2}=\sqrt{\left(3+\sqrt{2}\right)^2}=3+\sqrt{2}\)

\(8,\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}-\sqrt{7+2\sqrt{7}+1}\)

                                                                    \(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}\)

                                                                    \(=\left(\sqrt{7}-1\right)-\left(\sqrt{7}+1\right)\)

                                                                      \(=-2\)

\(9,\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{5+2\sqrt{5}+1}+\sqrt{5-2\sqrt{5}+1}\)

                                                                    \(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)

                                                                    \(=\sqrt{5}+1+\sqrt{5}-1\)

                                                                    \(=2\sqrt{5}\)

25 tháng 7 2023

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left(2x+1\right)^2=6^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(\sqrt{4x^2-4\sqrt{7}x+7}=\sqrt{7}\)

\(\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left(2x-\sqrt{7}\right)^2=\left(\sqrt{7}\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt[]{7}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

25 tháng 7 2023

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(pt\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left|2x-\sqrt{7}\right|=\sqrt{7}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

 

6 tháng 8 2018

\(\sqrt{3-2\sqrt{2}}-\sqrt{6+4\sqrt{2}}\)

\(=\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(2+\sqrt{2}\right)^2}\)

\(=\sqrt{2}-1-\left(2+\sqrt{2}\right)\)

\(=-3\)