Cần gấp thì bạn cũng nên viết đầy đủ đề bài nhé.

** Bài toán rút gọn**

Lời giải:

\(\sqrt{17-12\sqrt{2}}=\sqrt{17-2\sqrt{72}}=\sqrt{9-2\sqrt{8.9}+8}=\sqrt{(\sqrt{9}-\sqrt{8})^2}\)

\(=\sqrt{9}-\sqrt{8}=3-2\sqrt{2}\)

\(\sqrt{24-8\sqrt{8}}=\sqrt{24-2\sqrt{128}}=\sqrt{16-2\sqrt{16.8}+8}=\sqrt{(\sqrt{16}-\sqrt{8})^2}\)

\(=\sqrt{16}-\sqrt{8}=4-2\sqrt{2}\)

\(\Rightarrow \sqrt{17-12\sqrt{2}}-\sqrt{24-8\sqrt{8}}=(3-2\sqrt{2})-(4-2\sqrt{2})=-1\)

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\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{17-12\sqrt{2}}+\sqrt{17+12\sqrt{2}}\)

\(=\sqrt{8-2\sqrt{8.9}+9}+\sqrt{8+2\sqrt{8.9}+9}\)

\(=\sqrt{(\sqrt{8}-\sqrt{9})^2}+\sqrt{(\sqrt{8}+\sqrt{9})^2}\)

\(=|\sqrt{8}-\sqrt{9}|+|\sqrt{8}+\sqrt{9}|=3-2\sqrt{2}+3+2\sqrt{2}=6\)

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\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{9+2\sqrt{9.2}+2}-\sqrt{9-2\sqrt{9.2}+2}\)

\(=\sqrt{(\sqrt{9}+\sqrt{2})^2}-\sqrt{(\sqrt{9}-\sqrt{2})^2}\)

\(=|\sqrt{9}+\sqrt{2}|-|\sqrt{9}-\sqrt{2}|=3+\sqrt{2}-(3-\sqrt{2})=2\sqrt{2}\)

\(\sqrt{17-12\sqrt{2}}-\sqrt{24-8\sqrt{8}}=\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(4-2\sqrt{2}\right)^2}\)

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

\(=-1\)

\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}\)

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

\(=6\)

\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)

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

\(=2\sqrt{2}\)

c) Ta có: \(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)

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

\(=2\sqrt{2}\)

\(2\sqrt{20}-\sqrt{45}+3\sqrt{18}+3\sqrt{32}-\sqrt{50}\\ =4\sqrt{5}-3\sqrt{5}+9\sqrt{2}+12\sqrt{2}-5\sqrt{2}\\ =\sqrt{5}+16\sqrt{2}\)

\(\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}-\dfrac{11\left(4-\sqrt{5}\right)}{16-5}=\sqrt{5}-4+\sqrt{5}=2\sqrt{5}-4\)

\(=\sqrt{5}-4+\sqrt{5}=2\sqrt{5}-4\)

a: \(A=\left(1-\sqrt{7}\right)\cdot\left(1+\sqrt{7}\right)=1-7=-6\)

b: \(B=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}=-4\sqrt{3}\)

c: \(C=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)

a) Ta có: \(A=\sqrt{20}-10\sqrt{\dfrac{1}{5}}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=2\sqrt{5}-2\sqrt{5}+\sqrt{5}-1\)

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

b) Ta có: \(B=2\sqrt{32}+5\sqrt{8}-4\sqrt{32}\)

\(=8\sqrt{2}+10\sqrt{2}-16\sqrt{2}\)

\(=2\sqrt{2}\)

6: Ta có: \(\left(3\sqrt{2}-\sqrt{3}\right)\left(3\sqrt{2}+\sqrt{3}\right)\)

=18-3

=15

7: Ta có: \(\sqrt{72}+\sqrt{4\dfrac{1}{2}}-\sqrt{32}-\sqrt{162}\)

\(=6\sqrt{2}+\dfrac{3}{2}\sqrt{2}-4\sqrt{2}-9\sqrt{2}\)

\(=-\dfrac{11}{2}\sqrt{2}\)

\(\dfrac{\sqrt{8}+3}{\sqrt{17-3\sqrt{32}}}-\dfrac{3-2\sqrt{5}}{\sqrt{29-12\sqrt{5}}}-\dfrac{1}{\sqrt{12+2\sqrt{35}}}\) 

\(=\dfrac{2\sqrt{2}+3}{\sqrt{17-12\sqrt{2}}}-\dfrac{3-2\sqrt{5}}{\sqrt{29-12\sqrt{5}}}-\dfrac{1}{\sqrt{12+2\sqrt{35}}}\)

\(=\dfrac{2\sqrt{2}+3}{\sqrt{3^2-2\cdot3\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2}}-\dfrac{3-2\sqrt{5}}{\sqrt{3^2-2\cdot3\cdot2\sqrt{5}+\left(2\sqrt{5}\right)^2}}-\dfrac{1}{\sqrt{\left(\sqrt{5}\right)^2+2\sqrt{5}\cdot\sqrt{7}+\left(\sqrt{7}\right)^2}}\)

\(=\dfrac{2\sqrt{2}+3}{\sqrt{\left(2\sqrt{2}-3\right)^2}}-\dfrac{3-2\sqrt{5}}{\sqrt{\left(3-2\sqrt{5}\right)^2}}-\dfrac{1}{\sqrt{\left(\sqrt{5}+\sqrt{7}\right)^2}}\)

\(=\dfrac{2\sqrt{2}+3}{2\sqrt{2}-3}+\dfrac{3-2\sqrt{5}}{3-2\sqrt{5}}-\dfrac{1}{\sqrt{5}+\sqrt{7}}\)

\(=\dfrac{\left(2\sqrt{2}+3\right)^2}{\left(2\sqrt{2}+3\right)\left(2\sqrt{2}-3\right)}+1-\dfrac{\sqrt{5}-\sqrt{7}}{\left(\sqrt{5}+\sqrt{7}\right)\left(\sqrt{5}-\sqrt{7}\right)}\)

\(=17-12\sqrt{2}+1-\dfrac{\sqrt{5}-\sqrt{7}}{2}\)

\(=\dfrac{2\cdot\left(18-12\sqrt{2}\right)}{2}-\dfrac{\sqrt{5}-\sqrt{7}}{2}\)

\(=\dfrac{36-24\sqrt{2}-\sqrt{5}+\sqrt{7}}{2}\)

\(a,\dfrac{1}{2-\sqrt{3}}-3\sqrt{\dfrac{1}{3}}+\sqrt{12}\\ =\dfrac{2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}-\dfrac{\sqrt{3^2}}{\sqrt{3}}+\sqrt{2^2.3}\\ =\dfrac{2+\sqrt{3}}{4-3}-\sqrt{3}+2\sqrt{3}\\ =2+\sqrt{3}-\sqrt{3}+2\sqrt{3}\\ =2+2\sqrt{3}\)

\(b,\dfrac{2}{1+\sqrt{2}}-\sqrt{9-\sqrt{32}}\\ =\dfrac{2\left(1-\sqrt{2}\right)}{\left(1+\sqrt{2}\right)\left(1-\sqrt{2}\right)}-\sqrt{9-4\sqrt{2}}\\ =\dfrac{2-2\sqrt{2}}{1-2}-\sqrt{\left(2\sqrt{2}\right)^2-2.2\sqrt{2}+1}\\ =-2+2\sqrt{2}-\sqrt{\left(2\sqrt{2}-1\right)^2}\\ =-2+2\sqrt{2}-\left|2\sqrt{2}-1\right|\\ =-2+2\sqrt{2}-2\sqrt{2}+1\\ =-1\)

Bài làm của: Phùng Khánh Linh

c)\(\sqrt{17-12\sqrt{2}}-\sqrt{24-8\sqrt{8}}\)

= \(\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}\) \(-\) \(\sqrt{4^2-2.4.\sqrt{8}+\left(\sqrt{8}\right)^2}\)

= \(\sqrt{\left(3-2\sqrt{2}\right)^2}\) \(-\) \(\sqrt{\left(4-\sqrt{8}\right)^2}\)

= \(\left|3-2\sqrt{2}\right|-\left|4-\sqrt{8}\right|\)

= (3 - 2\(\sqrt{2}\)) - (4 - \(\sqrt{8}\))

= 3 - 2\(\sqrt{2}\) - 4 + \(\sqrt{8}\)

= -1

\(a.\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{3+2\sqrt{3}.1+1}-\sqrt{3-2\sqrt{3}.1+1}=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}=\text{|}\sqrt{3}+1\text{|}-\text{|}\sqrt{3}-1\text{|}=2\)\(b.\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}=\sqrt{5-4\sqrt{5}+4}-\sqrt{5+4\sqrt{5}+4}=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}=\text{|}\sqrt{5}-2\text{|}-\text{|}\sqrt{5}+2\text{|}=-4\) Còn lại tương tự nhé .