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1. \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{2}\)
\(=2\sqrt{2}\)
\(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}=\sqrt{25-2\cdot5\cdot2\sqrt{6}+24}+\sqrt{25-2\cdot5\cdot2\sqrt{6}+24}=\sqrt{\left(5+2\sqrt{6}\right)^2}+\sqrt{\left(5-2\sqrt{6}\right)^2}=5+2\sqrt{6}+5-2\sqrt{6}=10\) ---
\(\sqrt{13-\sqrt{160}}+\sqrt{53+4\sqrt{90}}=\sqrt{8-2\sqrt{5}\cdot\sqrt{8}+5}+\sqrt{45+2\cdot3\sqrt{5}\cdot\sqrt{8}+8}=\sqrt{\left(\sqrt{8}-\sqrt{5}\right)^2}+\sqrt{\left(3\sqrt{5}+\sqrt{8}\right)^2}=\sqrt{8}-\sqrt{5}+3\sqrt{5}+\sqrt{8}=2\sqrt{8}+2\sqrt{5}\)
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\(\sqrt{11-6\sqrt{2}}+\sqrt{3-2\sqrt{2}}=\sqrt{9-2\cdot3\cdot\sqrt{2}+2}+\sqrt{2-2\sqrt{2}+1}=\sqrt{\left(3-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}-1\right)^2}=3-\sqrt{2}+\sqrt{2}-1=2\)
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\(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}=\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{27-2\cdot\sqrt{27}\cdot\sqrt{8}+8}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3\sqrt{3}-2\sqrt{2}\right)^2}=3-\sqrt{6}+3\sqrt{3}-2\sqrt{2}\)
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\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{9-2\cdot3\cdot2\sqrt{2}+8}+\sqrt{9+2\cdot2\cdot2\sqrt{2}+8}=\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}=3-2\sqrt{2}+3+2\sqrt{2}=6\)
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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 .
f, \(\sqrt{\sqrt{5}+\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}+\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=\sqrt{\sqrt{5}+\sqrt{3-2\sqrt{5}+3}}=\sqrt{\sqrt{5}+\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}+\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}+\sqrt{5}-1}=\sqrt{2\sqrt{5}-1}\)
mik sửa lại câu f , tí nhé :
f , \(\sqrt{\sqrt{5}+\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=|2+\sqrt{3}|-|2-\sqrt{3}|\)
\(=2+\sqrt{3}-2+\sqrt{3}\)
\(=2\sqrt{3}\)
\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=|3+\sqrt{2}|-|3-\sqrt{2}|\)
\(=3+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}\)
\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(=\sqrt{\left(3+2\sqrt{2}\right)^2}+\sqrt{\left(3-2\sqrt{2}\right)^2}\)
\(=|3+2\sqrt{2}|+|3-2\sqrt{2}|\)
\(=3+2\sqrt{2}+3-2\sqrt{2}\)
\(=6\)
\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(=\sqrt{\left(2+\sqrt{5}\right)^2}-\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(=|2+\sqrt{5}|-|2-\sqrt{5}|\)
\(=2+\sqrt{5}-\sqrt{5}+2\)
\(=4\)
\(E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{\left(1-\sqrt{5}\right)^2}\)
\(=|1+\sqrt{5}|-|1-\sqrt{5}|\)
\(=1+\sqrt{5}-\sqrt{5}+1\)
\(=2\)
\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)
\(A=\sqrt{3}+2+2-\sqrt{3}\)
A = 2 + 2
A = 4
\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)
\(B=\sqrt{2}+3+3-\sqrt{2}\)
B = 3 + 3
B = 6
\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)
\(C=3+2\sqrt{2}+3-2\sqrt{2}\)
C = 3 + 3
C = 6
\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)
\(D=\sqrt{5}+2-\sqrt{5}+2\)
D = 2 + 2
D = 4
\(E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(E=\sqrt{5}+1-\sqrt{5}+1\)
E = 1 + 1
E = 2
a, \(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)
= \(\sqrt{3^2-2.3.\sqrt{6}+\left(\sqrt{6}\right)^2}+\sqrt{6^2-2.6.\sqrt{6}+\left(\sqrt{6}\right)^2}\)
= \(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(6-\sqrt{6}\right)^2}\)
= \(\left|3-\sqrt{6}\right|+\left|6-\sqrt{6}\right|\)
= \(3-\sqrt{6}+6-\sqrt{6}\)
= \(9-2\sqrt{6}\)
b. Đặt B = \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
Nhận xét : B > 0 , bình phương hai vế ta được :
\(B^2=\left(\sqrt{17-3\sqrt{32}}\right)^2+\left(\sqrt{17+3\sqrt{32}}\right)^2\)
\(B^2=17-3\sqrt{32}+17+3\sqrt{32}+2\sqrt{\left(17-3\sqrt{32}\right)\left(17+3\sqrt{32}\right)}\)
\(B^2=34+2\sqrt{17^2-\left(3\sqrt{32}\right)^2}\)
\(B^2=34+2\sqrt{289-288}\)
\(B^2=34+2=36\)
=> \(B=\pm\sqrt{36}\) mà B > 0 nên \(B=\sqrt{36}=6\)
c, Đặt C = \(\sqrt{49-5\sqrt{96}}+\sqrt{49+5\sqrt{96}}\)
Nhận xét : C > 0 , bình phương hai vế ta đươc :
\(C^2=\left(\sqrt{49-5\sqrt{96}}\right)^2+\left(\sqrt{49+5\sqrt{96}}\right)^2\)
\(C^2=49-5\sqrt{96}+49+5\sqrt{96}+2\sqrt{\left(49-5\sqrt{96}\right)\left(49+5\sqrt{96}\right)}\)
\(C^2=98+2\sqrt{49^2-\left(5\sqrt{96}\right)^2}\)
\(C^2=98+2\sqrt{2401-2400}\)
\(C^2=98+2=100\)
=> \(C=\pm\sqrt{100}\) mà C > 0 nên \(C=\sqrt{100}=10\)
a) Ta có: \(\sqrt{15-6\sqrt{6}}+\sqrt{35-12\sqrt{6}}\)
\(=\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{27-2\cdot3\sqrt{3}\cdot2\sqrt{2}+8}\)
\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3\sqrt{3}-2\sqrt{2}\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|3\sqrt{3}-2\sqrt{2}\right|\)
\(=3-\sqrt{6}+3\sqrt{3}-2\sqrt{2}\)(Vì \(\left\{{}\begin{matrix}3>\sqrt{6}\\3\sqrt{3}>2\sqrt{2}\end{matrix}\right.\))
b) Ta có: \(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}\)
\(=\frac{\sqrt{34-6\sqrt{32}}+\sqrt{34+6\sqrt{32}}}{\sqrt{2}}\)
\(=\frac{\sqrt{18-2\cdot3\sqrt{2}\cdot4+16}+\sqrt{18+2\cdot3\sqrt{2}\cdot4+16}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(3\sqrt{2}-4\right)^2}+\sqrt{\left(3\sqrt{2}+4\right)^2}}{\sqrt{2}}\)
\(=\frac{\left|3\sqrt{2}-4\right|+\left|3\sqrt{2}+4\right|}{\sqrt{2}}\)
\(=\frac{3\sqrt{2}-4+3\sqrt{2}+4}{\sqrt{2}}\)(Vì \(3\sqrt{2}>4>0\))
\(=\frac{6\sqrt{2}}{\sqrt{2}}=6\)
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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\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{9+2\sqrt{9.2}+2}-\sqrt{9-2\sqrt{9.2}+2}\)
\(=\sqrt{(\sqrt{9}+\sqrt{2})^2}-\sqrt{(\sqrt{9}-\sqrt{2})^2}\)
\(=|\sqrt{9}+\sqrt{2}|-|\sqrt{9}-\sqrt{2}|=3+\sqrt{2}-(3-\sqrt{2})=2\sqrt{2}\)
\(\sqrt{17-12\sqrt{2}}-\sqrt{24-8\sqrt{8}}=\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(4-2\sqrt{2}\right)^2}\)
\(=\left|3-2\sqrt{2}\right|-\left|4-2\sqrt{2}\right|=3-2\sqrt{2}-4+2\sqrt{2}\)
\(=-1\)
\(\sqrt{17-3\sqrt{32}}+\sqrt{17+3\sqrt{32}}=\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}\)
\(=\left|3-2\sqrt{2}\right|+\left|3+2\sqrt{2}\right|=3-2\sqrt{2}+3+2\sqrt{2}\)
\(=6\)
\(\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}=\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(=\left|3+\sqrt{2}\right|-\left|3-\sqrt{2}\right|=3+\sqrt{2}-3+\sqrt{2}\)
\(=2\sqrt{2}\)