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LN
2
NC
29 tháng 8 2020
Bài làm:
Đặt \(A=\sqrt{7-\sqrt{13}}-\sqrt{7+\sqrt{13}}\)
\(\Leftrightarrow A^2=\left(\sqrt{7-\sqrt{13}}-\sqrt{7+\sqrt{13}}\right)^2\)
\(=7-\sqrt{13}-2\sqrt{\left(7-\sqrt{13}\right)\left(7+\sqrt{13}\right)}+7+\sqrt{13}\)
\(=14-2\sqrt{49-13}\)
\(=14-2\sqrt{36}=14-2.6=14-12=2\)
\(\Rightarrow A=\sqrt{2}\)
Thay vào ta được:
\(\sqrt{7-\sqrt{13}}-\sqrt{7+\sqrt{13}}+\sqrt{2}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)
TD
2
19 tháng 10 2017
=> \(A^2=13+\sqrt{7+\sqrt{13+\sqrt{7+\sqrt{13+\sqrt{7+....}}}}}\)
=>\(\left(A^2-13\right)^2=7+\sqrt{13+\sqrt{7+\sqrt{13+\sqrt{7...}}}}\)
=>\(\left(A^2-13\right)^2=7+A\)
Đến đây tách ra giải PT bậc 4 nha!
22 tháng 6 2018
\(1.\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}=\sqrt{5-2.\sqrt{2}.\sqrt{5}+2}-\sqrt{5+2.\sqrt{5}.\sqrt{2}+2}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}=\text{|}\sqrt{5}-\sqrt{2}\text{|}-\text{|}\sqrt{5}+\sqrt{2}\text{|}=-2\sqrt{2}\)\(2.\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}=\sqrt{8+2.2\sqrt{2}.\sqrt{5}+5}+\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5}=\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}=\text{|}2\sqrt{2}+\sqrt{5}\text{|}+\text{|}2\sqrt{2}-\sqrt{5}\text{|}=4\sqrt{2}\)\(3.\left(\sqrt{3}+\sqrt{5}\right)\sqrt{7-2\sqrt{10}}=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{5-2.\sqrt{5}.\sqrt{2}+2}=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\left(\sqrt{3}+\sqrt{5}\right)\text{|}\sqrt{5}-\sqrt{2}\text{|}=\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)\)
AH
Akai Haruma
Giáo viên
18 tháng 6 2019
Lời giải:
a)
\((\sqrt{5-2\sqrt{5}}+\sqrt{5+2\sqrt{5}})^2=5-2\sqrt{5}+5+2\sqrt{5}+2\sqrt{(5-2\sqrt{5})(5+2\sqrt{5})}\)
\(=10+2\sqrt{5^2-(2\sqrt{5})^2}=10+2\sqrt{5}\)
\(\Rightarrow \sqrt{5-2\sqrt{5}}+\sqrt{5+2\sqrt{5}}=\sqrt{10+2\sqrt{5}}\)
b)
\(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}=\sqrt{2^2+3-2.2\sqrt{3}}+\sqrt{2^2+3+2.2\sqrt{3}}\)
\(=\sqrt{(2-\sqrt{3})^2}+\sqrt{(2+\sqrt{3})^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}=4\)
c)
\(\sqrt{13-4\sqrt{3}}+\sqrt{13+4\sqrt{3}}=\sqrt{13-2\sqrt{12}}+\sqrt{13+2\sqrt{12}}\)
\(=\sqrt{12+1-2\sqrt{12}}+\sqrt{12+1+2\sqrt{12}}=\sqrt{(\sqrt{12}-1)^2}+\sqrt{(\sqrt{12}+1)^2}\)
\(=\sqrt{12}-1+\sqrt{12}+1=2\sqrt{12}=4\sqrt{3}\)
AH
Akai Haruma
Giáo viên
26 tháng 6 2019
Lời giải:
a)
\((\sqrt{5-2\sqrt{5}}+\sqrt{5+2\sqrt{5}})^2=5-2\sqrt{5}+5+2\sqrt{5}+2\sqrt{(5-2\sqrt{5})(5+2\sqrt{5})}\)
\(=10+2\sqrt{5^2-(2\sqrt{5})^2}=10+2\sqrt{5}\)
\(\Rightarrow \sqrt{5-2\sqrt{5}}+\sqrt{5+2\sqrt{5}}=\sqrt{10+2\sqrt{5}}\)
b)
\(\sqrt{7-4\sqrt{3}}+\sqrt{7+4\sqrt{3}}=\sqrt{2^2+3-2.2\sqrt{3}}+\sqrt{2^2+3+2.2\sqrt{3}}\)
\(=\sqrt{(2-\sqrt{3})^2}+\sqrt{(2+\sqrt{3})^2}\)
\(=2-\sqrt{3}+2+\sqrt{3}=4\)
c)
\(\sqrt{13-4\sqrt{3}}+\sqrt{13+4\sqrt{3}}=\sqrt{13-2\sqrt{12}}+\sqrt{13+2\sqrt{12}}\)
\(=\sqrt{12+1-2\sqrt{12}}+\sqrt{12+1+2\sqrt{12}}=\sqrt{(\sqrt{12}-1)^2}+\sqrt{(\sqrt{12}+1)^2}\)
\(=\sqrt{12}-1+\sqrt{12}+1=2\sqrt{12}=4\sqrt{3}\)
ND
10 tháng 8 2018
*\(A=2\sqrt{80\sqrt{7}}-2\sqrt{45\sqrt{7}}-5\sqrt{20\sqrt{7}}\)
\(A=16\sqrt{5\sqrt{7}}-6\sqrt{5\sqrt{7}}-10\sqrt{5\sqrt{7}}\)
\(A=\left(16-6-10\right)\sqrt{5\sqrt{7}}=0\)
* \(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
\(B^3=5+2\sqrt{13}+5-2\sqrt{13}+3\left(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\right).\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}\)
\(B^3=10-9B\)
\(\Rightarrow B^3+9B-10=0\)
\(\Rightarrow B^3-B^2+B^2-B+10B-10=0\)
\(\Rightarrow B^2\left(B-1\right)+B\left(B-1\right)+10\left(B-1\right)=0\)
\(\Rightarrow\left(B-1\right)\left(B^2+B+10\right)=0\)
\(\Rightarrow B=1\)
24 tháng 5 2022
b: \(=\dfrac{\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{2}}=-\sqrt{2}\)
c: \(=\dfrac{\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}-1-\sqrt{5}-1}{\sqrt{2}}=-\sqrt{2}\)
d: \(=\dfrac{\sqrt{18-2\sqrt{17}}-\sqrt{18+2\sqrt{17}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{17}-1-\sqrt{17}-1}{\sqrt{2}}=-\sqrt{2}\)
29 tháng 7 2018
sữa lại câu cuối cho Nhã Doanh
\(\sqrt{22-2\sqrt{21}-\sqrt{22+2\sqrt{21}}}=\sqrt{22-2\sqrt{21}-\sqrt{\left(\sqrt{21}+1\right)^2}}\)
\(=\sqrt{22-2\sqrt{21}-\sqrt{21}-1}=\sqrt{21-3\sqrt{21}}\)
ND
29 tháng 7 2018
\(a.\sqrt{8+2\sqrt{7}}-\sqrt{7}=\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{7}=\sqrt{7}+1-\sqrt{7}=1\)
\(b.\sqrt{7+4\sqrt{3}}-2\sqrt{3}=\sqrt{\left(2+\sqrt{3}\right)^2}-2\sqrt{3}=2+\sqrt{3}-2\sqrt{3}=2-\sqrt{3}\)
\(c.\sqrt{14-2\sqrt{13}}+\sqrt{14+2\sqrt{13}}=\sqrt{\left(\sqrt{13}-1\right)^2}+\sqrt{\left(\sqrt{13}+1\right)^2}=\sqrt{13}-1+\sqrt{13}+1=2\sqrt{13}\)\(d.\sqrt{22-2\sqrt{21}-\sqrt{22+2\sqrt{21}}}=\sqrt{\left(\sqrt{21}-1\right)^2-\sqrt{\left(\sqrt{21}+1\right)^2}}=\sqrt{21}-1-\sqrt{\sqrt{21}+1}\)
DH
1
MT
10 tháng 7 2015
đặt
\(A=\sqrt{7+\sqrt{13}}+\sqrt{7-\sqrt{13}}\)
=>\(\sqrt{2}A=\sqrt{2}\sqrt{7+\sqrt{13}}+\sqrt{2}\sqrt{7-\sqrt{13}}\)
\(=\sqrt{14+2\sqrt{13}}+\sqrt{14-2\sqrt{13}}\)
\(=\sqrt{13+2\sqrt{13}+1}+\sqrt{13-2\sqrt{13}+1}\)
\(=\sqrt{\left(\sqrt{13}+1\right)^2}+\sqrt{\left(\sqrt{13}-1\right)^2}\)
\(=\sqrt{13}+1+\sqrt{13}-1=2\sqrt{13}\)
=>\(A=\frac{2\sqrt{13}}{\sqrt{2}}=\frac{\sqrt{2}\sqrt{2}\sqrt{13}}{\sqrt{2}}=\sqrt{2}\sqrt{13}=\sqrt{26}\)
suy ra : ĐPCM
Đặt \(A=\sqrt{7+\sqrt{13}}+\sqrt{7-\sqrt{13}}\Rightarrow A^2=7+\sqrt{13}+7-\sqrt{13}+2\sqrt{\left(7+\sqrt{13}\right)\left(7-\sqrt{13}\right)}=14+2\sqrt{49-13}=14+2\sqrt{36}=14+12=26\Rightarrow A=\pm\sqrt{26}\)Mà \(\left\{{}\begin{matrix}\sqrt{7+\sqrt{13}}>0\\\sqrt{7-\sqrt{13}}>0\end{matrix}\right.\)⇒\(\sqrt{7+\sqrt{13}}+\sqrt{7-\sqrt{13}}>0\Rightarrow A>0\)
Vậy \(A=\sqrt{26}\Rightarrow\sqrt{7+\sqrt{13}}+\sqrt{7-\sqrt{13}}=\sqrt{26}\)
Mình thấy nhân cả 2 vế với \(\sqrt{2}\) nhanh hơn?