=> \(A^2=13+\sqrt{7+\sqrt{13+\sqrt{7+\sqrt{13+\sqrt{7+....}}}}}\)

=>\(\left(A^2-13\right)^2=7+\sqrt{13+\sqrt{7+\sqrt{13+\sqrt{7...}}}}\)

=>\(\left(A^2-13\right)^2=7+A\)

Đến đây tách ra giải PT bậc 4 nha!

Bạn giải giúp mình với bấm không ra nghiệm nơi

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

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:

\(\sqrt{7-\sqrt{13}}-\sqrt{7+\sqrt{13}}+\sqrt{2}=\sqrt{2}+\sqrt{2}=2\sqrt{2}\)

thanks bạn nha

\(1.\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}=\sqrt{5-2.\sqrt{2}.\sqrt{5}+2}-\sqrt{5+2.\sqrt{5}.\sqrt{2}+2}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}=\text{|}\sqrt{5}-\sqrt{2}\text{|}-\text{|}\sqrt{5}+\sqrt{2}\text{|}=-2\sqrt{2}\)\(2.\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}=\sqrt{8+2.2\sqrt{2}.\sqrt{5}+5}+\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5}=\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}=\text{|}2\sqrt{2}+\sqrt{5}\text{|}+\text{|}2\sqrt{2}-\sqrt{5}\text{|}=4\sqrt{2}\)\(3.\left(\sqrt{3}+\sqrt{5}\right)\sqrt{7-2\sqrt{10}}=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{5-2.\sqrt{5}.\sqrt{2}+2}=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\left(\sqrt{3}+\sqrt{5}\right)\text{|}\sqrt{5}-\sqrt{2}\text{|}=\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)\)

*\(A=2\sqrt{80\sqrt{7}}-2\sqrt{45\sqrt{7}}-5\sqrt{20\sqrt{7}}\)

\(A=16\sqrt{5\sqrt{7}}-6\sqrt{5\sqrt{7}}-10\sqrt{5\sqrt{7}}\)

\(A=\left(16-6-10\right)\sqrt{5\sqrt{7}}=0\)

* \(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)

\(B^3=5+2\sqrt{13}+5-2\sqrt{13}+3\left(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\right).\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}\)

\(B^3=10-9B\)

\(\Rightarrow B^3+9B-10=0\)

\(\Rightarrow B^3-B^2+B^2-B+10B-10=0\)

\(\Rightarrow B^2\left(B-1\right)+B\left(B-1\right)+10\left(B-1\right)=0\)

\(\Rightarrow\left(B-1\right)\left(B^2+B+10\right)=0\)

\(\Rightarrow B=1\)

bình lên cậu ạ

em nghĩ là 0

\(\sqrt{2}A=\sqrt{14+2\sqrt{13}}-\sqrt{14-2\sqrt{13}}\)

\(\sqrt{\left(\sqrt{13}+1\right)^2}-\sqrt{\left(\sqrt{13}-1\right)^2}=\sqrt{13}+1-\sqrt{13}+1=2\)

\(\Rightarrow A=\sqrt{2}\)  

a)\(A=\sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\)

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

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

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

b)\(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)

\(\Leftrightarrow B^3=5+2\sqrt{13}+3\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}\left(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5+2\sqrt{13}}\right)+5-2\sqrt{13}\)

\(\Leftrightarrow B^3=10+3.\sqrt[3]{-27}.B\)

\(\Leftrightarrow B^3+9B-10=0\)

\(\Leftrightarrow\left(B-1\right)\left(B^2+B+10\right)=0\)

\(\Leftrightarrow B=1\) (vì \(B^2+B+10>0\))

c)\(C=\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\)

\(\Leftrightarrow2C=\sqrt[3]{8\sqrt{5}+16}-\sqrt[3]{8\sqrt{5}-16}=\sqrt[3]{1+3\sqrt{5}+3\sqrt{5^2}+5\sqrt{5}}-\sqrt[3]{5\sqrt{5}-3\sqrt{5^2}+3\sqrt{5}-1}\)

\(=\sqrt[3]{\left(1+\sqrt{5}\right)^3}-\sqrt[3]{\left(\sqrt{5}-1\right)^3}\)

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

\(\Rightarrow C=1\)

d) \(D=\dfrac{10}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}\left(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\right)\)

\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{\left(\sqrt[3]{3}+\sqrt[3]{2}\right)\left(\sqrt[3]{9^2}-\sqrt[3]{6}+\sqrt[3]{2^2}\right)}\left(\dfrac{1+\sqrt{2}}{\sqrt{\left(1-\sqrt{3}\right)^2}}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\right)\)

\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{5}.\dfrac{1+\sqrt{2}}{\left|1-\sqrt{3}\right|}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)

\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(\sqrt{2}\right)^2-1}{\left(\sqrt{3}\right)^2-1}\)

\(=\sqrt[3]{3}+\sqrt[3]{2}\)

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