\(\sqrt{7-\sqrt{13}}\)-\(\sqrt{7+\sqrt{13}}\)+\(\sqrt{2}\)
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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
\(1.\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}=\sqrt{5-2.\sqrt{2}.\sqrt{5}+2}-\sqrt{5+2.\sqrt{5}.\sqrt{2}+2}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}=\text{|}\sqrt{5}-\sqrt{2}\text{|}-\text{|}\sqrt{5}+\sqrt{2}\text{|}=-2\sqrt{2}\)\(2.\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}=\sqrt{8+2.2\sqrt{2}.\sqrt{5}+5}+\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5}=\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}=\text{|}2\sqrt{2}+\sqrt{5}\text{|}+\text{|}2\sqrt{2}-\sqrt{5}\text{|}=4\sqrt{2}\)\(3.\left(\sqrt{3}+\sqrt{5}\right)\sqrt{7-2\sqrt{10}}=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{5-2.\sqrt{5}.\sqrt{2}+2}=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\left(\sqrt{3}+\sqrt{5}\right)\text{|}\sqrt{5}-\sqrt{2}\text{|}=\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)\)
24 tháng 6 2021
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) \(D=\dfrac{10}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}\left(\dfrac{1+\sqrt{2}}{\sqrt{4-2\sqrt{3}}}:\dfrac{\sqrt{3}+1}{\sqrt{2}-1}\right)\)
\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{\left(\sqrt[3]{3}+\sqrt[3]{2}\right)\left(\sqrt[3]{9^2}-\sqrt[3]{6}+\sqrt[3]{2^2}\right)}\left(\dfrac{1+\sqrt{2}}{\sqrt{\left(1-\sqrt{3}\right)^2}}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\right)\)
\(=\dfrac{10\left(\sqrt[3]{3}+\sqrt[3]{2}\right)}{5}.\dfrac{1+\sqrt{2}}{\left|1-\sqrt{3}\right|}.\dfrac{\sqrt{2}-1}{\sqrt{3}+1}\)
\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=2\left(\sqrt[3]{3}+\sqrt[3]{2}\right).\dfrac{\left(\sqrt{2}\right)^2-1}{\left(\sqrt{3}\right)^2-1}\)
\(=\sqrt[3]{3}+\sqrt[3]{2}\)
Vậy...
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 6 2023
g: \(=\left|\sqrt{6}-1\right|=\sqrt{6}-1\)
h: \(=\left|2\sqrt{3}-1\right|=2\sqrt{3}-1\)
l: \(=\left|2-\sqrt{3}\right|-2=2-\sqrt{3}-2=-\sqrt{3}\)
j: \(=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|\)
\(=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)
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)
\(\sqrt{13-4\sqrt{3}}+\sqrt{13+4\sqrt{3}}=\sqrt{13-2\sqrt{12}}+\sqrt{13+2\sqrt{12}}\)
\(=\sqrt{12+1-2\sqrt{12}}+\sqrt{12+1+2\sqrt{12}}=\sqrt{(\sqrt{12}-1)^2}+\sqrt{(\sqrt{12}+1)^2}\)
\(=\sqrt{12}-1+\sqrt{12}+1=2\sqrt{12}=4\sqrt{3}\)
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
\(a.\sqrt{8+2\sqrt{7}}-\sqrt{7}=\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{7}=\sqrt{7}+1-\sqrt{7}=1\)
\(b.\sqrt{7+4\sqrt{3}}-2\sqrt{3}=\sqrt{\left(2+\sqrt{3}\right)^2}-2\sqrt{3}=2+\sqrt{3}-2\sqrt{3}=2-\sqrt{3}\)
\(c.\sqrt{14-2\sqrt{13}}+\sqrt{14+2\sqrt{13}}=\sqrt{\left(\sqrt{13}-1\right)^2}+\sqrt{\left(\sqrt{13}+1\right)^2}=\sqrt{13}-1+\sqrt{13}+1=2\sqrt{13}\)\(d.\sqrt{22-2\sqrt{21}-\sqrt{22+2\sqrt{21}}}=\sqrt{\left(\sqrt{21}-1\right)^2-\sqrt{\left(\sqrt{21}+1\right)^2}}=\sqrt{21}-1-\sqrt{\sqrt{21}+1}\)
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