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*\(A=2\sqrt{80\sqrt{7}}-2\sqrt{45\sqrt{7}}-5\sqrt{20\sqrt{7}}\)
\(A=16\sqrt{5\sqrt{7}}-6\sqrt{5\sqrt{7}}-10\sqrt{5\sqrt{7}}\)
\(A=\left(16-6-10\right)\sqrt{5\sqrt{7}}=0\)
* \(B=\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\)
\(B^3=5+2\sqrt{13}+5-2\sqrt{13}+3\left(\sqrt[3]{5+2\sqrt{13}}+\sqrt[3]{5-2\sqrt{13}}\right).\sqrt[3]{\left(5+2\sqrt{13}\right)\left(5-2\sqrt{13}\right)}\)
\(B^3=10-9B\)
\(\Rightarrow B^3+9B-10=0\)
\(\Rightarrow B^3-B^2+B^2-B+10B-10=0\)
\(\Rightarrow B^2\left(B-1\right)+B\left(B-1\right)+10\left(B-1\right)=0\)
\(\Rightarrow\left(B-1\right)\left(B^2+B+10\right)=0\)
\(\Rightarrow B=1\)
sữa lại câu cuối cho Nhã Doanh
\(\sqrt{22-2\sqrt{21}-\sqrt{22+2\sqrt{21}}}=\sqrt{22-2\sqrt{21}-\sqrt{\left(\sqrt{21}+1\right)^2}}\)
\(=\sqrt{22-2\sqrt{21}-\sqrt{21}-1}=\sqrt{21-3\sqrt{21}}\)
\(a.\sqrt{8+2\sqrt{7}}-\sqrt{7}=\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{7}=\sqrt{7}+1-\sqrt{7}=1\)
\(b.\sqrt{7+4\sqrt{3}}-2\sqrt{3}=\sqrt{\left(2+\sqrt{3}\right)^2}-2\sqrt{3}=2+\sqrt{3}-2\sqrt{3}=2-\sqrt{3}\)
\(c.\sqrt{14-2\sqrt{13}}+\sqrt{14+2\sqrt{13}}=\sqrt{\left(\sqrt{13}-1\right)^2}+\sqrt{\left(\sqrt{13}+1\right)^2}=\sqrt{13}-1+\sqrt{13}+1=2\sqrt{13}\)\(d.\sqrt{22-2\sqrt{21}-\sqrt{22+2\sqrt{21}}}=\sqrt{\left(\sqrt{21}-1\right)^2-\sqrt{\left(\sqrt{21}+1\right)^2}}=\sqrt{21}-1-\sqrt{\sqrt{21}+1}\)
a, \(\sqrt{8-2\sqrt{15}}\)
= \(\sqrt{3-2\sqrt{15}+5}\)
= \(\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}\)
= |\(\sqrt{3}-\sqrt{5}\)| = \(\sqrt{5}-\sqrt{3}\) (Do \(\sqrt{5}>\sqrt{3}\))
b, \(\sqrt{9-4\sqrt{5}}\)
= \(\sqrt{5-4\sqrt{5}+4}\)
= \(\sqrt{\left(\sqrt{5}-2\right)^2}\)
= \(\sqrt{5}-2\) (Lười quá bỏ trị tuyệt đối cũng được :v)
Phần c sao sao ý (chắc do mk ngu :v)
d, \(\sqrt{7-2\sqrt{10}}+\sqrt{20}+\frac{1}{2}\sqrt{8}\)
= \(\sqrt{5-2\sqrt{10}+2}+\sqrt{20}+\sqrt{2}\)
= \(\sqrt{5}-\sqrt{2}+\sqrt{20}+\sqrt{2}\)
= \(\sqrt{5}+\sqrt{20}\)
= \(\sqrt{5}\left(1+\sqrt{4}\right)\) = \(3\sqrt{5}\)
Chúc bn học tốt! (Sorry phần c mk thấy sao sao ý nên chịu :v)
a) Sửa đề: \(A=\sqrt{8+2\sqrt{7}}-\sqrt{7}\)
Ta có: \(A=\sqrt{8+2\sqrt{7}}-\sqrt{7}\)
\(=\sqrt{7+2\cdot\sqrt{7}\cdot1+1}-\sqrt{7}\)
\(=\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{7}\)
\(=\left|\sqrt{7}+1\right|-\sqrt{7}\)
\(=\sqrt{7}+1-\sqrt{7}\)
=1
b) Ta có: \(B=\sqrt{7+4\sqrt{3}}-2\sqrt{3}\)
\(=\sqrt{4+2\cdot2\cdot\sqrt{3}+3}-2\sqrt{3}\)
\(=\sqrt{\left(2+\sqrt{3}\right)^2}-2\sqrt{3}\)
\(=\left|2+\sqrt{3}\right|-2\sqrt{3}\)
\(=2+\sqrt{3}-2\sqrt{3}\)
\(=2-\sqrt{3}\)
c) Ta có: \(C=\sqrt{14-2\sqrt{13}}+\sqrt{14+2\sqrt{13}}\)
\(=\sqrt{13-2\cdot\sqrt{13}\cdot1+1}+\sqrt{13+2\cdot\sqrt{13}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{13}-1\right)^2}+\sqrt{\left(\sqrt{13}+1\right)^2}\)
\(=\left|\sqrt{13}-1\right|+\left|\sqrt{13}+1\right|\)
\(=\sqrt{13}-1+\sqrt{13}+1\)
\(=2\sqrt{13}\)
d) Ta có: \(D=\sqrt{22-2\sqrt{21}}-\sqrt{22+2\sqrt{21}}\)
\(=\sqrt{21-2\cdot\sqrt{21}\cdot1+1}-\sqrt{21+2\cdot\sqrt{21}\cdot1+1}\)
\(=\sqrt{\left(\sqrt{21}-1\right)^2}-\sqrt{\left(\sqrt{21}+1\right)^2}\)
\(=\left|\sqrt{21}-1\right|-\left|\sqrt{21}+1\right|\)
\(=\sqrt{21}-1-\left(\sqrt{21}+1\right)\)
\(=\sqrt{21}-1-\sqrt{21}-1\)
=-2
\(x=\sqrt[3]{13-7\sqrt{6}}+\sqrt[3]{13+7\sqrt{6}}\Rightarrow x^3=26-15x\)
\(x^3+15x-25=1\Rightarrow\left(x^3+15x-25\right)^{2013}=1\)
Vậy P(x)=1 với .....
\(a.\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}=\sqrt{49-2.7.3\sqrt{5}+45}-\sqrt{49+2.7.3\sqrt{5}+45}=7-3\sqrt{5}-7-3\sqrt{5}=-6\sqrt{5}\) \(b.\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\dfrac{\sqrt{7+2\sqrt{7}+1}-\sqrt{7-2\sqrt{7}+1}}{\sqrt{2}}=\dfrac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}=\sqrt{2}\) \(c.\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{5-\sqrt{12+2.2\sqrt{3}+1}}=\sqrt{3-2\sqrt{3}+1}=\sqrt{3}-1\)
=> \(A^2=13+\sqrt{7+\sqrt{13+\sqrt{7+\sqrt{13+\sqrt{7+....}}}}}\)
=>\(\left(A^2-13\right)^2=7+\sqrt{13+\sqrt{7+\sqrt{13+\sqrt{7...}}}}\)
=>\(\left(A^2-13\right)^2=7+A\)
Đến đây tách ra giải PT bậc 4 nha!
Bạn giải giúp mình với bấm không ra nghiệm nơi