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- \(\sqrt{\sqrt{5}^2-2.2\sqrt{5}+4}-\sqrt{5}=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}=\sqrt{5}-2-\sqrt{5}=-2\left(dpcm\right)\)
- \(\sqrt{23+8\sqrt{7}}-\sqrt{7}=\sqrt{\sqrt{7}^2+2.4\sqrt{7}+16}-\sqrt{7}\)\(=\sqrt{\left(\sqrt{7}+4\right)^2}-\sqrt{7}=\sqrt{7}+4-\sqrt{7}=4\left(DPCM\right)\)
a) \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}=\sqrt{1+2\sqrt{5}+\left(\sqrt{5}\right)^2}+\sqrt{1-2\sqrt{5}+\left(\sqrt{5}\right)^2}\)\(=\sqrt{\left(1+\sqrt{5}\right)^2}+\sqrt{\left(1-\sqrt{5}\right)^2}=1+\sqrt{5}-\left(1-\sqrt{5}\right)=1+\sqrt{5}-1+\sqrt{5}=2\sqrt{5}\)
a) \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}+1\right)^2}+\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}+1+\sqrt{5}-1=2\sqrt{5}\)
b) \(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}\)
\(=\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}\)
\(=2\sqrt{2}+\sqrt{5}+2\sqrt{2}-\sqrt{5}=4\sqrt{2}\)
c) \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
\(=2\sqrt{2\sqrt{3}}-10\sqrt{2\sqrt{3}}+8\sqrt{2\sqrt{3}}=0\)
Câu b, c tương tự câu a. Mình làm câu a coi như tượng trưng nha !!!!!!
a) Đặt: \(A=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)
<=> \(A^3=2+\sqrt{5}+2-\sqrt{5}+3\sqrt[3]{\left(2-\sqrt{5}\right)\left(2+\sqrt{5}\right)}.\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)
<=> \(A^3=4+3\sqrt[3]{4-5}.A\)
<=> \(A^3=4-3A\)
<=> \(A^3+3A-4=0\)
<=> \(\left(A-1\right)\left(A^2+A+4\right)=0\)
Có: \(A^2+A+4=\left(A+\frac{1}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}>0\)
=> \(A-1=0\)
<=> \(A=1\)
=> \(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=1\)
VẬY TA CÓ ĐPCM
Sửa đề: \(\sqrt{5-2\sqrt{6}}-\sqrt{5+2\sqrt{6}}\)
Ta có: \(\sqrt{5-2\sqrt{6}}-\sqrt{5+2\sqrt{6}}\)
\(=\sqrt{2-2\cdot\sqrt{2}\cdot\sqrt{3}+3}-\sqrt{2+2\cdot\sqrt{2}\cdot\sqrt{3}+3}\)
\(=\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}\)
\(=\left|\sqrt{2}-\sqrt{3}\right|-\left|\sqrt{2}+\sqrt{3}\right|\)
\(=\sqrt{3}-\sqrt{2}-\left(\sqrt{2}+\sqrt{3}\right)\)
\(=\sqrt{3}-\sqrt{2}-\sqrt{2}-\sqrt{3}\)
\(=-2\sqrt{2}\)(đpcm)
CM : \(\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\)
Giải :
VT= \(\sqrt{9-4\sqrt{5}}-\sqrt{5}=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}=\left|\sqrt{5}-2\right|-\sqrt{5}=\sqrt{5}-2-\sqrt{5}=-2\)
Thấy VT = VP = - 2
=> \(\sqrt{9-4\sqrt{5}}-\sqrt{5}=-2\) ( đpcm )