Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Sửa đề
\(A=\left(2-\sqrt{3}\right)\sqrt[3]{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt[3]{26-15\sqrt{3}}\)
\(=\left(2-\sqrt{3}\right)\sqrt[3]{8+12\sqrt{3}+18+3\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt[3]{8-12\sqrt{3}+18-3\sqrt{3}}\)
\(=\left(2-\sqrt{3}\right)\sqrt[3]{\left(2+\sqrt{3}\right)^3}-\left(2+\sqrt{3}\right)\sqrt[3]{\left(2-\sqrt{3}\right)^3}\)
\(=\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)-\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)=0\)
Thu gọn:
\(\left(2-\sqrt{3}\right)\sqrt{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}\)
Ta có: \(\left(2-\sqrt{3}\right)\sqrt{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}\)
\(=\frac{\left(2-\sqrt{3}\right)\sqrt{52+30\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{52-30\sqrt{3}}}{\sqrt{2}}\)
\(=\frac{\left(2-\sqrt{3}\right)\cdot\sqrt{27+2\cdot3\sqrt{3}\cdot5+25}-\left(2+\sqrt{3}\right)\sqrt{27-2\cdot3\sqrt{3}\cdot5+25}}{\sqrt{2}}\)
\(=\frac{\left(2-\sqrt{3}\right)\sqrt{\left(3\sqrt{3}+5\right)^2}-\left(2+\sqrt{3}\right)\cdot\sqrt{\left(3\sqrt{3}-5\right)^2}}{\sqrt{2}}\)
\(=\frac{\left(2-\sqrt{3}\right)\left|3\sqrt{3}+5\right|-\left(2+\sqrt{3}\right)\left|3\sqrt{3}-5\right|}{\sqrt{2}}\)
\(=\frac{\left(2-\sqrt{3}\right)\left(3\sqrt{3}+5\right)-\left(2+\sqrt{3}\right)\left(3\sqrt{3}-5\right)}{\sqrt{2}}\)(Vì \(3\sqrt{3}>5>0\))
\(=\frac{6\sqrt{3}+10-9-5\sqrt{3}-\left(6\sqrt{3}-10+9-5\sqrt{3}\right)}{\sqrt{2}}\)
\(=\frac{\sqrt{3}+1-\left(\sqrt{3}-1\right)}{\sqrt{2}}\)
\(=\frac{\sqrt{3}+1-\sqrt{3}+1}{\sqrt{2}}\)
\(=\frac{2}{\sqrt{2}}=\sqrt{2}\)
a)
\(A=\sqrt{26+15\sqrt{3}}=\sqrt{\frac{52+30\sqrt{3}}{2}}=\sqrt{\frac{27+25+2\sqrt{27.25}}{2}}\)
\(=\sqrt{\frac{(\sqrt{27}+\sqrt{25})^2}{2}}=\frac{\sqrt{27}+\sqrt{25}}{\sqrt{2}}=\frac{3\sqrt{3}+5}{\sqrt{2}}=\frac{3\sqrt{6}+5\sqrt{2}}{2}\)
b)
\(B\sqrt{2}=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}-2\)
\(=\sqrt{7+1+2\sqrt{7}}-\sqrt{7+1-2\sqrt{7}}-2\)
\(=\sqrt{(\sqrt{7}+1)^2}-\sqrt{(\sqrt{7}-1)^2}-2=\sqrt{7}+1-(\sqrt{7}-1)-2=0\)
\(\Rightarrow B=0\)
c)
\(C=\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}=\sqrt{3+5-2\sqrt{3.5}}-\sqrt{3+5+2\sqrt{3.5}}\)
\(=\sqrt{(\sqrt{5}-\sqrt{3})^2}-\sqrt{(\sqrt{5}+\sqrt{3})^2}=(\sqrt{5}-\sqrt{3})-(\sqrt{5}+\sqrt{3})=-2\sqrt{3}\)
d)
\(D=(\sqrt{6}-2)(5+2\sqrt{6})\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{2}(\sqrt{3}-\sqrt{2})(2+3+2\sqrt{2.3})\sqrt{2+3-2\sqrt{2.3}}\)
\(=\sqrt{2}(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})^2\sqrt{(\sqrt{3}-\sqrt{2})^2}\)
\(=\sqrt{2}(\sqrt{3}-\sqrt{2})^2(\sqrt{3}+\sqrt{2})^2=\sqrt{2}[(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})]^2\)
\(=\sqrt{2}.1^2=\sqrt{2}\)
e)
\(E=(\sqrt{10}-\sqrt{2})\sqrt{3+\sqrt{5}}=(\sqrt{5}-1).\sqrt{2}.\sqrt{3+\sqrt{5}}\)
\(=(\sqrt{5}-1)\sqrt{6+2\sqrt{5}}=(\sqrt{5}-1)\sqrt{5+1+2\sqrt{5.1}}\)
\(=(\sqrt{5}-1)\sqrt{(\sqrt{5}+1)^2}=(\sqrt{5}-1)(\sqrt{5}+1)=4\)
f)
\(F=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20+9-2\sqrt{20.9}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{(\sqrt{20}-3)^2}}}=\sqrt{\sqrt{5}-\sqrt{3-(\sqrt{20}-3)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{5+1-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{(\sqrt{5}-1)^2}}=\sqrt{\sqrt{5}-(\sqrt{5}-1)}=\sqrt{1}=1\)
\(=\frac{\left(2-\sqrt{3}\right)\cdot\sqrt{2}\cdot\sqrt{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\cdot\sqrt{2}\cdot\sqrt{26-15\sqrt{3}}}{\sqrt{2}}\)
\(=\frac{\left(2-\sqrt{3}\right)\cdot\sqrt{52+2\sqrt{675}}-\left(2+\sqrt{3}\right)\cdot\sqrt{52-2\sqrt{675}}}{\sqrt{2}}\)
\(=\frac{\left(2-\sqrt{3}\right)\cdot\sqrt{27+2\cdot\sqrt{27\cdot25}+25}-\left(2+\sqrt{3}\right)\cdot\sqrt{27-2\sqrt{27\cdot25}+25}}{\sqrt{2}}\)
\(=\frac{\left(2-\sqrt{3}\right)\left(3\sqrt{3}+5\right)-\left(2+\sqrt{3}\right)\left(3\sqrt{3}-5\right)}{\sqrt{2}}\)
\(=\frac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{\sqrt{2}}=\frac{2}{\sqrt{2}}=\sqrt{2}\)
Xét: \(A=\sqrt{26+15\sqrt{3}}\) dễ thấy A > 0
\(\Leftrightarrow A^2=52-2\sqrt{26^2-15^2.3}=50\Leftrightarrow A=\sqrt{50}\)
Vậy: \(A=2+\sqrt{3}.\sqrt{26+15\sqrt{3}}-2\sqrt{3}.\sqrt{26-15\sqrt{3}}\)
\(=2+\sqrt{3}.A=2+\sqrt{3}.\sqrt{50}=5\sqrt{6}+10\sqrt{2}\)
a) \(\sqrt{26+15\sqrt{3}}\)
\(=\frac{\sqrt{52+30\sqrt{3}}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(3\sqrt{3}\right)^2+2.3\sqrt{3}.5+5^2}}{\sqrt{2}}\)
\(=\frac{\sqrt{\left(3\sqrt{3}+5\right)^2}}{\sqrt{2}}=\frac{3\sqrt{3}+5}{\sqrt{2}}\)
b) \(\)\(\sqrt{2-\sqrt{3}}=\frac{\sqrt{4-2\sqrt{3}}}{\sqrt{2}}=\frac{\sqrt{\left(\sqrt{3}-1\right)^2}}{\sqrt{2}}\)
\(=\frac{\left|\sqrt{3}-1\right|}{\sqrt{2}}=\frac{\sqrt{3}-1}{\sqrt{2}}\)
c) \(\left(\sqrt{10}-\sqrt{2}\right).\left(\sqrt{3+5}\right)\)
\(=\sqrt{10}.\sqrt{8}-\sqrt{2}.\sqrt{8}\)
\(=\sqrt{80}-\sqrt{16}=4\sqrt{5}-4\)
d) \(\left(\sqrt{6}-2\right)\left(5+\sqrt{24}\right)\sqrt{5-\sqrt{24}}\)
\(=\left(\sqrt{6}-2\right)\left(\sqrt{5+\sqrt{24}}\right).\sqrt{5-\sqrt{24}}.\left(\sqrt{5+\sqrt{24}}\right)\)
\(=\left(\sqrt{6}-2\right)\left(\sqrt{5+\sqrt{24}}\right).1\)
\(=\left(\sqrt{6}-2\right).\left(\sqrt{5+\sqrt{24}}\right)\)
\(=\sqrt{2}.\left(\sqrt{3}-\sqrt{2}\right).\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=\sqrt{2}.\left(3-2\right)=\sqrt{2}\)
\(B^2=\left(2-\sqrt{3}\right)^2.\left(26+15\sqrt{3}\right)+\left(2+\sqrt{3}\right)^2.\left(26-15\sqrt{3}\right)-2\left(4-3\right)\sqrt{26^2-3.15^2}\)
\(B^2=\left(7-4\sqrt{3}\right).\left(26+15\sqrt{3}\right)+\left(7+4\sqrt{3}\right)\left(26-15\sqrt{3}\right)-2\)
\(B^2+2=\left(a-b\right)\left(c+d\right)+\left(a+b\right)\left(c-d\right)=ac+ad-bc-bd+ac-ad+bc-bd=2\left(ac-bd\right)\)\(B^2+2=2.\left(7.26-4.3.15\right)=2\left(182-180\right)\Rightarrow B^2=2\)
\(B>0\Rightarrow B=\sqrt{2}\)