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Chứng minh rằng số \sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}} là bình phương của một số nguyên.
a = \(\sqrt[3]{26+15\sqrt{3}}\)+\(\sqrt[3]{26-15\sqrt{3}}\)=\(\sqrt[3]{8+2.3.3+3.4.\sqrt{3}+3\sqrt{3}}+\sqrt[3]{8-3.4.\sqrt{3}+2.3.3-3\sqrt{3}}\)
=\(\sqrt[3]{2+\sqrt{3}}^3\)+\(\sqrt[3]{2-\sqrt{3}}^3\)
=2+\(\sqrt{3}\)+2-\(\sqrt{3}\)
=4=\(2^2\)
Ta có \(a=\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}}=\sqrt[3]{8+12\sqrt{3}+18+3\sqrt{3}}+\sqrt[3]{8-12\sqrt{3}+18-3\sqrt{3}}=\sqrt[3]{2^3+3.2^2.\sqrt{3}+3.2.\left(\sqrt{3}\right)^2+\left(\sqrt{3}\right)^3}+\sqrt[3]{2^3-3.2^2.\sqrt{3}+3.2.\left(\sqrt{3}\right)^2-\left(\sqrt{3}\right)^3}=\sqrt[3]{\left(2+\sqrt{3}\right)^3}+\sqrt[3]{\left(2-\sqrt{3}\right)^3}=2+\sqrt{3}+2-\sqrt{3}=4=2^2\)
Vậy a là bình phương của một số nguyên
a) Ta có: \(M=\dfrac{2}{\sqrt{7}-\sqrt{6}}-\sqrt{28}+\sqrt{54}\)
\(=\dfrac{2\left(\sqrt{7}+\sqrt{6}\right)}{\left(\sqrt{7}-\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right)}-2\sqrt{7}+3\sqrt{6}\)
\(=2\sqrt{7}+2\sqrt{6}-2\sqrt{7}+3\sqrt{6}\)
\(=5\sqrt{6}\)
b) Ta có: \(N=\left(2-\sqrt{3}\right)\left(\sqrt{26+15\sqrt{3}}\right)-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{52+30\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{52-30\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{27+2\cdot3\sqrt{3}\cdot5+25}-\left(2+\sqrt{3}\right)\sqrt{27-2\cdot3\sqrt{3}\cdot5+25}}{\sqrt{2}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{\left(3\sqrt{3}+5\right)^2}-\left(2+\sqrt{3}\right)\sqrt{\left(3\sqrt{3}-5\right)^2}}{\sqrt{2}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\left(3\sqrt{3}+5\right)-\left(2+\sqrt{3}\right)\left(3\sqrt{3}-5\right)}{\sqrt{2}}\)
\(=\dfrac{6\sqrt{3}+10-9-5\sqrt{3}-\left(6\sqrt{3}-10+9-5\sqrt{3}\right)}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}+1-\sqrt{3}+1}{\sqrt{2}}\)
\(=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
\(M^3=\sqrt{5}+2-\left(\sqrt{5}-2\right)-3\sqrt[3]{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}\left(\sqrt[3]{\sqrt{5}+2}-\sqrt[3]{\sqrt{5}-2}\right)\)
\(\Rightarrow M^3=4-3M\)
\(\Leftrightarrow\left(M-1\right)\left(M^2+M+4\right)\)
\(\Leftrightarrow M=1\text{ (do }M^2+M+4=\left(M+\frac{1}{2}\right)^2+\frac{15}{4}>0\text{)}\)
Vậy M = 1
b/ làm tương tự
\(=\sqrt{\left(2-\sqrt{3}\right)^2\left(26+15\sqrt{3}\right)}-\sqrt{\left(2+\sqrt{3}\right)^2\left(26-15\sqrt{3}\right)}=\)
\(=\sqrt{\left(7-4\sqrt{3}\right)\left(26+15\sqrt{3}\right)}-\sqrt{\left(7+4\sqrt{3}\right)\left(26-15\sqrt{3}\right)=}\)
\(=\sqrt{7.26+7.15\sqrt{3}-4.26\sqrt{3}-180}-\sqrt{7.26-7.15\sqrt{3}+4.26\sqrt{3}-180}=\)
\(=\sqrt{4+\sqrt{3}}-\sqrt{4-\sqrt{3}}\)