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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)\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=1\)\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=1\)
\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}=1\)
\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=1\)
\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=1\)
\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)
\(\Leftrightarrow\sqrt{1}=1\) (đpcm)
Đặt \(x=\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt[]{3}}\)
\(\Rightarrow x^3=52+3\sqrt[3]{\left(26+15\sqrt[]{3}\right)\left(26-15\sqrt[]{3}\right)}.x\)
\(\Leftrightarrow x^3=52+3x\)
\(\Leftrightarrow x^3-3x-52=0\)
\(\Leftrightarrow\left(x-4\right)\left(x^2+4x+13\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left[\left(x+2\right)^2+9\right]=0\)
\(\Leftrightarrow x=4\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2\left(26+15\sqrt{3}\right)}-\sqrt{\left(2+\sqrt{3}\right)^2\left(26-15\sqrt{3}\right)}=\)
\(=\sqrt{\left(7-4\sqrt{3}\right)\left(26+15\sqrt{3}\right)}-\sqrt{\left(7+4\sqrt{3}\right)\left(26-15\sqrt{3}\right)=}\)
\(=\sqrt{7.26+7.15\sqrt{3}-4.26\sqrt{3}-180}-\sqrt{7.26-7.15\sqrt{3}+4.26\sqrt{3}-180}=\)
\(=\sqrt{4+\sqrt{3}}-\sqrt{4-\sqrt{3}}\)
Ta có: \(\hept{\begin{cases}\left(2-\sqrt{3}\right)^2.\left(26+15\sqrt{3}\right)=2+\sqrt{3}\\\left(2+\sqrt{3}\right)^2.\left(26-15\sqrt{3}\right)=2-\sqrt{3}\end{cases}}\)
Sửa đề:
\(\left(2-\sqrt{3}\right)\sqrt{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}=\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)
\(=\frac{1}{\sqrt{2}}\left(\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\right)\)
\(=\frac{1}{\sqrt{2}}\left(\sqrt{3}+1-\sqrt{3}+1\right)\)
\(=\sqrt{2}\)
Lời giải:
Đặt \(\sqrt[3]{26+15\sqrt{3}}=a; \sqrt[3]{26-15\sqrt{3}}=b\)
\(\Rightarrow \left\{\begin{matrix} a^3+b^3=52\\ ab=\sqrt[3]{(26+15\sqrt{3})(26-15\sqrt{3})}=\sqrt[3]{1}=1\end{matrix}\right.\)
Khi đó:
\((\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}})^3=(a+b)^3=a^3+b^3+3ab(a+b)\)
\(\Leftrightarrow (a+b)^3=52+3(a+b)\). Đặt \(a+b=t\)
\(\Leftrightarrow t^3-3t-52=0\)
\(\Leftrightarrow t(t^2-16)+13(t-4)=0\)
\(\Leftrightarrow (t-4)(t^2+4t+13)=0\)
Thấy rằng \(t^2+4t+13=(t+2)^2+9>0\forall t\) nên \(t-4=0\Leftrightarrow t=4\)
Vậy giá trị của biểu thức là $4$
\(a=\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}}\)
\(a^3=26+15\sqrt{3}+26-15\sqrt{3}+3\sqrt[3]{\left(26+15\sqrt{3}\right)\left(26-15\sqrt{3}\right)}a\)
\(a^3=52+3.\sqrt{26^2-3.15^2}.a\)
\(a^3=52+3a\)
\(a^3-64=3a-12\)
\(a^3-4^3=3\left(a-4\right)\)
\(\left(a-4\right)\left(a^2+4a+16\right)=3\left(a-4\right)\)
\(\left(a-4\right)\left(a^2+4a+13\right)=0\) \(\Delta_{a^2+4a+13}=4-13< 0=>vn\)
a=4 là duy nhất
\(B^2=\left(2-\sqrt{3}\right)^2.\left(26+15\sqrt{3}\right)+\left(2+\sqrt{3}\right)^2.\left(26-15\sqrt{3}\right)-2\left(4-3\right)\sqrt{26^2-3.15^2}\)
\(B^2=\left(7-4\sqrt{3}\right).\left(26+15\sqrt{3}\right)+\left(7+4\sqrt{3}\right)\left(26-15\sqrt{3}\right)-2\)
\(B^2+2=\left(a-b\right)\left(c+d\right)+\left(a+b\right)\left(c-d\right)=ac+ad-bc-bd+ac-ad+bc-bd=2\left(ac-bd\right)\)\(B^2+2=2.\left(7.26-4.3.15\right)=2\left(182-180\right)\Rightarrow B^2=2\)
\(B>0\Rightarrow B=\sqrt{2}\)