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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)\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=1\)\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}=1\)
\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}=1\)
\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=1\)
\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=1\)
\(\Leftrightarrow\sqrt{\sqrt{5}-\sqrt{5}+1}=1\)
\(\Leftrightarrow\sqrt{1}=1\) (đpcm)
\(A=\sqrt[3]{26+15\sqrt{3}}-\sqrt[3]{26-15\sqrt{3}}=\sqrt[3]{8+3.4.\sqrt{3}+3.2.3+3\sqrt{3}}-\sqrt[3]{8-3.4.\sqrt{3}+3.2.3-3\sqrt{3}}=\sqrt[3]{\left(2+\sqrt{3}\right)^3}-\sqrt[3]{\left(2-\sqrt{3}\right)^3}=2+\sqrt{3}-2+\sqrt{3}=2\sqrt{3}\)
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: \(A=2+\sqrt{3}.\sqrt{26+15\sqrt{3}}-2\sqrt{3}.\sqrt{26-15\sqrt{3}}\)
\(=2+\sqrt{3}.A=2+\sqrt{3}.\sqrt{50}=5\sqrt{6}+10\sqrt{2}\)
Đặt \(x=\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt[]{3}}\)
\(\Rightarrow x^3=52+3\sqrt[3]{\left(26+15\sqrt[]{3}\right)\left(26-15\sqrt[]{3}\right)}.x\)
\(\Leftrightarrow x^3=52+3x\)
\(\Leftrightarrow x^3-3x-52=0\)
\(\Leftrightarrow\left(x-4\right)\left(x^2+4x+13\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left[\left(x+2\right)^2+9\right]=0\)
\(\Leftrightarrow x=4\)
a)\(A=^3\sqrt{20+14\sqrt{2}}+^3\sqrt{20-14\sqrt{2}}\)
=> \(A^3=\left(\sqrt[3]{20+14\sqrt{2}}+\sqrt[3]{20-14\sqrt{2}}\right)^3\)
\(=20+14\sqrt{2}+20-14\sqrt{2}\)
\(+3\left(\text{}^3\sqrt{20+14\sqrt{2}}+^3\sqrt{20-14\sqrt{2}}\right)\left(^3\sqrt{20+14\sqrt{2}}.^3\sqrt{20-14\sqrt{2}}\right)\)
\(=40+3A.^3\sqrt{\left(20+14\sqrt{2}\right)\left(20+14\sqrt{2}\right)}\)
\(\Rightarrow A^3=40+3.A.2\)
=> \(A^3-6A-40=0\)
<=> \(A^3-16A+10A-40=0\)
<=> \(A\left(A-4\right)\left(A+4\right)+10\left(A-4\right)=0\)
<=> \(\left(A-4\right)\left(A^2+4A+10\right)=0\)
<=> A = 4 ( vì \(A^2+4A+10=\left(A+2\right)^2+6>0\))
Vậy A = 4.
b/ \(B=^3\sqrt{26+15\sqrt{3}}-^3\sqrt{26-15\sqrt{3}}\)
=> \(B^3=\left(^3\sqrt{26+15\sqrt{3}}-^3\sqrt{26-15\sqrt{3}}\right)^3\)
\(=26+15\sqrt{3}-26+15\sqrt{3}\)
\(-3\left(^3\sqrt{26+15\sqrt{3}}-^3\sqrt{26-15\sqrt{3}}\right).^3\sqrt{26+15\sqrt{3}}.^3\sqrt{26-15\sqrt{3}}\)
\(=30\sqrt{3}-3B.1\)
=> \(B^3+3B-30\sqrt{3}=0\)
<=> \(B^3-12B+15B-30\sqrt{3}=0\)
<=> \(B\left(B-2\sqrt{3}\right)\left(B+2\sqrt{3}\right)+15\left(B-2\sqrt{3}\right)=0\)
<=> \(\left(B-2\sqrt{3}\right)\left(B^2+2\sqrt{3}B+15\right)=0\)
<=> \(B-2\sqrt{3}=0\)( vì \(B^2+2\sqrt{3}B+15=\left(B+\sqrt{3}\right)^2+12>0\))
<=> \(B=2\sqrt{3}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2\left(26+15\sqrt{3}\right)}-\sqrt{\left(2+\sqrt{3}\right)^2\left(26-15\sqrt{3}\right)}=\)
\(=\sqrt{\left(7-4\sqrt{3}\right)\left(26+15\sqrt{3}\right)}-\sqrt{\left(7+4\sqrt{3}\right)\left(26-15\sqrt{3}\right)=}\)
\(=\sqrt{7.26+7.15\sqrt{3}-4.26\sqrt{3}-180}-\sqrt{7.26-7.15\sqrt{3}+4.26\sqrt{3}-180}=\)
\(=\sqrt{4+\sqrt{3}}-\sqrt{4-\sqrt{3}}\)
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