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Trả lời:
\(E=\sqrt[3]{\sqrt{5}-2}+\sqrt[3]{\sqrt{5}+2}\)
\(2E=2.\sqrt[3]{\sqrt{5}-2}+2.\sqrt[3]{\sqrt{5}+2}\)
\(2E=\sqrt[3]{8\sqrt{5}-16}+\sqrt[3]{8\sqrt{5}+16}\)
\(2E=\sqrt[3]{5\sqrt{5}-15+3\sqrt{5}-1}+\sqrt[3]{5\sqrt{5}+15+3\sqrt{5}+1}\)
\(2E=\sqrt[3]{\left(\sqrt{5}-1\right)^3}+\sqrt[3]{\left(\sqrt{5}+1\right)^3}\)
\(2E=\sqrt{5}-1+\sqrt{5}+1\)
\(2E=2\sqrt{5}\)
\(E=\sqrt{5}\)
\(F=\sqrt[3]{182+\sqrt{33125}}+\sqrt[3]{182-\sqrt{33125}}\)
\(F=\sqrt[3]{182+25\sqrt{53}}+\sqrt[3]{182-25\sqrt{53}}\)
\(2F=2.\sqrt[3]{182+25\sqrt{53}}+2.\sqrt[3]{182-25\sqrt{53}}\)
\(2F=\sqrt[3]{1456+200\sqrt{53}}+\sqrt[3]{1456-200\sqrt{53}}\)
\(2F=\sqrt[3]{343+147\sqrt{53}+1113+53\sqrt{53}}+\sqrt[3]{343-147\sqrt{53}+1113-53\sqrt{53}}\)
\(2F=\sqrt[3]{\left(7+\sqrt{53}\right)^3}+\sqrt[3]{\left(7-\sqrt{53}\right)^3}\)
\(2F=7+\sqrt{53}+7-\sqrt{53}\)
\(2F=14\)
\(F=7\)
a/ \(D\sqrt{2}=\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\sqrt{3}-1+\sqrt{3}+1=2\sqrt{3}\Rightarrow D=\frac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
b/\(2E=\sqrt[3]{8\sqrt{5}-16}+\sqrt[3]{8\sqrt{5}+16}\)
\(=\sqrt[3]{5\sqrt{5}-3.5.1+3\sqrt{5}-1}+\sqrt[3]{5\sqrt{5}+3.5.1+3\sqrt{5}+1}\)
\(=\sqrt[3]{\left(\sqrt{5}-1\right)^3}+\sqrt[3]{\left(\sqrt{5}+1\right)^3}=\sqrt{5}-1+\sqrt{5}+1=2\sqrt{5}\)
\(\Rightarrow E=\sqrt{5}\)
c/
\(F=\sqrt[3]{182+25\sqrt{53}}+\sqrt[3]{182-25\sqrt{53}}\)
\(F^3=364+3F\sqrt[3]{182^2-33125}=364-3F\)
\(\Leftrightarrow F^3+3F-364=0\)
\(\Leftrightarrow\left(F-7\right)\left(F^2+7F+52\right)=0\)
\(\Rightarrow F=7\)
Bài 2:
a/ \(C=\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+\frac{\sqrt{4}-\sqrt{3}}{\left(\sqrt{4}-\sqrt{3}\right)\left(\sqrt{4}+\sqrt{3}\right)}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}\)
\(=\sqrt{4}-1=2-1=1\)
Ta có \(A=\sqrt[3]{182+\sqrt{33125}}+\sqrt[3]{182-\sqrt{33125}}\)
\(\Rightarrow A^3=364+3.\sqrt[3]{182+\sqrt{33125}}.\sqrt[3]{182-\sqrt{33125}}.A\)
\(\Leftrightarrow A^3=364-3A\)
\(\Leftrightarrow\left(A-7\right)\left(A^2+7A+52\right)=0\)
Vì \(A^2+7A+52=\left(A^2+7A+\frac{49}{4}\right)+\frac{159}{4}=\left(A+\frac{7}{2}\right)^2+\frac{159}{4}>0\)
Do đó A - 7 = 0 => A = 7
a: Ta có: \((x - \sqrt{x^2+5})(x + \sqrt{x^2+5}) = x^2 - (x^2+5) = -5\)
=>\(x + \sqrt{x^2+5} = \frac{-5}{x - \sqrt{x^2+5}}\)
Ta có: \((x - \sqrt{x^2+5})(y - \sqrt{y^2+5}) = 5\)
=>\((x-\sqrt{x^2+5})(y-\sqrt{y^2+5})\left(x+\sqrt{x^2+5}\right)=5\left(x+\sqrt{x^2+5}\right)\)
=>\(-5 (y - \sqrt{y^2+5}) = 5 (x + \sqrt{x^2+5})\)
=>\(-(y-\sqrt{y^2+5})=x+\sqrt{x^2+5}\)
=>\(\sqrt{y^2+5}-y=x+\sqrt{x^2+5}\quad(1)\)
Ta có: \((x - \sqrt{x^2+5})(y - \sqrt{y^2+5}) = 5\)
=>\((x-\sqrt{x^2+5})(y-\sqrt{y^2+5})\left(y+\sqrt{y^2+5}\right)=5\left(y+\sqrt{y^2+5}\right)\)
=>\(-5 (x - \sqrt{x^2+5}) = 5 (y + \sqrt{y^2+5})\)
=>\(\sqrt{x^2+5}-x=y+\sqrt{y^2+5}\quad(2)\)
Từ (1),(2) suy ra -x-y=x+y
=>x+y=0
=>x=-y
\(M=x^{2015}+y^{2015}=\left(-y\right)^{2015}+y^{2015}=0\)
b: Đặt \(u=\sqrt[3]{5 + 2\sqrt{13}};v=\sqrt[3]{5-2\sqrt{13}}\)
=>\(uv = \sqrt[3]{(5 + 2\sqrt{13})(5 - 2\sqrt{13})} = \sqrt[3]{5^2 - (2\sqrt{13})^2} = \sqrt[3]{25 - 52} = \sqrt[3]{-27} = -3\)
x=u+v
=>\(x^3 = (u + v)^3 = u^3 + v^3 + 3uv(u + v)\)
=>\(x^3=(5+2\sqrt{13})+(5-2\sqrt{13})+3(-3)x\)
=>\(x^3=10-9x\)
=>\(x^3+9x-10=0\)
=>(x-1)(x^2+x+10)=0
=>x-1=0
=>x=1
\(A=x^{2015}-x^{2016}+2017\)
\(=1^{2015}-1^{2016}+2017\)
=2017
\(E^3=182+\sqrt{33125}+182-\sqrt{33125}+3\sqrt[3]{182^2-33125}\left(E\right)\)
=\(364-3E\)
\(\Rightarrow E^3+3E-364=0\)
\(\Leftrightarrow E^3-7E^2+7E^2-49E+52E-364=0\)
\(\Leftrightarrow\left(E-7\right)\left(E^2+7E+52\right)=0\)
\(\Rightarrow E=7\)
ta có \(E^3=\left(\sqrt[3]{182+\sqrt{33125}}+\sqrt[3]{182-\sqrt{33125}}\right)^3\)
\(E^3=\left(182+\sqrt{33125}\right)+\left(182-\sqrt{33125}\right)+3\cdot E\cdot\sqrt[3]{33124-33125}\)
\(E^3=364-3E\)
giải phương trình \(E^3+3E-364=0\)
suy ra E= 7
Áp dụng hằng đẳng thức (a + b)3 = a3 + b3 + 3ab(a + b)
\(A^3=182+\sqrt{33125}+182-\sqrt{33125}+3\sqrt[3]{182+\sqrt{33125}}.\sqrt[3]{182-\sqrt{33125}}.\left(\sqrt[3]{182+\sqrt{33125}}+\sqrt[3]{182-\sqrt{33125}}\right)\)
\(A^3=364+3\sqrt[3]{182^2-33125}.A\)
A3 = 364 - 3A
<=> A3 + 3A - 364 = 0
<=> A3 - 7A2 + 7A2 - 49A + 52A 364 = 0
<=> A2. (A - 7) + 7A.(A - 7) + 52.(A - 7)= 0
<=> (A - 7).(A2 + 7A + 52) = 0
<=> A = 7 hoặc A2 + 7A + 52 = 0 (*)
Giải (*) <=> (A+ 3,5)2 + 39,75 = 0 (vô nghiệm)
Vậy A = 7