Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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/ \(C=\frac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}+\frac{\sqrt{4}-\sqrt{3}}{\left(\sqrt{4}-\sqrt{3}\right)\left(\sqrt{4}+\sqrt{3}\right)}\)
\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}\)
\(=\sqrt{4}-1=2-1=1\)
Đặt \(x=\sqrt[3]{182+\sqrt[]{33125}}+\sqrt[3]{182-\sqrt[]{33125}}\)
\(\Rightarrow x^3=364+3\sqrt[3]{182^2-33125}.\left(\sqrt[3]{182+\sqrt[]{33125}}+\sqrt[]{182-\sqrt[]{33125}}\right)\)
\(\Rightarrow x^3=364+3.\left(-1\right).x\)
\(\Rightarrow x^3+3x-364=0\)
\(\Rightarrow\left(x-7\right)\left(x^2+7x+52\right)=0\)
\(\Rightarrow x-7=0\) (do \(x^2+7x+52>0;\forall x\))
\(\Rightarrow x=7\)
a) Ta có: \(A^3=\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)^3\)
\(=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)
\(=4-3\cdot A\)
\(\Leftrightarrow A^3+3A-4=0\)
\(\Leftrightarrow A^3-A+4A-4=0\)
\(\Leftrightarrow A\left(A-1\right)\left(A+1\right)+4\left(A-1\right)=0\)
\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)
\(\Leftrightarrow A=1\)
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
\(E^3=182+\sqrt{33125}+182-\sqrt{33125}+3\sqrt[3]{182^2-33125}\left(E\right)\)
=\(364-3E\)
\(\Rightarrow E^3+3E-364=0\)
\(\Leftrightarrow E^3-7E^2+7E^2-49E+52E-364=0\)
\(\Leftrightarrow\left(E-7\right)\left(E^2+7E+52\right)=0\)
\(\Rightarrow E=7\)
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
Bấm máy tinh ta được \(A=7\)nên sẽ dự đoán như sau (lưu ý \(\sqrt{33125}=25\sqrt{53}\)):
\(\hept{\begin{cases}\sqrt[3]{182-25\sqrt{53}}=\frac{7}{2}-a\sqrt{53}\\\sqrt[3]{182+25\sqrt{53}}=\frac{7}{2}+a\sqrt{53}\end{cases}}\)
Khi đó cộng lại sẽ được 7
Tìm a thì quá đơn giản: \(a=\frac{\sqrt[3]{182+25\sqrt{53}}-\frac{7}{2}}{\sqrt{53}}\)
Bấm máy tính, ta được ngay \(a=\frac{1}{2}\)
Vậy \(\sqrt[3]{182\pm25\sqrt{53}}=\frac{7}{2}\pm\frac{\sqrt{53}}{2}\)
Muốn chứng minh thì lập phương 2 vế là được.
\(A=\sqrt[3]{182-\sqrt{33125}}+\sqrt[3]{182+\sqrt{33125}}\)
\(A^3=182-\sqrt{33125}+182+\sqrt{33125}+3\sqrt[3]{182^2-\left(\sqrt{33125}\right)^2}.A\)
\(A^3=364+3\sqrt[3]{-1}.A\)
\(A^3=364-3A\)
\(A^3+3A-364=0\)
......................................
......................................
......................................
~~~~~~~~~~~~~~~~~~~
Đến đây bạn tự giải phương trình tiếp rồi sẽ ra nha! Chúc bạn học giỏi nhé!