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\(C=\sqrt{x}+\frac{\sqrt[3]{2-\sqrt{3}}.\sqrt[6]{7+4\sqrt{3}}-x}{\sqrt[4]{9-4\sqrt{5}}.\sqrt{2+\sqrt{5}}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{\sqrt[6]{\left(7-4\sqrt{3}\right).\left(7+4\sqrt{3}\right)}-x}{\sqrt[4]{\left(9+4\sqrt{5}\right).\left(9-4\sqrt{5}\right)}+\sqrt{x}}\)
\(=\sqrt{x}+\frac{1-x}{1+\sqrt{x}}=\sqrt{x}+\frac{\left(1+\sqrt{x}\right).\left(1-\sqrt{x}\right)}{1+\sqrt{x}}\)
\(=\sqrt{x}+1-\sqrt{x}=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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mik ko biết
mong bn thông cảm
nha ................
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![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(a=\sqrt[3]{9+4\sqrt{5}},b=\sqrt[3]{9-4\sqrt{5}}\)
\(\Rightarrow\hept{\begin{cases}a+b=x\\ab=1\end{cases}}\)
Ta có: \(x^3=\left(a+b\right)^3=a^3+b^3+3ab\left(a+b\right)\)
\(\Rightarrow x^3=\left(9+4\sqrt{5}\right)+\left(9-4\sqrt{5}\right)+3.1.x\)
\(\Leftrightarrow x^3=18+3x\)
\(\Leftrightarrow x^3-3x-18=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+6\right)=0\)
Vì \(x^2+3x+6=\left(x+\frac{3}{2}\right)^2+\frac{15}{4}>0\)
\(\Rightarrow x-3=0\Leftrightarrow x=3\)
Thay x=3 vào \(x^5-3x-18=0\), thấy không thoả mãn.
KL: Đề sai !
![](https://rs.olm.vn/images/avt/0.png?1311)
Điều kiện là x lớn hơn 0 , x không bằng 4 và x không bằng 9 ạ Mong nhận đc câu trả lời ạ
\(A=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)
\(\Leftrightarrow A^3=9+4\sqrt{5}+9-4\sqrt{5}\)
\(+3\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\sqrt[3]{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)}\)
\(\Leftrightarrow A^3=18+3A\Leftrightarrow A^3-3A-18=0\)
\(\Leftrightarrow\left(A-3\right)\left(A^2+3A+6\right)=0\)
Dễ thấy : \(A^2+3A+6=\left(A+\frac{3}{2}\right)^2+\frac{15}{4}\ge0\forall A\)
\(\Leftrightarrow A=3\)
Chúc bạn học tốt !!!
\(A=\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\)
\(\Leftrightarrow A^3=9+4\sqrt{5}+9-4\sqrt{5}\)
\(+3\left(\sqrt[3]{9+4\sqrt{5}}+\sqrt[3]{9-4\sqrt{5}}\right)\sqrt[3]{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)}\)
\(\Leftrightarrow A^3+18+3A\Leftrightarrow A^3-3A-18=0\)
\(\Leftrightarrow\left(A-3\right)\left(A^2+3A+6\right)=0\)
Dễ thấy : \(A^2+3A+6=\left(A+\frac{3}{2}\right)^2+\frac{15}{4}\ge0\forall A\)
\(\Leftrightarrow A=3\)
Chúc bạn học tốt !!!