Bài 9: Căn bậc ba

\(\sqrt[3]{7+5\sqrt{2}}+\dfrac{1}{\sqrt[3]{7+5\sqrt{2}}}=\sqrt[3]{2\sqrt{2}+6+3\sqrt{2}+1}+\dfrac{1}{\sqrt[3]{2\sqrt{2}+6+3\sqrt{2}+1}}=\sqrt[3]{\left(\sqrt{2}+1\right)^3}+\dfrac{1}{\sqrt[3]{\left(\sqrt{2}+1\right)^3}}=\sqrt{2}+1+\dfrac{1}{\sqrt{2}+1}=\sqrt{2}+1+\dfrac{\sqrt{2}-1}{2-1}=2\sqrt{2}\)

b: \(A=\dfrac{1}{\sqrt[3]{4-\sqrt{15}}}+\sqrt[3]{4-\sqrt{15}}\)

\(=\sqrt[3]{4+\sqrt{15}}+\sqrt[3]{4-\sqrt{15}}\)

\(\Leftrightarrow A^3=4+\sqrt{15}+4-\sqrt{15}+3\cdot A\cdot1\)

\(\Leftrightarrow A^3-3A-8=0\)

hay \(A\simeq2.49\)

a: \(B=\sqrt[3]{5-\sqrt{17}}+\sqrt[3]{5+\sqrt{17}}\)

\(\Leftrightarrow B^3=5-\sqrt{17}+5+\sqrt{17}+3\cdot B\cdot2=10+6B\)

\(\Leftrightarrow B^3-6B-10=0\)

hay \(B\simeq3.05\)

Lời giải: Đặt \(\sqrt[3]{x+1}=a; \sqrt[3]{7-x}=b\). Khi đó ta có: \(\left\{\begin{matrix} a^3+b^3=8\\ a+b=2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} (a+b)^3-3ab(a+b)=8\\ a+b=2\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} 8-6ab=8\\ a+b=2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} ab=0\\ a+b=2\end{matrix}\right.\)

\(\Rightarrow (a,b)=(2,0); (0,2)\)

\(\Rightarrow x=7\) hoặc \(x=-1\)

Mọi người ơi giải giúp mình với😥😥

a)\(\left\{{}\begin{matrix}x+3\ne0\\x^2+x-6\ne0\\2-x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-3\\\left(x-3\right)\left(x-2\right)\ne\\2\ne x\end{matrix}\right.0\Leftrightarrow\left\{{}\begin{matrix}x\ne-3\\x\ne2\end{matrix}\right.\)

b)\(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^3-x-6}+\dfrac{1}{2-X}\)

\(\sqrt[3]{7+\sqrt{50}}+\sqrt[3]{7-\sqrt{50}}\)

\(=\sqrt[3]{2\sqrt{2}+3.2+3\sqrt{2}+1}+\sqrt[3]{-2\sqrt{2}+3.2-3\sqrt{2}+1}\)

\(=\sqrt[3]{\left(\sqrt{2}+1\right)^3}+\sqrt[3]{\left(-\sqrt{2}+1\right)^3}\)

\(=\sqrt{2}+1-\sqrt{2}+1=2\in N\)

Đặt \(\sqrt[3]{6x-9}=a\).

Ta có hpt:

\(\left\{{}\begin{matrix}x^3=6a-9\\a^3=6x-9\end{matrix}\right.\).

Nếu x > a thì \(x^3>a^3\Rightarrow6a-9>6x-9\Rightarrow a>x\) (vô lí).

Nếu x < a thì tương tự ta cũng có điều vô lí.

Do đó x = a

\(\Leftrightarrow x^3-6x+9=0\Leftrightarrow\left(x+3\right)\left(x^2-3x+3\right)=0\Leftrightarrow x=-3\).

Vậy nghiệm của pt đã cho x = -3.

`x=\root{3}{4(\sqrt5+1)}-\root{3}{4(\sqrt5-1)}`

`<=>x^3=4(sqrt5+1)-4(\sqrt5-1)-3\root{3}{16(5-1)}(\root{3}{4(\sqrt5+1)}-\root{3}{4(\sqrt5-1)})`

`<=>x^3=4\sqrt5+4-4sqrt5+4-3\root{3}{64}x`

`<=>x^3=8-12x`

`<=>x^3+12x-8=0`

`=>P=(x^3+12-8-1)^2021=(-1)^2021=-1`

*Có gì khum hiểu comment bên dưới.