Bài 9: Căn bậc ba

Đặt \(\sqrt[3]{b}=x\Rightarrow b=x^3\). Khi đó biểu thức B được biến đổi về dạng :

\(B=\left(\dfrac{x^3}{x^3+8}-\dfrac{4x^3}{\left(x+2\right)^3}\right).\left(\dfrac{1+\dfrac{2}{x}}{1-\dfrac{2}{x}}\right)^2-\dfrac{24}{x^3+8}\)

ĐK: \(x\ne0;x\ne\pm2\)

\(B=\left(\dfrac{x^3}{x^3+8}-\dfrac{4x^3}{\left(x+2\right)^3}\right).\left(\dfrac{x+2}{x-2}\right)^2-\dfrac{24}{x^3+8}\)

\(B=\dfrac{x^3\left(x+2\right)^2-4x^3\left(x^2-2x+4\right)}{\left(x+2\right)^3\left(x^2-2x+4\right)}.\dfrac{\left(x+2\right)^2}{\left(x-2\right)^2}-\dfrac{24}{x^3+8}\)

\(=\dfrac{-3x^5+12x^4-12x^3}{\left(x+2\right)^3\left(x^2-2x+4\right)}.\dfrac{\left(x+2\right)^2}{\left(x-2\right)^2}-\dfrac{24}{x^3+8}\)

\(=\dfrac{-3x^3\left(x-2\right)^2}{\left(x+2\right)^3\left(x^2-2x+4\right)}.\dfrac{\left(x+2\right)^2}{\left(x-2\right)^2}-\dfrac{24}{x^3+8}\)

\(=\dfrac{-3x^3}{\left(x+2\right)\left(x^2-2x+4\right)}-\dfrac{24}{x^3+8}=\dfrac{-3x^3}{x^3+8}-\dfrac{24}{x^3+8}\)

\(=\dfrac{-3\left(x^3+8\right)}{x^3+8}=-3\)

Bấm thử máy tính nó có ra số nguyên đâu em? Xem lại đề bài nhé.

\(\text{Y}=\sqrt[3]{\left(10+6\sqrt{3}\right)}+\sqrt[3]{\left(10-6\sqrt{3}\right)}\)

\(\text{Y}=\sqrt[3]{\left(\sqrt{3}\right)^2+3.\left(\sqrt{3}\right)^2.1+3\sqrt{3}.1^2+1^3}-\sqrt[3]{\left(\sqrt{3}\right)^3-3.\left(\sqrt{3}\right)^2.1+3\sqrt{3.1^2-1^3}}\)

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

\(\text{Y}=\left(\sqrt{3}+1\right)-\left(\sqrt{3}-1\right)\)

\(\Rightarrow\text{Y}=2\)

\(x^3+x^{^{ }2}-x+1=0\)

\(\Leftrightarrow x^2\left(x+1\right)-\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2-1\right)=0\)

\(\left[{}\begin{matrix}x+1=0\Leftrightarrow x=-1\\x^2-1=0\Leftrightarrow x^2=1\Leftrightarrow x=1\end{matrix}\right.\)

Chúc bạn học tốt!

\(\sqrt[3]{x^3+8}=x+2\)

\(\Leftrightarrow x^3+8=\left(x+2\right)^3\)

\(\Leftrightarrow x^3+8=x^3+6x^2+12x+8\)

\(\Leftrightarrow6x^2+12x=0\)

\(\Leftrightarrow6x\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)

Vậy \(S=\left\{0;-2\right\}\)

a) \(x^3=2\\ x=\Leftrightarrow x=\sqrt[3]{2}\)

b)\(27x^3=-81\\ \Leftrightarrow x^3=-3\\ x=\Leftrightarrow x=\sqrt[3]{-3}\)

c) \(\sqrt[3]{3x+1}=4\\ \Leftrightarrow3x+1=64\\ \Leftrightarrow3x=64-1\\ \Leftrightarrow3x=63\\ \Leftrightarrow x=21\)

d)\(\sqrt[3]{x-2}+2=x\\ \Leftrightarrow\sqrt[3]{x-2}=x-2\\ \Leftrightarrow x-2=\left(x-2\right)^3\\ \Leftrightarrow\left(x-2\right)-\left(x-2\right)^3=0\\ \Leftrightarrow\left(x-2\right)\left[1-\left(x-2\right)^2\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\1-\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\\x=1\end{matrix}\right.\)

Giải:

a) \(x^3=2\)

\(\Leftrightarrow x^3=\left(\sqrt[3]{2}\right)^3\)

\(\Leftrightarrow x=\sqrt[3]{2}\)

Vậy ...

b) \(27x^3=-81\)

\(\Leftrightarrow x^3=-3\)

\(\Leftrightarrow x^3=\left(\sqrt[3]{-3}\right)^3\)

\(\Leftrightarrow x=\sqrt[3]{-3}\)

Vậy ...

c) \(\sqrt[3]{3x+1}=4\)

\(\Leftrightarrow\sqrt[3]{3x+1}=\sqrt[3]{64}\)

\(\Leftrightarrow3x+1=64\)

\(\Leftrightarrow3x=63\)

\(\Leftrightarrow x=21\)

Vậy ...

d) \(\sqrt[3]{x-2}+2=x\)

\(\Leftrightarrow\sqrt[3]{x-2}=x-2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\\x=1\end{matrix}\right.\)

Vậy ...

a) \(\sqrt[3]{27}-\sqrt[3]{-8}-\sqrt[3]{125}\\ =3-\left(-2\right)-5\\ =3+2-5\\ =0\)

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

\(x=\dfrac{\left(\sqrt{5}+2\right)\sqrt[3]{\left(\sqrt{5}-2\right)^3}}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}=\dfrac{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}{\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}}=\dfrac{5-4}{\sqrt{5}+3-\sqrt{5}}=\dfrac{1}{3}\)A=\(\left(3\left(\dfrac{1}{3}\right)^3+8\left(\dfrac{1}{3}\right)^2+2\right)^{2009}-3^{2009}=3^{2009}-3^{2009}=0\)

Bài 2: 

a: \(=\sqrt[3]{3}\left(\sqrt[3]{5}-\sqrt[3]{7}\right)\)

b: \(=\sqrt[3]{3}\left(1-\sqrt[3]{9}\right)\)

c: \(=\sqrt[3]{x}\left(\sqrt[3]{a^2}+\sqrt[3]{b^2}\right)\)