\(x=\sqrt[3]{5-\sqrt{17}}+\sqrt[3]{5+\sqrt{17}}\)

\(\Leftrightarrow x^3=5-\sqrt{17}+5+\sqrt{17}+3\left(\sqrt[3]{5-\sqrt{17}}+\sqrt[3]{5+\sqrt{17}}\right)\sqrt[3]{5-\sqrt{17}}\sqrt[3]{5+\sqrt{17}}\)

\(\Leftrightarrow x^3=10+3x\sqrt[3]{\left(5-\sqrt{17}\right)\left(5+\sqrt{17}\right)}\)

\(\Leftrightarrow x^3=10+3x\sqrt[3]{8}\)\(\Leftrightarrow x^3=10+6x\)

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

Hay ta co DPCM

\(x^3=10+3x\sqrt[3]{\left(5-\sqrt{17}\right)\left(5+\sqrt{17}\right)}=10+6x\)

Thay vào -> dpcm

\(x=\sqrt[3]{5-\sqrt{17}}+\sqrt[3]{5+\sqrt{17}}\)

\(\Leftrightarrow x^3=5-\sqrt{17}+5+\sqrt{17}\)

\(+3\left(\sqrt[3]{5-\sqrt{17}}+\sqrt[3]{5+\sqrt{17}}\right)\sqrt[3]{5-\sqrt{17}}\sqrt[3]{5+\sqrt{17}}\)

\(\Leftrightarrow x^3=10+3x\sqrt[3]{\left(5-\sqrt{17}\right)\left(5+\sqrt{17}\right)}\)

\(\Leftrightarrow x^3=10+3x\sqrt[3]{8}\Leftrightarrow x^3=10+6x\)

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

\(\Rightarrow\) Đpcm

Chúc bạn học tốt !!!

đây là toán lớp 9 mà

trả lời chỉ để lấy tích thời mọi người tích giùm hihi

\(x^3=3+\sqrt{17}+3-\sqrt{17}+3a.b\left(a+b\right)\) dài quá đặt a,b

a.b=-2

x^3=6-6(a+b)=6-6x

=>x^3+6x-5=6-5=1

KL: P(x)=1²⁰¹⁶ =1

Tìm P_{(a) với a = ..... nhé}

_{Nhầm đề tí!}

Thì nó cũng vậy thôi

\(x^3\)=\(3+\sqrt{17}+3-\sqrt{17}+3.\sqrt[3]{\left(3+\sqrt{17}\right)\left(3-\sqrt{17}\right)}.x\)

        =\(6+3\sqrt[3]{-8}x=6-6x\) 

\(\Rightarrow x^3+6x-6=0\)

M=\(x^3+6x-5=\left(x^3+6x-6\right)+1=0+1=1\)

\(a^3=6+3a\sqrt[3]{\left(3+\sqrt{17}\right)\left(3-\sqrt{17}\right)}\)

\(\Rightarrow a^3=6-6a\)

\(\Rightarrow a^3+6a-5=1\)

\(\Rightarrow f\left(a\right)=1^{2020}=1\)

\(x^3=6+3x.\sqrt[3]{3^2-17}=6-6x\)

\(\Leftrightarrow x^3+6x-6=0\)

\(\Rightarrow B=\left(x^3+6x-6+1\right)^{2012}=1^{2012}=1\)