2)[tex]\sqrt[3]{(x-29)^{2}}+\sqrt[3]{(x-1)^{2}}=7-\sqrt[3]{x^{2}-30x+29}\Leftrightarrow \sqrt[3]{(x-29)^{2}}+\sqrt[3]{(x-1)^{2}}=7-\sqrt[3]{(x-29)(x-1)}[/tex]
Đặt [tex]\sqrt[3]{x-29}=a;\sqrt[3]{x-1}=b[/tex], ta có hệ pt:
[tex]\left\{\begin{matrix} a^{3}+b^{3}=x-29-x+1=-28\\ a^{2}+b^{2}=7-ab \end{matrix}\right.[/tex]
[tex]\Leftrightarrow \left\{\begin{matrix} (a-b)(a^{2}+ab+b^{2})=-28\\ a^{2}+b^{2}+ab=7 \end{matrix}\right. \Leftrightarrow \left\{\begin{matrix} 7(a-b)=-28\\ a^{2}+b^{2}+ab=7 \end{matrix}\right. \Leftrightarrow \left\{\begin{matrix} a-b=-4\\ a^{2}+b^{2}+ab=7 \end{matrix}\right. \Leftrightarrow \left\{\begin{matrix} a=b-4\\ (b-4)^{2}+b^{2}+(b-4)b-7=0(1) \end{matrix}\right.[/tex]
Giải pt (1):
[tex](b-4)^{2}+b^{2}+(b-4)b-7=0\Leftrightarrow b^{2}-8b+16+b^{2}+b^{2}-4b-7=0\Leftrightarrow 3b^{2}-12b+9=0\Leftrightarrow b^{2}-4b+3=0\Leftrightarrow (b-1)(b-3)=0[/tex]
=>b=1 hoặc b=3
Nếu b=1 => a=1-4=-3
[tex]\Rightarrow \left\{\begin{matrix} \sqrt[3]{x-29}=-3\\ \sqrt[3]{x-1}=1 \end{matrix}\right. \Leftrightarrow \left\{\begin{matrix} x-29=-27\\ x-1=1 \end{matrix}\right. \Rightarrow x=2[/tex]
Nếu b=3 => a=3-4=-1
[tex]\Rightarrow \left\{\begin{matrix} \sqrt[3]{x-29}=-1\\ \sqrt[3]{x-1}=3 \end{matrix}\right. \Leftrightarrow \left\{\begin{matrix} x-29=-1\\ x-1=27 \end{matrix}\right. \Rightarrow x=28[/tex]
Vậy x=2 hoặc x=28
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