$\sqrt[3]{2-x}$+ `sqrt{x-1}=1` giải phương trình vô tỉ trên 03/09/2021 Bởi Eva $\sqrt[3]{2-x}$+ `sqrt{x-1}=1` giải phương trình vô tỉ trên
ĐKXD : `x≥1` $\sqrt[3]{2-x}+\sqrt{x-1}=1$ $⇔\sqrt{x-1}=1-\sqrt[3]{2-x}$ Đặt $\sqrt[3]{2-x}=a;\sqrt{x-1}=b$ `⇒b=1-a ` Ta thấy `x-2+x-1=1` `⇔a^3+b^2=1` `⇔a^3+(1-a)^2=1` `⇔a^3+1-2a+a^2=1` `⇔a^3+a^2-2a=0` `⇔a(a^2+a-2)=0` `⇔a(a-1)(a+2)=0` `⇔`\(\left[ \begin{array}{l}a=0\\a=-2\\a=1\end{array} \right.\) `⇔`\(\left[ \begin{array}{l}\sqrt[3]{2-x}=0\\\sqrt[3]{2-x}=1\\\sqrt[3]{2-x}=-2\end{array} \right.\) `⇔`\(\left[ \begin{array}{l}2-x=0\\2-x=1\\2-x=-8\end{array} \right.\) `⇔`\(\left[ \begin{array}{l}x=2\\x=1\\x=10\end{array} \right.\) Bình luận
ĐKXD : `x≥1`
$\sqrt[3]{2-x}+\sqrt{x-1}=1$
$⇔\sqrt{x-1}=1-\sqrt[3]{2-x}$
Đặt $\sqrt[3]{2-x}=a;\sqrt{x-1}=b$
`⇒b=1-a `
Ta thấy `x-2+x-1=1`
`⇔a^3+b^2=1`
`⇔a^3+(1-a)^2=1`
`⇔a^3+1-2a+a^2=1`
`⇔a^3+a^2-2a=0`
`⇔a(a^2+a-2)=0`
`⇔a(a-1)(a+2)=0`
`⇔`\(\left[ \begin{array}{l}a=0\\a=-2\\a=1\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}\sqrt[3]{2-x}=0\\\sqrt[3]{2-x}=1\\\sqrt[3]{2-x}=-2\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}2-x=0\\2-x=1\\2-x=-8\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=2\\x=1\\x=10\end{array} \right.\)