Đáp án `:` `↓↓↓`
Giải thích `:`
`x^2(x^2+4)-x^2-4=0`
`<=>x^2(x^2+4)-(x^2+4)=0`
`<=>(x^2+4)(x^2-1)=0`
`<=>`$\left[\begin{array}{l}x^2+4=0\\x^2-1=0\end{array}\right.$
`<=>`$\left[\begin{array}{l}x^2=-4\\x^2=1\end{array}\right.$
`<=>`$\left[\begin{array}{l}x=∅\\x=1;-1\end{array}\right.$
Vậy `...`
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`x^4-x^3+x^2-x=0`
`<=>(x^4-x^3)+(x^2-x)=0`
`<=>x^3(x-1)+x(x-1)=0`
`<=>(x-1)(x^2+x)=0`
`<=>x(x-1)(x^2+1)=0`
`=>x={0;1}`
Vậy `...`
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`(2x-1)^2-(4x^2-1)=0`
`<=>(2x-1)^2-(2x-1)(2x+1)=0`
`<=>(2x-1)(2x-1-2x-1)=0`
`<=>-2(2x-1)=0`
`<=>2x-1=0`
`<=>2x=1`
`<=>x=1/2`
`@Ki`
