Câu hỏi của pansak9

Question

Giải PT sau: $\left(2-\sqrt{5}\right)$x2 + $\left(6-\sqrt{5}\right)$x $-$ $8$ + $2\sqrt{5}$ = 0

YangSu · Accepted Answer

$\left(2-\sqrt{5}\right)x^2+\left(6-\sqrt{5}\right)x-8+2\sqrt{5}=0$$\Leftrightarrow\left(2-\sqrt{5}\right)x^2-\left(2-\sqrt{5}\right)x+\left(8-2\sqrt{5}\right)x-(8-2\sqrt{5})=0$$\Leftrightarrow\left(2-\sqrt{5}\right)x\left(x-1\right)+\left(8-2\sqrt{5}\right)\left(x-1\right)=0$$\Leftrightarrow\left(x-1\right)\left[\left(2-\sqrt{5}\right)x+\left(8-2\sqrt{5}\right)\right]=0$$\Leftrightarrow\left[{}\begin{matrix}x-1=0\\left(2-\sqrt{5}\right)x=-8+2\sqrt{5}\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}x=1\x=\dfrac{-8+2\sqrt{5}}{2-\sqrt{5}}\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}x=1\x=6+4\sqrt{5}\end{matrix}\right.$Vậy $S=\left\{1;6+4\sqrt{5}\right\}$Câu trả lời: $\left(2-\sqrt{5}\right)x^2+\left(6-\sqrt{5}\right)x-8+2\sqrt{5}=0$$\Leftrightarrow\left(2-\sqrt{5}\right)x^2-\left(2-\sqrt{5}\right)x+\left(8-2\sqrt{5}\right)x-(8-2\sqrt{5})=0$$\Leftrightarrow\left(2-\sqrt{5}\right)x\left(x-1\right)+\left(8-2\sqrt{5}\right)\left(x-1\right)=0$$\Leftrightarrow\left(x-1\right)\left[\left(2-\sqrt{5}\right)x+\left(8-2\sqrt{5}\right)\right]=0$$\Leftrightarrow\left[{}\begin{matrix}x-1=0\\left(2-\sqrt{5}\right)x=-8+2\sqrt{5}\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}x=1\x=\dfrac{-8+2\sqrt{5}}{2-\sqrt{5}}\end{matrix}\right.$$\Leftrightarrow\left[{}\begin{matrix}x=1\x=6+4\sqrt{5}\end{matrix}\right.$Vậy $S=\left\{1;6+4\sqrt{5}\right\}$

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