Câu hỏi của Phạm Thị Thu Hoài

Question

$3\left(x^2-x+1\right)=\left(x+\sqrt{x-1}\right)^2$

Ma Sói · Accepted Answer

đk: $x\ge1$

$3\left(x^2-x+1\right)=\left(x+\sqrt{x-1}\right)^2$

$\Leftrightarrow3\left(x^2-x+1\right)=x^2+x-1+2x\sqrt{x-1}$

$\Leftrightarrow2x^2-4x+4=2x\sqrt{x-1}$

$\Leftrightarrow x^2-2x+2=x\sqrt{x-1}$

$\Rightarrow x^4+4x^2+4-4x^3+4x^2-8x=x^3-x^2$

$\Leftrightarrow x^4-5x^3+9x^2-8x+4=0$

$\Leftrightarrow x^4-2x^3-3x^3+6x^2+3x^2-6x-2x+4=0$

$\Leftrightarrow x^3\left(x-2\right)-3x^2\left(x-2\right)+3x\left(x-2\right)-2\left(x-2\right)=0$

$\Leftrightarrow\left(x-2\right)\left(x^3-3x^2+3x-2\right)=0$

$\Leftrightarrow\left(x-2\right)^2\left(x^2-x+1\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x=2\x^2-x+1=0\left(vl\right)\end{matrix}\right.$

Vậy S={2}Câu trả lời: đk: $x\ge1$

$3\left(x^2-x+1\right)=\left(x+\sqrt{x-1}\right)^2$

$\Leftrightarrow3\left(x^2-x+1\right)=x^2+x-1+2x\sqrt{x-1}$

$\Leftrightarrow2x^2-4x+4=2x\sqrt{x-1}$

$\Leftrightarrow x^2-2x+2=x\sqrt{x-1}$

$\Rightarrow x^4+4x^2+4-4x^3+4x^2-8x=x^3-x^2$

$\Leftrightarrow x^4-5x^3+9x^2-8x+4=0$

$\Leftrightarrow x^4-2x^3-3x^3+6x^2+3x^2-6x-2x+4=0$

$\Leftrightarrow x^3\left(x-2\right)-3x^2\left(x-2\right)+3x\left(x-2\right)-2\left(x-2\right)=0$

$\Leftrightarrow\left(x-2\right)\left(x^3-3x^2+3x-2\right)=0$

$\Leftrightarrow\left(x-2\right)^2\left(x^2-x+1\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x=2\x^2-x+1=0\left(vl\right)\end{matrix}\right.$

Vậy S={2}

Violympic toán 9