Câu hỏi của Bánh Cúc Ki

Question

Tìm x :$\sqrt{x-1}=x-3$

Nguyễn Huy Tú · Accepted Answer

$\sqrt{\left(x-1\right)}=x-3$ $\Rightarrow\left|x-1\right|=\left(x-3\right)^2$ $\Rightarrow\left|x-1\right|=x^2-6x+9$ +) Xét $x\ge1$ có: $x-1=x^2-6x+9$ $\Leftrightarrow x^2-7x+10=0$ $\Leftrightarrow x^2-5x-2x+10=0$ $\Leftrightarrow x\left(x-5\right)-2\left(x-5\right)=0$ $\Leftrightarrow\left(x-2\right)\left(x-5\right)=0$ $\Rightarrow\left[{}\begin{matrix}x-2=0\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\x=5\end{matrix}\right.$ Thử lại: x = 5 ( t/m ), x = 2 ( không t/m ) +) Xét x < 1 có: $1-x=x^2-6x+9$ $\Leftrightarrow x^2+5x+8=0$ $\Leftrightarrow x^2+5x+\dfrac{25}{4}+\dfrac{7}{4}=0$ $\Leftrightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{7}{4}=0$ Do $\left(x+\dfrac{5}{2}\right)^2+\dfrac{7}{4}>0$ $\Rightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{7}{4}=0$ ( vô lí ) Vậy x = 5Câu trả lời: $\sqrt{\left(x-1\right)}=x-3$ $\Rightarrow\left|x-1\right|=\left(x-3\right)^2$ $\Rightarrow\left|x-1\right|=x^2-6x+9$ +) Xét $x\ge1$ có: $x-1=x^2-6x+9$ $\Leftrightarrow x^2-7x+10=0$ $\Leftrightarrow x^2-5x-2x+10=0$ $\Leftrightarrow x\left(x-5\right)-2\left(x-5\right)=0$ $\Leftrightarrow\left(x-2\right)\left(x-5\right)=0$ $\Rightarrow\left[{}\begin{matrix}x-2=0\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\x=5\end{matrix}\right.$ Thử lại: x = 5 ( t/m ), x = 2 ( không t/m ) +) Xét x < 1 có: $1-x=x^2-6x+9$ $\Leftrightarrow x^2+5x+8=0$ $\Leftrightarrow x^2+5x+\dfrac{25}{4}+\dfrac{7}{4}=0$ $\Leftrightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{7}{4}=0$ Do $\left(x+\dfrac{5}{2}\right)^2+\dfrac{7}{4}>0$ $\Rightarrow\left(x+\dfrac{5}{2}\right)^2+\dfrac{7}{4}=0$ ( vô lí ) Vậy x = 5

๖ۣۜĐặng♥๖ۣۜQuý · Answer

$\sqrt{x-1}=x-3\
\Leftrightarrow x-1=x^2-6x+9\
\Leftrightarrow x^2-7x+10=0\
\Leftrightarrow x^2-2.3,5.x+\left(3,5\right)^2=\left(3,5\right)^2-10\
\Leftrightarrow\left(x-3,5\right)^2=\dfrac{9}{4}\\Rightarrow
\left[{}\begin{matrix}x-3,5=\dfrac{3}{2}\x-3,5=-\dfrac{3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\x=2\end{matrix}\right.$vậy x={2,5}Câu trả lời: $\sqrt{x-1}=x-3\
\Leftrightarrow x-1=x^2-6x+9\
\Leftrightarrow x^2-7x+10=0\
\Leftrightarrow x^2-2.3,5.x+\left(3,5\right)^2=\left(3,5\right)^2-10\
\Leftrightarrow\left(x-3,5\right)^2=\dfrac{9}{4}\\Rightarrow
\left[{}\begin{matrix}x-3,5=\dfrac{3}{2}\x-3,5=-\dfrac{3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\x=2\end{matrix}\right.$vậy x={2,5}

Serena chuchoe · Answer

$\sqrt{x-1}=x-3$

ĐKXĐ: x $\ge1$

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

$\Leftrightarrow x-1=x^2-6x+9$

$\Leftrightarrow x-1-x^2+6x-9=0$

$\Leftrightarrow-x^2+7x-10=0$

$\Leftrightarrow-x^2+2x+5x-10=0$

$\Leftrightarrow x\left(-x+2\right)-5\left(-x+2\right)=0$

$\Leftrightarrow\left(x-5\right)\left(2-x\right)=0$

$\Leftrightarrow\left[{}\begin{matrix}x-5=0\2-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\x=2\end{matrix}\right.$

Vậy pt có 2 nghiệm là.....Câu trả lời: $\sqrt{x-1}=x-3$