\(\sqrt{43-x}=x-1\left(đk:x\le43\right)\)
\(\Leftrightarrow\left|43-x\right|=\left(x-1\right)^2\)
\(\Leftrightarrow43-x=x^2-2x+1\)
\(\Leftrightarrow x^2-x-42=0\)
\(\Delta=\left(-1\right)^2-4.\left(-42\right)=169>0\)
Do \(\Delta\) > 0 nên pt có 2 nghiệm phân biện:
\(x_1=\dfrac{1+\sqrt{169}}{2}=7\left(TM\right)\)
\(x_2=\dfrac{1-\sqrt{169}}{2}=-6\left(TM\right)\)
\(ĐK:x\le43\)
\(\sqrt{43-x}=x-1\)
\(\Leftrightarrow\left(\sqrt{43-x}\right)^2=\left(x-1\right)^2\)
\(\Leftrightarrow43-x=x^2-2x+1\)
\(\Leftrightarrow x^2-x-42=0\)
\(\Delta=\left(-1\right)^2-4.\left(-42\right)=1+168=169>0\)
\(\rightarrow\left\{{}\begin{matrix}x_1=\dfrac{1+\sqrt{169}}{2}=7\left(tm\right)\\x_2=\dfrac{1-\sqrt{169}}{2}=-6\left(tm\right)\end{matrix}\right.\)
Vậy \(S=\left\{7;-6\right\}\)
Có em nhé, cho anh thêm:
\(đk:x\ge-1\) vì căn luôn lớn hơn 0
