Toán Lớp 9: tìm x
$\sqrt{x^{2}-10x+25}$ =x+1
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Toán Lớp 9: tìm x
$\sqrt{x^{2}-10x+25}$ =x+1
Comments ( 2 )
$⇔(\sqrt{x^2-10x+25})^2 =(x+1)^2$
$⇔x^2-10x+25 =x^2+2x+1$
$⇔-12x =-24$
$⇔-x =2$
$\sqrt[]{(x-5)^2}=x+1$
$|x-5|=x+1$
\(\left[ \begin{array}{l}x-5=x+1(x≥5)\\5-x=x+1(x≤5)\end{array} \right.\)
\(\left[ \begin{array}{l}0+6(vô lí)\\x=2(2≤5 TM)\end{array} \right.\)