$Bài_{}$ $3:_{}$
$a)_{}$ $x+1=(x+1)^2_{}$
$⇔x+1=x^2+2x+1_{}$
$⇔x=x^2+2x_{}$
$⇔x-x^2-2x=0_{}$
$⇔-x-x^2=0_{}$
$⇔-x.(1+x)=0_{}$
$⇔x.(1+x)=0_{}$
$⇔_{}$ \(\left[ \begin{array}{l}x=0\\1+x=0\end{array} \right.\) $⇔_{}$ \(\left[ \begin{array}{l}x=0\\x=-1\end{array} \right.\)
$b)_{}$ $2x.(x-3)+x^2-9=0_{}$
$⇔2x^2-6x+x^2-9=0_{}$
$⇔3x^2-6x-9=0_{}$
$⇔3.(x^2-2x-3)=0_{}$
$⇔x^2-2x-3=0_{}$
$⇔x^2-3x+x-3=0_{}$
$⇔x.(x-3)+(x-3)=0_{}$
$⇔(x+1)(x-3)=0_{}$
$⇔_{}$ \(\left[ \begin{array}{l}x+1=0\\x-3=0\end{array} \right.\) $⇔_{}$ \(\left[ \begin{array}{l}x=-1\\x=3\end{array} \right.\)
$c)_{}$ $(x+5)^2-(2x-1)^2=0_{}$
$⇔x^2+10x+25-(4x^2-4x+1)=0_{}$
$⇔x^2+10x+25-4x^2+4x-1=0_{}$
$⇔-3x^2+14x+24=0_{}$
$⇔3x^2-14x-24=0_{}$
$⇔3x^2+4x-18x-24=0_{}$
$⇔x.(3x+4)-6.(3x+4)=0_{}$
$⇔(3x+4)(x-6)=0_{}$
$⇔_{}$ \(\left[ \begin{array}{l}3x+4=0\\x-6=0\end{array} \right.\) $⇔_{}$ \(\left[ \begin{array}{l}x=-\frac{4}{3}\\x=6\end{array} \right.\)
$d)_{}$ $3x.(x-2)+2x-4=0_{}$
$⇔3x.(x-2)+2.(x-2)=0_{}$
$⇔(x-2)(3x+2)=0_{}$
$⇔_{}$ \(\left[ \begin{array}{l}x-2=0\\3x+2=0\end{array} \right.\) $⇔_{}$ \(\left[ \begin{array}{l}x=2\\x=-\frac{2}{3}\end{array} \right.\)
$Bài_{}$ $4:_{}$
$x^2+4x+4_{}$ $tại_{}$ $x_{}=98$
$=x^2+2.x.2+2^2_{}$
$=(x+2)^2_{}$
$+)_{}$ $Thay_{}$ $x=98_{}$ $ta_{}$ $được_{}$
$(98+2)^2_{}$
$A=(100)^2_{}$
$A=10000_{}$
Vậy A = 10000 khi x = 98
