a/\(\left(x+2\right)^2=\left(2x-1\right)^2\Leftrightarrow x^2+4x+4=4x^2-4x+1\Leftrightarrow3x^2-8x-3=0\)
Giải hệ pt ta đc:\(\left[{}\begin{matrix}x=3\\x=-\frac{1}{3}\end{matrix}\right.\)
b/\(\left(x+2\right)^2-x+4=0\Leftrightarrow x^2+4x+4-x+4=0\Leftrightarrow x^2-3x+8=0\Leftrightarrow\left[{}\begin{matrix}\frac{3-\sqrt{23}}{2}\\\frac{3+\sqrt{23}}{2}\end{matrix}\right.\)
a) (x+2)2 = (2x-1)2
⇔ x2 + 4x + 4 = 4x2 -4x + 1
⇔ x2 + 4x + 4 - 4x2 +4x -1 = 0
⇔ -3x2 + 8x + 3 = 0
⇔3x2 - 8x -3 =0
⇔ 3x2 + x - 9x -3 = 0
⇔x(3x+1) - 3(3x+1) = 0
⇔ (3x+1) (x-3) = 0
⇔\(\left[{}\begin{matrix}3x+1=0\\x-3=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}x=-\frac{1}{3}\\x=3\end{matrix}\right.\)
S={ \(-\frac{1}{3};3\)}
b) (x+2)2-x+4=0
⇔ x2 + 4x +4 - x + 4 = 0
⇔ x2 + 3x + 8 = 0
Do lớp 8 chưa học giải phương trình bậc 2
S= ∅
