\( (x-1)^2=x^2 \)
\( x=\frac{1}{2} \)
\( x=0 \)
\( (x-4)^2=(x+2)(x-1) \)
\( x=2 \)
\( (x+3)^2=(x-3)^2 \)
\( x=3 \)
\( \)\( x^2+(x-2)^2=2(x+1)^2 \)
\( x=\frac{1}{4} \)
Solve the following equation:
\( \frac{(2x-1)^2}{x-2}+\frac{(x-2)^2}{2x-1}=4.5x \)
\( \)\( -1\pm\sqrt{3} \)
\( 1\pm\sqrt{12} \)
(x−1)2=x2 (x-1)^2=x^2 (x−1)2=x2
x=12 x=\frac{1}{2} x=21
(x−4)2=(x+2)(x−1) (x-4)^2=(x+2)(x-1) (x−4)2=(x+2)(x−1)
x=2 x=2 x=2
(x+3)2=(x−3)2 (x+3)^2=(x-3)^2 (x+3)2=(x−3)2
x=0 x=0 x=0
x2+(x−2)2=2(x+1)2 x^2+(x-2)^2=2(x+1)^2 x2+(x−2)2=2(x+1)2
x=14 x=\frac{1}{4} x=41
(2x−1)2x−2+(x−2)22x−1=4.5x \frac{(2x-1)^2}{x-2}+\frac{(x-2)^2}{2x-1}=4.5x x−2(2x−1)2+2x−1(x−2)2=4.5x
−1±3 -1\pm\sqrt{3} −1±3