Odpowiedź:
[tex]\huge\boxed{\boxed{\text{A}, \ \text{B}, \ \text{D}}}[/tex]
Szczegółowe wyjaśnienie:
Równanie A
[tex]x(x+1)-5=x^2\\\\x^2+x-5=x^2\\\\x^2+x-x^2=5\\\\x=5\in\text{C}[/tex]
Równanie B
[tex]3x^2+24x=3(x+2)^2 \ \ /:3\\\\x^2+8x=(x+2)^2\\\\x^2+8x=x^2+2\cdot x\cdot2+2^2\\\\x^2+8x=x^2+4x+4\\\\x^2+8x-x^2-4x=4\\\\4x=4 \ \ /:4\\\\x=1\in\text{C}[/tex]
Równanie C
[tex]\frac{1}{2}x+\frac{3}{4}x+5=25x \ \ /\cdot4\\\\\frac{1}{2}x\cdot4+\frac{3}{4}x\cdot4+5\cdot4=25x\cdot4\\\\2x+3x+20=100x\\\\5x+20=100x\\\\5x-100x=-20\\\\-95x=-20 \ \ /:(-95)\\\\x=\frac{20}{95}=\frac{4}{19}\notin\text{C}[/tex]
Równanie D
[tex]\frac{x^2-1}{5}=\frac{x(2x+1)}{10} \ \ /\cdot10\\\\\frac{x^2-1}{5}\cdot10=\frac{2x^2+x}{10}\cdot10\\\\2(x^2-1)=2x^2+x\\\\2x^2-2=2x^2+x\\\\2x^2-2x^2-x=2\\\\-x=2 \ \ /:(-1)\\\\x=-2\in\text{C}[/tex]
