Rozwiąż równanie
a)
\frac{2x-4}{x+1}=\frac{x-1}{x-2}
x\ne-1 , x\ne 2
(2x-4)(x-2)=(x-1)(x+1)
2x^2-4x-4x+8=x^2-1
2x^2-8x+8-x^2+1=0
x^2-8x+9=0
\Delta=b^2-4ac=64-4*1*9=64-36=28
\sqrt\Delta=\sqrt{28}=\sqrt{4*7}=2\sqrt7
x_1=\frac{-b-\sqrt\Delta}{2a}=\frac{8-2\sqrt7}{2}=\frac{2(4-\sqrt7)}{2}=4-\sqrt7
x_2=\frac{-b+\sqrt\Delta}{2a}=\frac{8+2\sqrt7}{2}=\frac{2(4+\sqrt7)}{2}=4+\sqrt7
b)
x=\frac{7x-4}{2x-2}
2x-2\ne0 => x\ne1
\frac{x}{1}=\frac{7x-4}{2x-2}
x(2x-2)=7x-4
2x^2-2x-7x+4=0
2x^2-9x+4=0
\Delta=b^2-4ac=81-4*2*4=81-32=49
\sqrt\Delta=7
x_1=\frac{-b-\sqrt\Delta}{2a}=\frac{9-7}{2*2}=\frac{2}{4}=\frac{1}{2}
x_2=\frac{-b+\sqrt\Delta}{2a}=\frac{9+7}{4}=\frac{16}{4}=4