Question

2В39( а, б) помогите решить!

Answer 1

A
[(x+3)(x-1)+(x+1)√(x+3)(x-3)]/[(x-3)(x+1)+(x-1)√(x-3)(x+3)]=
=√(x+3)[(x-1)√(x+3)+(x+1)√(x-3)]/√(x-3)[(x+1)√(x-3)+(x-1)√(x+3)]=
=√(x+3)/√(x-3)=√[(x+3)/(x-3)]
б
[(t-3)(t+2)+(t+3)√(t-2)(t+2)]/[(t+3)(t-2)-(t-3)√(y-2)(t+2)]=
=√(t+2)[(t-3)√(t+2)-(t+3)√(t-2)]/√(t+2)[(t+3)√(t-2)-(t-3)√(t+2)]=
=-√(t+2)/√(t-2)=-√[(t+2)/(t-2)]