Question

证明根号(a^2+1/b^2+a^2/(ab+1)^2=绝对值(a+1/b-a/ab+1)

Answer 1

√(a^2+1/(b^2)+a^2/(ab+1)^2)

=√(a^2+1)(ab+1)^2/(b^2)(ab+1)^2+a^2*b^2/(ab+1)^2*b^2)

=√(a^2+1)(ab+1)^2+(ab)^2/(b^2)(ab+1)^2)

=√[(a^2+1)(a^2*b^2+2ab+1)+(ab)^2]/(b^2)(ab+1)^2)

=√[a^4*b^2+a^2*b^2+a^2+1+2a^3*b+2ab+(ab)^2]/(b^2)(ab+1)^2)

=√[a^4*b^2+a^2+1+2a^3*b+2ab+2(ab)^2]/(b^2)(ab+1)^2)

=√[(ab+1)^2-ab]^2/(b^2)(ab+1)^2)

=|[(ab+1)^2-ab]/b(ab+1)|

=|a+1/b-a/(ab+1)|