a) Ta có: A=√x+1x+4√x+4:(xx+2√x+x√x+2)A=x+1x+4x+4:(xx+2x+xx+2)

=√x+1(√x+2)2:(x√x(√x+2)+x√x+2)=x+1(x+2)2:(xx(x+2)+xx+2)

=√x+1(√x+2)2:(√x√x+2+x√x+2)=x+1(x+2)2:(xx+2+xx+2)

=√x+1(√x+2)2⋅√x+2√x(√x+1)=x+1(x+2)2⋅x+2x(x+1)

=1x+2√x=1x+2x

b) Để A≥13√xA≥13x thì A−13√x≥0A−13x≥0

⇔1x+2√x−13√x≥0⇔1x+2x−13x≥0

⇔1√x(√x+2)−13√x≥0⇔1x(x+2)−13x≥0

⇔33√x(√x+2)−√x+23√x(√x+2)≥0⇔33x(x+2)−x+23x(x+2)≥0

⇔1−√x3√x(√x+2)≥0⇔1−x3x(x+2)≥0

mà 3√x(√x+2)>0∀x3x(x+2)>0∀x thỏa mãn ĐKXĐ

nên 1−√x≥01−x≥0

⇔√x≤1