Question

[Bảo Yến Nguyễn]

giúp vs ạ ! mk chỉ cần p2 thôi ạ !

Answer 1

[Nguyễn Lê Phước Thịnh]

1: \(A=\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-\sqrt{x}+1}-\dfrac{x+2}{x\sqrt{x}+1}\)

\(=\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{x-\sqrt{x}+1}-\dfrac{x+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\dfrac{x-\sqrt{x}+1+2\left(\sqrt{x}+1\right)-x-2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\dfrac{-\sqrt{x}-1+2\sqrt{x}+2}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}\)

\(=\dfrac{1}{x-\sqrt{x}+1}\)

2: A>=4/3

=>\(A-\dfrac{4}{3}>=0\)

=>\(\dfrac{1}{x-\sqrt{x}+1}-\dfrac{4}{3}>=0\)

=>\(\dfrac{3-4x+4\sqrt{x}-4}{3\left(x-\sqrt{x}+1\right)}>=0\)

=>\(-4x+4\sqrt{x}-1>=0\)

=>\(4x-4\sqrt{x}+1< =0\)

=>\(\left(2\sqrt{x}-1\right)^2< =0\)

mà \(\left(2\sqrt{x}-1\right)^2>=0\forall x>=0\)

nên \(2\sqrt{x}-1=0\)

=>\(\sqrt{x}=\dfrac{1}{2}\)

=>\(x=\dfrac{1}{4}\left(nhận\right)\)

Answer 2

[Trần Vũ Minh Huy]

kết quả phần 1 đâu bn