Q=\(\frac{\sqrt{x}-1}{x-\sqrt{x}+1}+\frac{x+2}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}+1}\) điều kiện x>=0

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

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

ta thấy cả tử và mẫu đề >=0=> Q>=0

dấu = xảy ra khi x=0

=> Q=0 khi x=0

A=\(\frac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

A= \(\frac{2x-3\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)=\(\frac{2x-2\sqrt{x}-\sqrt{x}+1}{x-1}=\frac{2\sqrt{x}-1}{x+1}\)

 Để A=1/2 thì 

\(\frac{2\sqrt{x}-1}{x+1}=\frac{1}{2}\)

nhân chéo ta đc pt \(x-4\sqrt{x}+3=0\)

giải pt ta đc x=1 (loại)  hoặc x= 9

vậy x=9 TM

Để A<1 thì \(\frac{2\sqrt{x}-1}{\sqrt{x}+1}< 1\Leftrightarrow2\sqrt{x}-1< \sqrt{x}+1\Leftrightarrow\sqrt{x}< 2\)

                                                                                               =>  x<4   

vậy vs 0\(\le x< 4\) và x khác 1 TM

Mình nghĩ thế này ạ

a) Với \(x\ge0,x\ne1\)ta có: \(\frac{\sqrt{x}+1}{\sqrt{x}-1x}+\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{3\sqrt{x}+1}{x-1}\)

\(=\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x-1}\right)}-\frac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

=\(\frac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

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

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

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

\(=\frac{2\sqrt{x}-1}{\sqrt{x}+1}\)

Kết luận :

b) Với \(x\ge0,x\ne1\)ta có: \(A=\frac{2\sqrt{x}-1}{\sqrt{x}+1}\)

\(A=\frac{1}{2}\Leftrightarrow\frac{2\sqrt{x}-1}{\sqrt{x}+1}=\frac{1}{2}\)

\(\Leftrightarrow\frac{4\sqrt{x}-2}{2\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{2\left(\sqrt{x}+1\right)}=0\)

\(\Leftrightarrow\frac{4\sqrt{x}-2-\sqrt{x}-1}{2\left(\sqrt{x}+1\right)}=0\)

\(\Leftrightarrow\frac{3\sqrt{x}-3}{2\left(\sqrt{x}+1\right)}=0\)

\(\Leftrightarrow3\sqrt{x}-3=0\)

\(\Leftrightarrow3\sqrt{x}=3\)

\(\Leftrightarrow\sqrt{x}=1\)

\(\Leftrightarrow x=1\)( không tm đkxđ)

Vậy không có gtri nào của x để A = \(\frac{1}{2}\)

Em gửi ảnh ạ !

Em gửi ảnh trên ạ !!!!!

a, Ta có \(x=49\Rightarrow\sqrt{x}=7\)

Thay vào biểu thức A ta được : 

\(A=\frac{7.4}{7-1}=\frac{28}{6}=\frac{14}{3}\)

b, Với \(x\ge0;x\ne1\)

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

\(=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)( đpcm )

1) Khi x = 49 thì:

\(A=\frac{4\sqrt{49}}{\sqrt{49}-1}=\frac{4\cdot7}{7-1}=\frac{28}{6}=\frac{14}{3}\)

2) Ta có:

\(B=\frac{1}{\sqrt{x}+1}+\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-1}\)

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

\(B=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(B=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

c) \(P=A\div B=\frac{4\sqrt{x}}{\sqrt{x}-1}\div\frac{\sqrt{x}+1}{\sqrt{x}-1}=\frac{4\sqrt{x}}{\sqrt{x}+1}\)

Ta có: \(P\left(\sqrt{x}+1\right)=x+4+\sqrt{x-4}\)

\(\Leftrightarrow\frac{4\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}=x+4+\sqrt{x-4}\)

\(\Leftrightarrow4\sqrt{x}=x+4+\sqrt{x-4}\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)^2+\sqrt{x-4}=0\)

Mà \(VT\ge0\left(\forall x\ge0,x\ne1\right)\)

\(\Rightarrow\hept{\begin{cases}\left(\sqrt{x}-2\right)^2=0\\\sqrt{x-4}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}=2\\x-4=0\end{cases}}\Rightarrow x=4\)

Vậy x = 4

Bạn vt đề bài rõ ra nhé, mk RG trc rùi phần câu hỏi xem sau( P là j z?)

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

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

\(=x-\sqrt{x}-3\)

\(\frac{4+\sqrt{X}}{7}\)