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 )

Bài 1: Sửa đề: \(B=\left(\frac{x-2}{x+2\sqrt{x}}+\frac{1}{\sqrt{x}+2}\right)\cdot\frac{\sqrt{x}+1}{\sqrt{x}-1}\)

a) Thay x=49 vào biểu thức \(A=\frac{\sqrt{x}+3}{\sqrt{x}-1}\), ta được:

\(A=\frac{\sqrt{49}+3}{\sqrt{49}-1}=\frac{7+3}{7-1}=\frac{10}{6}=\frac{5}{3}\)

Vậy: Khi x=49 thì \(A=\frac{5}{3}\)

b) Sửa đề: Rút gọn biểu thức B

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

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

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

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

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

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

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

c) Ta có: \(\frac{B}{A}=\frac{\sqrt{x}+1}{\sqrt{x}}:\frac{\sqrt{x}+3}{\sqrt{x}-1}\)

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

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

Để \(\frac{B}{A}< \frac{3}{4}\) thì \(\frac{x-1}{\sqrt{x}\left(\sqrt{x}+3\right)}-\frac{3}{4}< 0\)

\(\Leftrightarrow\frac{4\left(x-1\right)-3\sqrt{x}\left(\sqrt{x}+3\right)}{4\sqrt{x}\left(\sqrt{x}+3\right)}< 0\)

mà \(4\sqrt{x}\left(\sqrt{x}+3\right)>0\forall x\) thỏa mãn ĐKXĐ

nên \(4\left(x-1\right)-3\sqrt{x}\left(\sqrt{x}+3\right)< 0\)

\(\Leftrightarrow4x-4-3x-9\sqrt{x}< 0\)

\(\Leftrightarrow x-9\sqrt{x}-4< 0\)

\(\Leftrightarrow x^2-9x-4< 0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\frac{9}{2}+\frac{81}{4}-\frac{97}{4}< 0\)

\(\Leftrightarrow\left(x-\frac{9}{2}\right)^2< \frac{97}{4}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-\frac{9}{2}>-\frac{\sqrt{97}}{2}\\x-\frac{9}{2}< \frac{\sqrt{97}}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>\frac{9-\sqrt{97}}{2}\\x< \frac{9+\sqrt{97}}{2}\end{matrix}\right.\)

Kết hợp ĐKXĐ, ta được:

\(3< x< \frac{9+\sqrt{97}}{2}\)

a, - Thay x = 25 vào biểu thức B ta được :

\(B=\frac{\sqrt{25}-3}{\sqrt{25}+1}=\frac{1}{3}\)

b, Ta có : \(A=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{3-11\sqrt{x}}{9-x}\)

=> ​​​\(A=\frac{2\sqrt{x}}{\sqrt{x}+3}+\frac{\sqrt{x}+1}{\sqrt{x}-3}+\frac{11\sqrt{x}-3}{x-9}\)​

=> \(A=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{11\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

=> \(A=\frac{2\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)+11\sqrt{x}-3}{x-9}\)

=> \(A=\frac{2x-6\sqrt{x}+x+\sqrt{x}+3\sqrt{x}+3+11\sqrt{x}-3}{x-9}\)

=> \(A=\frac{3x+9\sqrt{x}}{x-9}=\frac{3\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{3\sqrt{x}}{\sqrt{x}-3}\)

c, Ta có : \(P=AB+1=\frac{3\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+1\)

=> \(P=\frac{3\sqrt{x}}{\sqrt{x}+1}+1\)

=> \(P=\frac{3\sqrt{x}+3-3}{\sqrt{x}+1}+1\)

=> \(P=4-\frac{3}{\sqrt{x}+1}\)

Ta thấy : \(\sqrt{x}\ge0\)

=> \(4-\frac{3}{\sqrt{x}+1}\ge1\)

Vậy Min_P = 1 khi x = 0

a) Ta có:

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

\(A=\frac{\sqrt{4}-3}{4-\sqrt{4}+1}\)

\(A=\frac{2-3}{4-2+1}=-\frac{1}{3}\)

b) đk: \(\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)

\(B=\left(\frac{3\sqrt{x}+6}{x-9}-\frac{2}{\sqrt{x}-3}\right):\frac{1}{\sqrt{x}+3}\)

\(B=\frac{3\sqrt{x}+6-2\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\left(\sqrt{x}+3\right)\)

\(B=\frac{3\sqrt{x}+6-2\sqrt{x}-6}{\sqrt{x}-3}\)

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

c) \(P=AB\)

\(P=\frac{\sqrt{x}-3}{x-\sqrt{x}+1}\cdot\frac{\sqrt{x}}{\sqrt{x}-3}\)

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

Vì \(\left|P\right|=P\Leftrightarrow\orbr{\begin{cases}P=P\\P=-P\end{cases}}\Leftrightarrow\orbr{\begin{cases}0=0\left(tm\right)\\\sqrt{x}=-\sqrt{x}\end{cases}}\Rightarrow\hept{\begin{cases}x\ge0\\x\ne9\end{cases}}\)

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