\(P=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)\(\left(ĐKXĐ:x\ne4\right)\)

\(P=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{2\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\frac{-2-5\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)

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

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

\(P=\frac{3\sqrt{x}}{\sqrt{x}+2}\)

b) Với  \(x=3\)( thỏa mãn ĐKXĐ ) ta có  \(P=\frac{3\sqrt{3}}{\sqrt{3}+2}=-9+6\sqrt{3}\)

c) A ở đâu ???? '-' 

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 : Với : \(x>0;x\ne1\)

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

Thay vào ta được : \(P=x=25\)

Bài 2 : 

a, Với \(x\ge0;x\ne1\)

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

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

Thay x = 9 vào A ta được : \(\frac{3}{3+1}=\frac{3}{4}\)

Bài 3 : \(x\ge0;x\ne1\)

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

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

b, Ta có : \(P=\frac{\sqrt{x}+2}{\sqrt{x}-1}=\frac{5}{4}\Rightarrow4\sqrt{x}+8=5\sqrt{x}-5\)

\(\Leftrightarrow\sqrt{x}=13\Leftrightarrow x=169\)(tmđk )

Bài 1 : 

a )\(A=\frac{3-\sqrt{3}}{\sqrt{3}-1}+\frac{\sqrt{35}-\sqrt{15}}{\sqrt{5}}-\sqrt{28}\)

\(A=\frac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+\frac{\sqrt{5}\left(\sqrt{7}-\sqrt{3}\right)}{\sqrt{5}}-\sqrt{28}\)

\(A=\sqrt{3}+\sqrt{7}-\sqrt{3}-\sqrt{28}\)

\(A=\sqrt{7}-\sqrt{28}\)

\(A=\sqrt{7}-2\sqrt{7}=-\sqrt{7}\)

Vậy \(A=-\sqrt{7}\)

b)\(B=\frac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\frac{\sqrt{a}+\sqrt{b}}{a-b}\left(a,b>0;a\ne b\right)\)

\(B=\frac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}:\frac{\sqrt{a}+\sqrt{b}}{a-b}\)

\(B=\left(\sqrt{a}+\sqrt{b}\right).\frac{a-b}{\sqrt{a}+\sqrt{b}}\)

\(B=a-b\)

Vậy \(B=a-b\left(a,b>0;a\ne b\right)\)

_Minh ngụy_

Bài 2 :

a )\(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{1-\sqrt{x}}{x+\sqrt{x}}\left(x>0\right)\)

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

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

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

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

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

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

Vậy \(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}\left(x>0\right)\)

b) \(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}\left(x>0\right)\)

Ta có : \(B>0\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}+1}>0\)

Vì : \(\sqrt{x}\ge0\forall x\Rightarrow\)để \(B>O\)cần \(\sqrt{x}-1>0\Leftrightarrow\sqrt{x}>1\Leftrightarrow x>1\)( thỏa mãn \(x>0\))

Vậy \(x>1\)thì \(B>0\)

_Minh ngụy_

Bài 3: 

a) \(K=\frac{5+\sqrt{5}}{\sqrt{5}+1}-\sqrt{6-2\sqrt{5}}=\frac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}-\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(=\sqrt{5}-\left|\sqrt{5}-1\right|=1\)

KL: K=1

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

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

\(=\frac{-4\sqrt{x}}{\left(1-x\right)}\cdot\left(1-x\right)x\sqrt{x}=-4x^2\)

b) Thay P = -4x^2 và K= 1 vào biểu thức P + 6K =2x , được: 

\(-4x^2+6=2x\Leftrightarrow2x^2+x-6=0\Leftrightarrow\left(2x-3\right)\left(x+2\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\left(n\right)\\x=-2\left(n\right)\end{cases}}\)

KL:.......

Trả lời:

b, \(B=\frac{\sqrt{x}-1}{\sqrt{x}}+\frac{2\sqrt{x}+1}{x+\sqrt{x}}\left(ĐK:x>0\right)\)

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

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

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

c, \(\frac{A}{B}>\frac{3}{2}\Leftrightarrow\frac{2+\sqrt{x}}{\sqrt{x}}:\frac{\sqrt{x}+2}{\sqrt{x}+1}>\frac{3}{2}\) \(\left(ĐK:x>0\right)\)

\(\Leftrightarrow\frac{\sqrt{x}+2}{\sqrt{x}}\cdot\frac{\sqrt{x}+1}{\sqrt{x}+2}>\frac{3}{2}\)

\(\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}}>\frac{3}{2}\Leftrightarrow\frac{\sqrt{x}+1}{\sqrt{x}}-\frac{3}{2}>0\)

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

\(\Rightarrow2\sqrt{x}+1-3\sqrt{x}>0\Leftrightarrow1-\sqrt{x}>0\)

\(\Leftrightarrow-\sqrt{x}>-1\Leftrightarrow\sqrt{x}< 1\Leftrightarrow x< 1\)

Vậy \(0< x< 1\) là giá trị cần tìm.