ĐKXĐ: x>=0; x<>1

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

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

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

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

ĐKXĐ: x>=0; x<>4

a: Thay x=2 vào A, ta được:

\(A=\frac{\sqrt2+2}{\sqrt2+1}=\frac{\sqrt2\left(\sqrt2+1\right)}{\sqrt2+1}=\sqrt2\)

b: \(B=\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{5\sqrt{x}+2}{4-x}\)

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

\(=\frac{x+3\sqrt{x}+2+2x-4\sqrt{x}-5\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{3x-6\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{3\sqrt{x}}{\sqrt{x}+2}\)

c: \(P=-A\cdot B=-\frac{\sqrt{x}+2}{\sqrt{x}+1}\cdot\frac{3\sqrt{x}}{\sqrt{x}+2}=\frac{-3\sqrt{x}}{\sqrt{x}+1}=\frac{-3\sqrt{x}-3+3}{\sqrt{x}+1}=-3+\frac{3}{\sqrt{x}+1}\)

Để P là số nguyên thì 3⋮\(\sqrt{x}+1\)

=>\(\sqrt{x}+1\in\left\lbrace1;3\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace0;2\right\rbrace\)

=>x∈{0;4}

Kết hợp ĐKXĐ, ta được: x=0

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

b: Để P=-1 thì \(\sqrt{x}-1=-\sqrt{x}\)

=>x=1/4(nhận)

a:Thay x=9 vào A, ta được:

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

a:

ĐKXĐ: x>=0; x<>1

Ta có: \(\frac{2}{\sqrt{x}-1}-\frac{5}{x+\sqrt{x}-2}\)

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

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

Ta có: \(1+\frac{3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

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

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

Ta có: \(P=\left(\frac{2}{\sqrt{x}-1}-\frac{5}{x+\sqrt{x}-2}\right):\left(1+\frac{3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)\)

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

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

b: Thay \(x=6-2\sqrt5=\left(\sqrt5-1\right)^2\) vào P, ta được:

\(P=\frac{2\cdot\sqrt{\left(\sqrt5-1\right)^2}-1}{\sqrt{\left(\sqrt5-1\right)^2}+1}\)

\(=\frac{2\left(\sqrt5-1\right)-1}{\sqrt5-1+1}=\frac{2\sqrt5-3}{\sqrt5}=2-\frac{3}{\sqrt5}=2-\frac{3\sqrt5}{5}=\frac{10-3\sqrt5}{5}\)

c: \(P=\frac{1}{\sqrt{x}}\)

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

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

=>\(2x-2\sqrt{x}-1=0\)

=>\(x-\sqrt{x}-\frac12=0\)

=>\(x-\sqrt{x}+\frac14-\frac34=0\)

=>\(\left(\sqrt{x}-\frac12\right)^2=\frac34\)

=>\(\left[\begin{array}{l}\sqrt{x}-\frac12=\frac{\sqrt3}{2}\\ \sqrt{x}-\frac12=-\frac{\sqrt3}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}\sqrt{x}=\frac{\sqrt3+1}{2}\\ \sqrt{x}=\frac{-\sqrt3+1}{2}\left(loại\right)\end{array}\right.\)

=>\(\sqrt{x}=\frac{\sqrt3+1}{2}\)

=>\(x=\left(\frac{\sqrt3+1}{2}\right)^2=\frac{4+2\sqrt3}{4}=\frac{2+\sqrt3}{2}\)

d: Để P là số nguyên thì \(2\sqrt{x}-1\) ⋮\(\sqrt{x}+1\)

=>\(2\sqrt{x}+2-3\) ⋮\(\sqrt{x}+1\)

=>-3⋮\(\sqrt{x}+1\)

=>\(\sqrt{x}+1\in\left\lbrace1;3\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace0;2\right\rbrace\)

=>x∈{0;4}

e: \(P<1-\sqrt{x}\)

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

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

=>\(2\sqrt{x}-1+x-1<0\)

=>\(x+2\sqrt{x}+1-3<0\)

=>\(\left(\sqrt{x}+1\right)^2<3\)

=>\(\sqrt{x}+1<\sqrt3\)

=>\(\sqrt{x}<\sqrt3-1\)

=>\(x<4-2\sqrt3\)

Kết hợp ĐKXĐ, ta được: 0<=x<\(4-2\sqrt3\)

a:

ĐKXĐ: x>=0; x<>1

Ta có: \(\frac{2}{\sqrt{x}-1}-\frac{5}{x+\sqrt{x}-2}\)

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

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

Ta có: \(1+\frac{3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\)

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

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

Ta có: \(P=\left(\frac{2}{\sqrt{x}-1}-\frac{5}{x+\sqrt{x}-2}\right):\left(1+\frac{3-x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\right)\)

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

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

b: Thay \(x=6-2\sqrt5=\left(\sqrt5-1\right)^2\) vào P, ta được:

\(P=\frac{2\cdot\sqrt{\left(\sqrt5-1\right)^2}-1}{\sqrt{\left(\sqrt5-1\right)^2}+1}\)

\(=\frac{2\left(\sqrt5-1\right)-1}{\sqrt5-1+1}=\frac{2\sqrt5-3}{\sqrt5}=2-\frac{3}{\sqrt5}=2-\frac{3\sqrt5}{5}=\frac{10-3\sqrt5}{5}\)

c: \(P=\frac{1}{\sqrt{x}}\)

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

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

=>\(2x-2\sqrt{x}-1=0\)

=>\(x-\sqrt{x}-\frac12=0\)

=>\(x-\sqrt{x}+\frac14-\frac34=0\)

=>\(\left(\sqrt{x}-\frac12\right)^2=\frac34\)

=>\(\left[\begin{array}{l}\sqrt{x}-\frac12=\frac{\sqrt3}{2}\\ \sqrt{x}-\frac12=-\frac{\sqrt3}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}\sqrt{x}=\frac{\sqrt3+1}{2}\\ \sqrt{x}=\frac{-\sqrt3+1}{2}\left(loại\right)\end{array}\right.\)

=>\(\sqrt{x}=\frac{\sqrt3+1}{2}\)

=>\(x=\left(\frac{\sqrt3+1}{2}\right)^2=\frac{4+2\sqrt3}{4}=\frac{2+\sqrt3}{2}\)

d: Để P là số nguyên thì \(2\sqrt{x}-1\) ⋮\(\sqrt{x}+1\)

=>\(2\sqrt{x}+2-3\) ⋮\(\sqrt{x}+1\)

=>-3⋮\(\sqrt{x}+1\)

=>\(\sqrt{x}+1\in\left\lbrace1;3\right\rbrace\)

=>\(\sqrt{x}\in\left\lbrace0;2\right\rbrace\)

=>x∈{0;4}

e: \(P<1-\sqrt{x}\)

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

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

=>\(2\sqrt{x}-1+x-1<0\)

=>\(x+2\sqrt{x}+1-3<0\)

=>\(\left(\sqrt{x}+1\right)^2<3\)

=>\(\sqrt{x}+1<\sqrt3\)

=>\(\sqrt{x}<\sqrt3-1\)

=>\(x<4-2\sqrt3\)

Kết hợp ĐKXĐ, ta được: 0<=x<\(4-2\sqrt3\)

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

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

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

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

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

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

b) \(A=\dfrac{2\sqrt{9}-1}{\sqrt{9}+1}=\dfrac{5}{4}\)

c) \(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}< 1\Rightarrow2\sqrt{x}-1< \sqrt{x}+1\Rightarrow\sqrt{x}< 2\Rightarrow x< 4\)

\(1,A=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ A=\dfrac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\\ 2,x=9\Leftrightarrow A=\dfrac{6-1}{3+1}=\dfrac{5}{4}\\ 3,A< 1\Leftrightarrow\dfrac{2\sqrt{x}-1-\sqrt{x}-1}{\sqrt{x}+1}< 0\\ \Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\Leftrightarrow\sqrt{x}-2< 0\left(\sqrt{x}+1>0\right)\\ \Leftrightarrow x< 4\Leftrightarrow0\le x< 4\)

ĐKXĐ: \(x\ge0\)

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

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

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

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

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

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

Mk ra đáp án khác với đáp án ủa bn nên bn bào sai chứ j, thật ra cả 2 đáp án đều giống nhau, do biến đổi dấu nên trở thành 2 đáp án khác nhau thôi :V

để mk lm lại phần đáp án của mk ra giống đáp án của bn nek :V

\(a,\)\(P=\dfrac{-x-1}{x-1}\)

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

Còn câu b thì hôm qua bn ghi là \(x=\dfrac{1}{\sqrt{2}}\) chứ có pk là \(1\sqrt{2}\) đou >:V

\(b,\)Thay \(x=1\sqrt{2}\) vào \(P\) ta có :

\(P=\dfrac{x-1}{1-x}\)

\(P=\dfrac{1\sqrt{2}-1}{1-1\sqrt{2}}=3+2\sqrt{2}\)