a, Với \(x\ge0;x\ne1\)

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

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

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

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

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

Bạn ghi chuẩn đề chưa vậy

đúng mà

a) rút gọn biểu thức Q

b) Tính giá trị của x để Q<1

\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)

\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)

\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)

\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)

\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)

_Minh ngụy_

\(\(c)\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)\)( tự ghi điều kiện )

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

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

\(\(=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)\)( phá ngoặc và tính )

\(\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)\)

_Minh ngụy_

\(\(d)\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\left(x\ge0\right)\)\)

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

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

\(\(=\frac{\sqrt{x}-1}{\sqrt{x}+1}\)\)( vì \(\(x\ge0\)\))

_Minh ngụy_

a, Thay x = 9 vào biểu thức \(A=\frac{\sqrt{x}-2}{\sqrt{x}-1}\)  ta được:

\(A=\frac{\sqrt{9}-2}{\sqrt{9}-1}=\frac{\sqrt{3^2}-2}{\sqrt{3^2}-1}=\frac{3-2}{3-1}=\frac{1}{2}\)

Vậy với x = 9 thì \(A=\frac{1}{2}\)

\(b,\left(\frac{\sqrt{x}}{\sqrt{x}+1}+\frac{1}{\sqrt{x}-1}\right).\frac{\sqrt{x}-1}{x+1}\left(x\ge0;x\ne1\right)\)

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

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

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

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

a, Thay x=9 ta có 

\(A=\frac{\sqrt{9}-2}{\sqrt{9}-1}=\frac{3-2}{3-1}=\frac{1}{2}\)

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

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

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

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

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

Giúp mình với mình đang cần gấp. Thk you các pạn

Ta có: \(B=\frac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)+5\left(\sqrt{x}+1\right)+4}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{x+2\sqrt{x}-3+5\sqrt{x}+5+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+6\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+6}{\sqrt{x}-1}\)

do đó \(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}.\frac{\sqrt{x}-6}{\sqrt{x}-1}=\frac{\sqrt{x}-6}{\sqrt{x}+1}=1-\frac{7}{\sqrt{x}+1}\)

Vì \(x\ge0\Rightarrow0< \frac{7}{\sqrt{x}+1}\le7\)

Để P nguyên thì \(\frac{7}{\sqrt{x}+1}\in Z\)

do đó \(\frac{7}{\sqrt{x}+1}\in\left\{1,2,3,4,5,6,7\right\}\)

Đến đây xét từng TH là  ra

rút gọn B ta có B=\(\frac{\sqrt{x}+6}{\sqrt{x}-1}\)\(\Rightarrow\)\(AB=\frac{\sqrt{x}+6}{\sqrt{x}+1}\in Z\)

=\(1+\frac{5}{\sqrt{x}+1}\)

Vì 1\(\in Z\) nên để P thuộc Z thì \(\frac{5}{\sqrt{x}+1}\in Z\)

\(\Rightarrow\left(\sqrt{x}+1\right)\inƯ\left(5\right)=\pm1;\pm5\)

Đến đây thì ez rồi

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

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

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

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

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

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

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

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

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

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

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

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

b/ Ta có: \(P=\frac{\sqrt{x}-1}{\sqrt{x}+1}=1-\frac{2}{\sqrt{x}+1}\)

    Để \(P\in Z\) thì \(\left(\sqrt{x}+1\right)\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)

    + Với \(\sqrt{x}+1=1\Rightarrow\sqrt{x}=0\Rightarrow x=0\)

    + Với \(\sqrt{x}+1=-1\Rightarrow\sqrt{x}=-2\left(vn\right)\)

    + Với \(\sqrt{x}+1=2\Rightarrow\sqrt{x}=1\Rightarrow x=1\)(loại)

    + Với \(\sqrt{x}+1=-2\Rightarrow\sqrt{x}=-3\left(vn\right)\)

                                         Vậy x = 0 thì P nguyên

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

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

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

Để P nguyên thì \(\sqrt{x}+1\in\left\{1;2\right\}\Leftrightarrow x\in\left\{0\right\}\) (Vì x khác 1 - điều kiện)

c) \(\sqrt{x}+1\ge1\Leftrightarrow\frac{2}{\sqrt{x}+1}\le\frac{1}{2}\Leftrightarrow1-\frac{2}{\sqrt{x}+1}\ge\frac{1}{2}\)

\(\Rightarrow P\ge\frac{1}{2}\). Dấu đẳng thức xảy ra khi x = 0

Vậy Min P = 1/2 <=> x = 0

Sửa lại câu c) 

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

Ta có \(\sqrt{x}+1\ge1\Leftrightarrow\frac{2}{\sqrt{x}+1}\le2\Leftrightarrow1-\frac{2}{\sqrt{x}+1}\le-1\)

Dấu "=" xảy ra khi x = 0

Vậy Min P = -1 khi và chỉ khi x = 0

a/ Ta có: \(x+2\sqrt{x}+1=\left(\sqrt{x}+1\right)^2\)

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

=> \(P=\left[\frac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\frac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right].\frac{\sqrt{x}+1}{\sqrt{x}}\)

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

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

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

b/ Thay \(x=\frac{\sqrt{3}}{2+\sqrt{3}}\)  => \(P=\frac{2}{\frac{\sqrt{3}}{2+\sqrt{3}}-1}=\frac{2\left(2+\sqrt{3}\right)}{\sqrt{3}-2-\sqrt{3}}\)

=> \(P=-\left(2+\sqrt{3}\right)\)

c/ \(P=\frac{2}{x-1}=-\frac{4}{\sqrt{x}+1}\) <=> \(\frac{1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=-\frac{2}{\sqrt{x}+1}\)

<=> \(\frac{1}{\sqrt{x}-1}=-2\)

<=> \(1=-2\sqrt{x}+2\)

<=> \(2\sqrt{x}=1=>\sqrt{x}=\frac{1}{2}=>x=\frac{1}{4}\)

\(a,A=\sqrt{27}+\frac{2}{\sqrt{3}-2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)

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

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

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

       \(=-3\)

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

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

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

    \(=\frac{\sqrt{x}-1}{\sqrt{x}}\)

b, Ta có \(B< A\)

\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}< -3\)

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

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

\(\Leftrightarrow\frac{4\sqrt{x}-1}{\sqrt{x}}< 0\)

\(\Leftrightarrow4\sqrt{x}-1< 0\left(Do\sqrt{x}>0\right)\)

\(\Leftrightarrow\sqrt{x}< \frac{1}{4}\)

\(\Leftrightarrow0< x< \frac{1}{2}\)(Kết hợp ĐKXĐ)

Vậy ...