a) ĐKXĐ : \(y\ne\pm1\)

 \(N=\left(\frac{1}{y-1}-\frac{y}{1-y^3}.\frac{y^2+y+1}{y+1}\right)\div\frac{1}{y^2-1}\)

\(=\left(\frac{1}{y-1}+\frac{y}{\left(y-1\right)\left(y^2+y+1\right)}.\frac{y^2+y+1}{y+1}\right)\div\frac{1}{y^2-1}\)

\(=\left(\frac{1}{y-1}+\frac{y}{\left(y-1\right)\left(y+1\right)}\right)\div\frac{1}{y^2-1}\)

\(=\frac{y+1+y}{\left(y-1\right)\left(y+1\right)}\div\frac{1}{\left(y-1\right)\left(y+1\right)}\)

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

\(=2y+1\)

Vậy \(N=2y+1\)khi \(y\ne\pm1\)

b) Với \(y=\frac{1}{2}\); phương trình N trở thành :

\(N=2.\frac{1}{2}+1=2\)

Vậy N=2 khi \(y=\frac{1}{2}\)

c) Để N luôn dương

\(\Leftrightarrow2y+1>0\)

\(\Leftrightarrow2y>-1\)

\(\Leftrightarrow y>\frac{-1}{2}\)

Kết hợp ĐKXĐ ta có : \(y>\frac{-1}{2};y\ne\pm1\)

Vậy N luôn dương khi \(y>\frac{-1}{2};y\ne\pm1\)
 

a., đk y khác cộng trừ 1

N=\(\left(\frac{1}{y-1}+\frac{y}{\left(y^3-1\right)}.\frac{y^2+y+1}{y+1}\right):\frac{1}{\left(y-1\right)\left(y+1\right)}\)

N=\(\left(\frac{1}{y-1}+\frac{y}{\left(y-1\right)\left(y+1\right)}\right).\left(y-1\right)\left(y+1\right)\)

N=\(\frac{y+1+y}{\left(y-1\right)\left(y+1\right)}.\left(y-1\right)\left(y+1\right)\)

N= \(2y+1\)

Vậy N=2y+1 với y khác cộng trừ 1

b, Thay y= \(\frac{1}{2}\) ( t/m đk y khác cộng trừ 1 )vào biểu thức N ta được:

N= \(2.\frac{1}{2}+1=1+1=2\) 

Vậy N=2 với y = 1/2

c, Để N luôn dương thì: 2y+1>0

<=> 2y>-1

<=>y>\(\frac{-1}{2}\)( t/ m đk y khác cộng trừ 1)

Vậy với y>-1/2 thì N luôn dương

a, \(N=\left(\frac{1}{y-1}-\frac{y}{1-y^3}.\frac{y^2+y+1}{y+1}\right):\frac{1}{y^2-1}\)

\(N=\left(\frac{1}{y-1}+\frac{y}{y^3-1}.\frac{y^2+y+1}{y+1}\right):\frac{1}{y^2-1}\)

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

\(N=\left(\frac{1}{y-1}+\frac{y}{\left(y-1\right)\left(y+1\right)}\right):\frac{1}{y^2-1}\)

\(N=\left(\frac{y+1}{\left(y-1\right)\left(y+1\right)}+\frac{y}{\left(y-1\right)\left(y+1\right)}\right):\frac{1}{\left(y-1\right)\left(y+1\right)}\)

\(N=\frac{y+1+y}{\left(y-1\right)\left(y+1\right)}:\frac{1}{\left(y-1\right)\left(y+1\right)}\)

\(N=\frac{2y+1}{\left(y-1\right)\left(y+1\right)}.\left(y-1\right)\left(y+1\right)\)

\(N=2y+1\)

b, Tại \(y=\frac{1}{2}\) ta có :

             \(N=2.\frac{1}{2}+1\)

\(\Rightarrow N=1+1=2\)

c, Để N luôn có giá trị dương thì \(y\in N\).

ĐKXĐ: \(\hept{\begin{cases}y>0\\y\ne1\end{cases}}\)

a/ Ta có: \(A=\left[\frac{\sqrt{y}^3-1}{\sqrt{y}\left(\sqrt{y}-1\right)}-\frac{\sqrt{y}^3+1}{\sqrt{y}\left(\sqrt{y}+1\right)}\right]:\frac{2\left(\sqrt{y}-1\right)^2}{\left(\sqrt{y}+1\right)\left(\sqrt{y}-1\right)}\)

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

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

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

b/ \(A=\frac{\sqrt{y}+1}{\sqrt{y}-1}=1+\frac{2}{\sqrt{y}-1}\)

    Để \(A\in Z\Rightarrow\left(\sqrt{y}-1\right)\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)

   Với \(\sqrt{y}-1=1\Rightarrow\sqrt{y}=2\Rightarrow y=4\)

   Với \(\sqrt{y}-1=-1\Rightarrow\sqrt{y}=0\Rightarrow y=0\)(loại)

   Với \(\sqrt{y}-1=2\Rightarrow\sqrt{y}=3\Rightarrow y=9\)

  Với \(\sqrt{y}-1=-2\Rightarrow\sqrt{y}=-1\) (loại)

      Vậy y = 4 , y = 9

1,

\(A=\left(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\right):\frac{a+2}{a-2}\left(đk:a\ne0;1;2;a\ge0\right)\)

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

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

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

Để \(A=1\)\(=>\frac{2a-4}{a+2}=1< =>2a-4-a-2=0< =>a=6\)

2, 

a, Điều kiện xác định của phương trình là \(x\ne4;x\ge0\)

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

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

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

c, Với \(x=3+2\sqrt{3}\)thì \(B=\frac{2}{3-2+2\sqrt{3}}=\frac{2}{1+2\sqrt{3}}\)