`a)D` xác định `<=>a-1 ne 0<=>a ne 1`

`b)` Với `a ne 1` có:

`D=([a-1]/[a^2+a+1]-[1-3a+a^2]/[(a-1)(a^2+a+1)]-1/[a-1]).[1-a]/[a^2+1]`

`D=[(a-1)^2-1+3a-a^2-a^2-a-1]/[(a-1)(a^2+a+1)].[-(a-1)]/[a^2+1]`

`D=[a^2-2a+1-1+3a-a^2-a^2-a-1]/[(-a^2-1)(a^2+a+1)]`

`D=[-a^2-1]/[(-a^2-1)(a^2+a+1)]=1/[a^2+a+1]`

`c)` Với `a ne 1` có:

`1/D=1/[1/[a^2+a+1]]=a^2+a+1=(a+1/2)^2+3/4`

Vì `(a+1/2)^2 >= 0 AA a ne 1`

   `=>(a+1/2)^2+3/4 >= 3/4 AA a ne 1`

  Hay `1/D >= 3/4 AA a ne 1=>1/D  _[mi n]=3/4`

Dấu "`=`" xảy ra `<=>a=-1/2` (t/m).

a,ĐKXĐ:\(\left\{{}\begin{matrix}x\ne\pm1\\x\ne\dfrac{1}{2}\end{matrix}\right.\)

\(A=\left(\dfrac{2}{x+1}-\dfrac{1}{x-1}+\dfrac{5}{x^2-1}\right):\dfrac{2x+1}{x^2-1}\\ =\left(\dfrac{2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}+\dfrac{5}{\left(x+1\right)\left(x-1\right)}\right).\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\\ =\dfrac{2x-2-x-1+5}{\left(x+1\right)\left(x-1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\\ =\dfrac{x+2}{2x+1}\)

\(b,A=3\\ \Leftrightarrow\dfrac{x+2}{2x+1}=3\\ \Leftrightarrow6x+3=x+2\\ \Leftrightarrow5x+1=0\\ \Leftrightarrow x=-\dfrac{1}{5}\left(tm\right)\)

\(c,\dfrac{1}{A}=\dfrac{2x+1}{x+2}=\dfrac{2x+4-3}{x+2}=\dfrac{2\left(x+2\right)-3}{x+2}=2-\dfrac{3}{x+2}\)

Để `1/A` là số nguyên thì `3/(x+2)` nguyên \(\Rightarrow x+2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

Ta có bảng:

Vậy \(x\in\left\{-5;-3\right\}\)

Đề sai rồi bạn

a) ĐKXĐ: a²-1 ≠0 ⇔ (a-1)(a+1)≠0 ⇔\(\left[{}\begin{matrix}a-1\ne0\\a+1\ne0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a\ne1\\a\ne-1\end{matrix}\right.\)

b) A=\(\dfrac{2a^2}{a^2-1}-\dfrac{a}{a+1}+\dfrac{a}{a-1}\) , a≠1, -1

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

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

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

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

vậy A =\(\dfrac{2a}{a-1}\) với a≠1,-1.

c) Có:A= \(\dfrac{2a}{a-1}\) = \(\dfrac{2a-2+2}{a-1}=\dfrac{2\left(a-1\right)+2}{a-1}=2+\dfrac{2}{a-1}\)

Để a∈Z thì a-1 ∈ Z ⇒ (a-1) ∈ Ư(2) =\(\left\{1;-1;2;-2\right\}\)

Ta có bảng sau:

Vậy để biểu thức A có giá trị nguyên thì a∈\(\left\{2;0;3\right\}\)

a) ĐKXĐ: \(a\notin\left\{1;-1\right\}\)

b) Ta có: \(A=\dfrac{2a^2}{a^2-1}-\dfrac{a}{a+1}+\dfrac{a}{a-1}\)

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

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

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

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

\(=\dfrac{2a}{a-1}\)

c) Để A nguyên thì \(2a⋮a-1\)

\(\Leftrightarrow2a-2+2⋮a-1\)

mà \(2a-2⋮a-1\)

nên \(2⋮a-1\)

\(\Leftrightarrow a-1\inƯ\left(2\right)\)

\(\Leftrightarrow a-1\in\left\{1;-1;2;-2\right\}\)

\(\Leftrightarrow a\in\left\{2;0;3;-1\right\}\)

Kết hợp ĐKXĐ, ta được: \(a\in\left\{0;2;3\right\}\)

Vậy: Để A nguyên thì \(a\in\left\{0;2;3\right\}\)

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

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

\(=\dfrac{a+\sqrt{a}+1}{\sqrt{a}-1}\)

b: Để P<1 thì P-1<0

\(\Leftrightarrow\dfrac{a+\sqrt{a}+1-\sqrt{a}+1}{\sqrt{a}-1}< 0\)

hay 0<a<1

\(A=\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)

\(\Rightarrow A=\left(\dfrac{x-2\left(x+2\right)+1\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\right)\)

\(\Rightarrow A=\left(\dfrac{-6}{x^2-4}\right):\left(\dfrac{6}{x+2}\right)\)

\(\Rightarrow A=-\dfrac{6}{x^2-4}.\dfrac{x+2}{6}=-\dfrac{6\left(x+2\right)}{\left(x-2\right)\left(x+2\right)6}=-\dfrac{1}{x-2}\)

để A<0 thì :

\(\left\{{}\begin{matrix}x-2\ne0\\x-2\notin Z-\end{matrix}\right.\)\(\Leftrightarrow x\in\left\{3;4;5;6;7;8;9;....n\right\}\)

( Z- là tập hợp số nguyên âm )

Để A có giá trị nguyên thì :

\(\left\{{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

a, ĐK: \(x\ge0,x\ne1\)

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

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

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

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

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

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

b, \(x=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)

Khi đó: 

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

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

\(=\dfrac{2\sqrt{3}-3}{\sqrt{3}}\)

\(=2-\sqrt{3}\)

c, \(A=\dfrac{1}{2}\)

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

\(\Leftrightarrow4\sqrt{x}-2=\sqrt{x}+1\)

\(\Leftrightarrow3\sqrt{x}=3\)

\(\Leftrightarrow x=1\left(l\right)\)

Vậy không tồn tại giá trị x thỏa mãn \(A=\dfrac{1}{2}\).

A đâu em?

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