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

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

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

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

Where is biểu thức ?

a) ĐK:\(\begin{cases} x + 2≠0\\ x - 2≠0 \end{cases}\)⇔\(\begin{cases} x ≠ -2\\ x≠ 2 \end{cases}\)

Vậy biểu thức P xác định khi x≠ -2 và x≠ 2

b) P= \(\dfrac{3}{x+2}\)-\(\dfrac{2}{2-x}\)-\(\dfrac{8}{x^2-4}\)

P=\(\dfrac{3}{x+2}\)+\(\dfrac{2}{x-2}\)-\(\dfrac{8}{(x-2)(x+2)}\)

P= \(\dfrac{3(x-2)}{(x-2)(x+2)}\)+\(\dfrac{2(x+2)}{(x-2)(x+2)}\)-\(\dfrac{8}{(x-2)(x+2)}\)

P= \(​​​​\dfrac{3x-6+2x+4-8}{(x-2)(x+2)}\)

P=\(\dfrac{5x-10}{(x-2)(x+2)}\)

P=\(\dfrac{5(x-2)}{(x-2)(x+2)}\)

P=\(\dfrac{5}{x+2}\)

Vậy P=\(\dfrac{5}{x+2}\)

a: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

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: ĐKXĐ: x<>4; x<>-4

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

c: Để A nguyên thì x+4-5 chia hết cho x+4

=>\(x+4\in\left\{1;-1;5;-5\right\}\)

=>\(x\in\left\{-3;-5;1;-9\right\}\)

a, ĐKXĐ: \(a\ne1;a\ne-1\) 

Ta có:

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

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

\(\Rightarrow P=\frac{2a}{a+1}\) 

b. Để P có giá trị nguyên \(\Rightarrow2a⋮a+1\Rightarrow2\left(a+1\right)-2a⋮a+1\Rightarrow2a+2-2a⋮a+1\)

\(\Rightarrow2⋮a+1\) vì \(a\in Z\Rightarrow a+1\in\left\{-2;-1;1;2\right\}\Rightarrow a\in\left\{-3;-2;0;1\right\}\)

Vậy \(a\in\left\{-3;-2;0;1\right\}\)

Giúp Mk vs mai thi rồi

a khác 1 mà bạn

a: \(A=\dfrac{x^2-2x+2x^2+4x-3x^2-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}=\dfrac{2}{x+2}\)

a, \(\dfrac{x}{x+2}\) + \(\dfrac{2x}{x-2}\) -\(\dfrac{3x^2-4}{x^2-4}\)

= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{x^2-4}\)

= \(\dfrac{x}{x+2}+\dfrac{2x}{x-2}-\dfrac{3x^2+4}{\left(x+2\right)\left(x-2\right)}\)

= \(\dfrac{x\left(x-2\right)+2x\left(x+2\right)-3x^2-4}{\left(x+2\right)\left(x-2\right)}\)

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

Có vài bước mình làm tắc á nha :>