a) x ≠ -5.

b) Ta có P = ( x + 5 ) 2 x + 5 = x + 5  

c) Ta có P = 1 Û x = -4 (TMĐK)

d) Ta có P = 0 Û x = -5 (loại). Do vậy x ∈ ∅ .

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c) tự làm, đkxđ: x≠1;x≠−1

ê k bn với mk ik

😘 😘 😘 😘

a. \(ĐKXĐ:\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)

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

c. Để \(A=-2\) thì \(\dfrac{3}{x-1}=-2=\dfrac{3}{\dfrac{-3}{2}}\\ \Leftrightarrow x-1=\dfrac{-3}{2}\\ \Leftrightarrow x=\dfrac{-1}{2}\left(\text{t/m ĐKXĐ}\right)\)

Vậy \(x=\dfrac{-1}{2}\) để phân thức nhận giá trị là -2.

a) Có: \(x^2-1=\left(x-1\right)\left(x+1\right)\)

ĐKXĐ là x ≠ 1; x ≠ -1

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

c) Theo đề ta có: \(\dfrac{3}{x-1}=2\)

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

\(\Rightarrow x=\dfrac{5}{2}\) (T/m ĐK)

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

b) Ta có: \(\dfrac{3x+3}{x^2-1}\)

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

\(=\dfrac{3}{x-1}\)

c) Để phân thức có giá trị bằng -2 thì \(\dfrac{3}{x-1}=-2\)

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

hay \(x=\dfrac{-1}{2}\)(nhận)

Vậy: Để phân thức có giá trị bằng -2 thì \(x=\dfrac{-1}{2}\)

a) ĐKXĐ: x\(\ne\)0, x\(\ne\)2 

Ta có: 
 A= 2x-4/ x²- 2x = 2(x-2)/ x(x-2) = 2/x

Vậy...

b) Ta thấy x=26 thỏa mãn ĐKXĐ

Thay x=26 vào bt A ta được
   A= 2/26 = 1/13

Vậy....

c) Với x\(\ne\)0, x\(\ne\)2 ta có A=12 \(\Leftrightarrow\) 2/x =12 \(\Leftrightarrow\) x=1/6

Vậy....

Lời giải:

a. ĐKXĐ: $x^2-16\neq 0\Leftrightarrow (x-4)(x+4)\neq 0$

$\Leftrightarrow x\neq \pm 4$

b. $A=\frac{x^2+8x+16}{x^2-16}=\frac{(x+4)^2}{(x-4)(x+4)}=\frac{x+4}{x-4}$

c. $A=3\Leftrightarrow \frac{x+4}{x-4}=3$

$\Rightarrow x+4=3(x-4)$

$\Leftrightarrow -2x+16=0$

$\Leftrightarrow x=8$ (tm) 

d. 

$A=0\Leftrightarrow \frac{x+4}{x-4}=0\Leftrightarrow x+4=0\Leftrightarrow x=-4$

Mà theo ĐKXĐ thì $x\neq \pm 4$ nên không tồn tại $x$ để $A=0$

giú mới ạ mái em noppj rồi

a) Phân thức  xác định

⇔ x + 2 ≠ 0

⇔ x ≠ -2

Vậy điều kiện xác định của phân thức là x ≠ -2.

c) A = 1 ⇔ x + 2 = 1 ⇔ x = -1 ≠ -2 (Thỏa mãn điều kiện)

Vậy với x = -1 thì A = 1.

d) A = 0 ⇔ x + 2 = 0 ⇔ x = -2 (Không thỏa mãn điều kiện)

Vậy không có giá trị nào của x để A = 0.

\(P=\dfrac{3x^2+6x+3}{x+1}\)

\(a,\) Điều kiện xác định: \(x+1\ne0\Leftrightarrow x\ne-1\)

\(b,P=\dfrac{3x^2+6x+3}{x+1}=\dfrac{3\left(x^2+2x+1\right)}{x+1}=\dfrac{3\left(x+1\right)^2}{x+1}=3\left(x+1\right)=3x+3\)

\(c,x=1\Rightarrow P=3.1+3=6\)

giúpp mình vs

a) \(\dfrac{3x+3}{x^2-1}\)

\(ĐKXĐ:x\ne1\)
b) \(\dfrac{3x+3}{x^2-1}=\dfrac{3\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{3}{x-1}\)

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

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

\(a,ĐK:x^2-1=\left(x-1\right)\left(x+1\right)\ne0\Leftrightarrow x\ne\pm1\\ \dfrac{3x+3}{x^2-1}=\dfrac{3\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{3}{x-1}=2\\ \Leftrightarrow x-1=\dfrac{3}{2}\Leftrightarrow x=\dfrac{5}{2}\left(tm\right)\\ b,\dfrac{3}{x-1}\in Z\\ \Leftrightarrow x-1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\\ \Leftrightarrow x\in\left\{-2;0;2;4\right\}\left(tm\right)\)