a,x^2- x # 0

hay x(x -1) # 0

<=> ĐKXĐ: x# 0

x -1 #0 => X# 1

bthay x= 0 vào A ta có

A= 2.0-1/ 0^2-0= -1

thay x= 3 vào A ta có

A= 2.3 -1/ 3^2-3 = 5/6

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c) tự làm, đkxđ: x≠1;x≠−1

ê k bn với mk ik

😘 😘 😘 😘

1.a)\(\frac{x^3}{x^2-4}-\frac{x}{x-2}-\frac{2}{x+2}\)

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

Để biểu thức được xác định thì:\(\left(x+2\right)\left(x-2\right)\ne0\)\(\Rightarrow x\ne\pm2\)

                                                      \(\left(x+2\right)\ne0\Rightarrow x\ne-2\)

                                                      \(\left(x-2\right)\ne0\Rightarrow x\ne2\)

                         Vậy để biểu thức xác định thì : \(x\ne\pm2\)

b) để C=0 thì ....

1, c , bn Nguyễn Hữu Triết chưa lm xong 

ta có : \(/x-5/=2\)

\(\Rightarrow\orbr{\begin{cases}x-5=2\\x-5=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=7\\x=3\end{cases}}\)

thay x = 7  vào biểu thứcC

\(\Rightarrow C=\frac{4.7^2\left(2-7\right)}{\left(7-3\right)\left(2+7\right)}=\frac{-988}{36}=\frac{-247}{9}\)KL :>...

thay x = 3 vào C 

\(\Rightarrow C=\frac{4.3^2\left(2-3\right)}{\left(3-3\right)\left(3+7\right)}\)

=> ko tìm đc giá trị C tại x = 3

chết mk nhìn nhầm phần c bài 2 :

\(2,\left(\frac{2+x}{2-x}+\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\frac{x^2-3x}{2x^2-x^3}\)

Để P xác định 

\(\Rightarrow2-x\ne0\Rightarrow x\ne2\)

\(2+x\ne0\Rightarrow x\ne-2\)

\(x^2-4\ne0\Rightarrow x\ne0\)

\(x^2-3x\ne0\Rightarrow x\ne3\)

b, \(P=\left(\frac{2+x}{2-x}+\frac{4x^2}{\left(2+x\right)\left(2-x\right)}+\frac{2-x}{2+x}\right):\frac{x\left(x-3\right)}{x^2\left(2-x\right)}\)

\(P=\left[\frac{4+4x+x^2}{\left(2-x\right)\left(2+x\right)}-\frac{4x^2}{\left(2+x\right)\left(2-x\right)}-\frac{4-4x+x^2}{\left(2+x\right)\left(2-x\right)}\right].\frac{x\left(2-x\right)}{x-3}\)

\(P=\left[\frac{8x-4x^2}{\left(2-x\right)\left(2+x\right)}\right].\frac{x\left(2-x\right)}{x-3}=\frac{4x\left(2-x\right)}{\left(2-x\right)\left(2+x\right)}.\frac{x\left(2-x\right)}{x-3}\)

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

d, ĐỂ \(p=\frac{8x^2-4x^3}{x^2-x-6}< 0\)

\(TH1:8x^2-4x^3< 0\)

\(\Rightarrow8x^2< 4x^3\)

\(\Rightarrow2< x\Rightarrow x>2\)

\(TH2:x^2-x-6< 0\Rightarrow x^2< x+6\)

giúp mik vs cần gấp ko làm vào vở nha

Đề sai rồi bạn

a, ĐỂ \(\frac{3x+3}{x^2-1}=\frac{3x+3}{\left(x+1\right)\left(x-1\right)}\)    Xác định 

\(\Rightarrow\left(x+1\right)\left(x-1\right)\ne0\)

\(\Rightarrow\hept{\begin{cases}x+1\ne0\\x-1\ne0\end{cases}\Rightarrow\hept{\begin{cases}x\ne-1\\x\ne1\end{cases}}}\)

KL : \(x\ne\pm1\)

b , 

\(\frac{3x+3}{x^2-1}\)xác định 

\(\Leftrightarrow x^2-1\ne0\Leftrightarrow x\ne\pm1\)

Vậy điều kiện xác định của \(\frac{3x+3}{x^2-1}\)là \(x\ne\pm1\)

\(\frac{3x+3}{x^2-1}=-2\)

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

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

\(\Leftrightarrow3=-2\left(x-1\right)\)

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

\(\Leftrightarrow x=\frac{-1}{2}\)

Vậy \(x=\frac{-1}{2}\)là giá trị cần tìm

phân thức được xác định ⇔ x² - 1 ≠ 0 ⇔ x ≠ \(\left\{-1;1\right\}\)

\(\dfrac{3x+3}{x^2-1}=-2\) 

=> 3x + 3 = -2x² + 2

=> 2x² + 3x + 1 = 0

=> (2x+1)(x+1) = 0

=> x = -1/2 (thỏa mãn) hoặc x = -1 (loại)

Vậy, để phân thức có giá trị bằng  –2 thì x = -1/2.

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

= \(\dfrac{3}{x-1}\)

=> Phân thức ban đầu có giá trị nguyên ⇔ 3 chia hết cho x-1

=> x-1 ∈\(\left\{-3;-1;1;3\right\}\)

=> x ∈\(\left\{-2;0;2;4\right\}\)

Vậy, để phân thức có giá trị là số nguyên.thì x ∈\(\left\{-2;0;2;4\right\}\).

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

b) Ta có: \(\dfrac{3x+3}{x^2-1}\)

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

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

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

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

hay \(x=-\dfrac{1}{2}\)(thỏa ĐK)

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

b) Ta có: \(\dfrac{3x+3}{x^2-1}\)

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

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

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

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

hay \(x=-\dfrac{1}{2}\)

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

c) Để phân thức có giá trị là số nguyên thì \(3⋮x-1\)

\(\Leftrightarrow x-1\inƯ\left(3\right)\)

\(\Leftrightarrow x-1\in\left\{1;-1;3;-3\right\}\)

\(\Leftrightarrow x\in\left\{2;0;4;-2\right\}\)

Kết hợp ĐKXĐ, ta được: 

\(x\in\left\{2;0;4;-2\right\}\)

Vậy: Để phân thức có giá trị là số nguyên thì \(x\in\left\{2;0;4;-2\right\}\)

\(\frac{3x^2+6x^2\left(xemlai\right)6x^2hay\left(6x\right)}{x^3+2x^2+x+2}=\frac{9x^2\left(culamtheode\right)}{x^3\left(x+2\right)+x+2=9}=\frac{9x^2}{\left(x^3+1\right)\left(x+2\right)}=\frac{9x^2}{\left(x+1\right)\left(x+2\right)\left(x^2-x+1\right)}\)

a)

\(x+1\ne0;x+2\ne0;x^2-x+1\ne0\)

\(x\ne-1;-2\)

b) khi 1<x<2 gia tri phan thuc <0 (-) xem lai

bangbangvp9 ha ha

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

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

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

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

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