a: ĐKXĐ: \(x\notin\left\{-3;2\right\}\)

b: \(A=\dfrac{x^2-4-5+x+3}{\left(x-2\right)\left(x+3\right)}=\dfrac{x^2+x-6}{\left(x-2\right)\left(x+3\right)}=\dfrac{x+2}{x-2}\)

c: Để A=3/4 thì 4x-8=3x+6

=>x=14

d: Để A nguyên thì \(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)

hay \(x\in\left\{3;1;4;0;6;-2\right\}\)

Câu 5:

a: Khi m=3 thì \(f\left(x\right)=\left(2\cdot3+1\right)x-3=7x-3\)

\(f\left(-3\right)=7\cdot\left(-3\right)-3=-21-3=-24\)

\(f\left(0\right)=7\cdot0-3=-3\)

b: Thay x=2 và y=3 vào f(x)=(2m+1)x-3, ta được:

\(2\left(2m+1\right)-3=3\)

=>2(2m+1)=6

=>2m+1=3

=>2m=2

=>m=1

c: Thay m=1 vào hàm số, ta được:

\(y=\left(2\cdot1+1\right)x-3=3x-3\)

*Vẽ đồ thị

d: Để hàm số y=(2m+1)x-3 là hàm số bậc nhất thì \(2m+1\ne0\)

=>\(2m\ne-1\)

=>\(m\ne-\dfrac{1}{2}\)

e: Để đồ thị hàm số y=(2m+1)x-3 song song với đường thẳng y=5x+1 thì \(\left\{{}\begin{matrix}2m+1=5\\-3\ne1\end{matrix}\right.\)

=>2m+1=5

=>2m=4

=>m=2

\(a)\) Ta có : 

\(M=\frac{2\left|x-3\right|}{x^2+2x-15}=\frac{2\left|x-3\right|}{\left(x^2+2x+1\right)-16}=\frac{2\left|x-3\right|}{\left(x+1\right)^2-16}=\frac{2\left|x-3\right|}{\left(x+1\right)^2-4^2}=\frac{2\left|x-3\right|}{\left(x+5\right)\left(x-3\right)}\)

+) Nếu \(x-3\ge0\) \(\Rightarrow\) \(x\ge3\) ta có : 

\(M=\frac{2\left|x-3\right|}{\left(x+5\right)\left(x-3\right)}=\frac{2\left(x-3\right)}{\left(x+5\right)\left(x-3\right)}=\frac{2}{x+5}\)

+) Nếu \(x-3< 0\)\(\Rightarrow\)\(x< 3\) ta có : 

\(M=\frac{2\left|x-3\right|}{\left(x+5\right)\left(x-3\right)}=\frac{-2\left(x-3\right)}{\left(x+5\right)\left(x-3\right)}=\frac{-2}{x+5}\)

Vậy : +) Nếu \(x\ge3\) thì \(M=\frac{2}{x+5}\) 

         +) Nếu \(x< 3\) thì \(M=\frac{-2}{x+5}\)

Chúc bạn học tốt ~ 

a: \(M=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)

a, ĐKXĐ: \(x\ne\pm3\)

\(A=\frac{x\left(x-3\right)+2x\left(x+3\right)-3x^2-12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)

\(=\frac{3x-12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}=\frac{3x-12}{3x+9}\)

b, \(x=-4\Rightarrow A=\frac{3.\left(-4\right)-12}{3.\left(-4\right)+9}=8\)

c, \(A\in Z\Rightarrow3x-12⋮\left(3x+9\right)\Rightarrow3x+9-21⋮\left(3x+9\right)\Rightarrow21⋮\left(3x+9\right)\)

\(\Rightarrow3x+9\inƯ\left(21\right)=\left\{\pm1;\pm3;\pm7;\pm21\right\}\)

Mà \(3x+9⋮3\Rightarrow3x+9\in\left\{-21;-3;3;21\right\}\Rightarrow x\in\left\{-10;-4;-2;4\right\}\) (thỏa mãn điều kiện)

a, ĐỂ A xác định : 

\(\Rightarrow\hept{\begin{cases}x+3\ne0\\x-3\ne0\\x^2-9\ne0\end{cases}}\Rightarrow x\ne\pm3.\)

\(A=\left(\frac{x}{x+3}+\frac{2x}{x-3}-\frac{3x^2+12}{\left(x+3\right)\left(x-3\right)}\right):\frac{3}{x-3}\)

\(A=\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3x^2+12}{\left(x-3\right)\left(x+3\right)}:\frac{3}{x-3}\)

\(A=\frac{x^2-3x+2x^2+6x-3x^2+12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)

\(A=\frac{3x+12}{\left(x-3\right)\left(x+3\right)}.\frac{x-3}{3}\)

\(A=\frac{x-4}{x+3}\)

b

a) \(A=\left(\frac{x}{x+3}+\frac{2x}{x-3}-\frac{3x^2+12}{x^2-9}\right):\frac{3}{x-3}\)

\(A=\left[\frac{x}{x+3}+\frac{2x}{x-3}-\frac{3x^2+12}{\left(x-3\right)\left(x+3\right)}\right]:\frac{3}{x-3}\)

A xác định \(\Leftrightarrow\hept{\begin{cases}x+3\ne0\\x-3\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne-3\\x\ne3\end{cases}}}\)

b) \(A=\left[\frac{x}{x+3}+\frac{2x}{x-3}-\frac{3x^2+12}{\left(x-3\right)\left(x+3\right)}\right]:\frac{3}{x-3}\)

\(A=\left[\frac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\frac{2x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{3x^2+12}{\left(x-3\right)\left(x+3\right)}\right]:\frac{3}{x-3}\)

\(A=\left[\frac{x^2-3x+2x^2+6x-3x^2-12}{\left(x+3\right)\left(x-3\right)}+\right]:\frac{3}{x-3}\)

\(A=\left[\frac{3x-12}{\left(x+3\right)\left(x-3\right)}\right].\frac{x-3}{3}\)

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

\(A=\frac{x-4}{x+3}\)

Với \(x=-4\)

\(\Rightarrow A=\frac{-4-4}{-4+3}=-\frac{8}{-1}=8\)

Vậy \(A=8\)tại \(x=-4\)

c) \(A=\frac{x-4}{x+3}=\frac{x+3-7}{x+3}=1-\frac{7}{x+3}\)

Có \(1\in Z\)

Để \(A\in Z\Rightarrow\frac{7}{x+3}\in Z\)

Có: \(x\in Z\Rightarrow x+3\in Z\Rightarrow\frac{7}{x+3}\in Z\Leftrightarrow\left(x+3\right)\in\text{Ư}\left(7\right)=\left\{\pm1;\pm7\right\}\)

b tự lập bảng nhé~