Question

Answer 1

\(a,ĐKXĐ:\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\\x^2-9\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\\x\ne\pm3\end{matrix}\right.\Leftrightarrow x\ne\pm3\)

\(b,M=\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}-\dfrac{48}{x^2-9}\)

\(\Rightarrow M=\dfrac{\left(x+3\right)^2-\left(x-3\right)^2-48}{\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow M=\dfrac{x^2+6x+9-x^2+6x-9-48}{\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow M=\dfrac{12x-48}{\left(x-3\right)\left(x+3\right)}\)

c,Thay x=0 vào M ta có:

\(M=\dfrac{12.0-48}{\left(0-3\right)\left(0+3\right)}=\dfrac{-48}{\left(-3\right).3}=\dfrac{-48}{-9}=\dfrac{16}{3}\)

Answer 2

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

b: \(M=\dfrac{x^2+6x+9-x^2+6x-9-48}{\left(x-3\right)\left(x+3\right)}=\dfrac{12x-48}{\left(x-3\right)\left(x+3\right)}\)

c: Thay x=0 vào M, ta được:

\(M=\dfrac{12\cdot0-48}{0^2-9}=\dfrac{-48}{-9}=\dfrac{16}{3}\)

Answer 3

Lời giải:
a. ĐKXĐ: 

\(\left\{\begin{matrix} x-3\neq 0\\ x+3\neq 0\\ x^2-9\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x-3\neq 0\\ x+3\neq 0\\ (x-3)(x+3)\neq 0\end{matrix}\right. \Leftrightarrow \left\{\begin{matrix} x-3\neq 0\\ x+3\neq 0\end{matrix}\right.\Leftrightarrow x\neq \pm 3\)

b.

\(M=\frac{(x+3)^2-(x-3)^2}{(x-3)(x+3)}-\frac{48}{(x-3)(x+3)}=\frac{12x-48}{(x-3)(x+3)}\)

c. Khi $x=0$ thì: $M=\frac{12.0-48}{(0-3)(0+3)}=\frac{-48}{-9}=\frac{16}{3}$

Answer 4

a)    M xác định ⇔ \(\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)

b) \(\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}-\dfrac{48}{x^2-9}\)

=  \(\dfrac{\left(x+3\right)\left(x-3\right)-\left(x-3\right)\left(x+3\right)-48}{x^2-9}\)

=  \(\dfrac{x^2-9-x^2+9-48}{x^2-9}\)

=  \(\dfrac{-48}{x^2-9}\)

c)  x = 0 ⇔ M = \(\dfrac{16}{3}\)