Question

[Nguyễn Phú Tùng Dương]

cho biểu thức A

Answer 1

[Yen Nhi]

Answer:

Bài 3:

a. ĐKXĐ: \(x\ne\pm2;x\ne0\)

Khi đó \(A=\frac{x^4}{x^2-4}.\left(\frac{x+x}{2\left(x-2\right)}+\frac{2-3x}{x\left(x-2\right)}\right)\)

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

\(=\frac{x^3}{x^2-4}.\frac{x^2+2x+4-6x}{2\left(x-2\right)}\)

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

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

b. Vì \(\left|2x-1\right|=3\Rightarrow\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\text{(Loại)}\\x=-1\end{cases}}\) 

Với x = - 1 thì \(A=\frac{\left(-1\right)^3}{2\left(-1+2\right)}\)\(=-\frac{1}{2}\)

Answer 2

[Yen Nhi]

Answer:

Bài 4:

a) \(M=\frac{2x^3-6x^2+x-8}{x-3}\)

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

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

\(=2x^2+\frac{x-3-5}{x-3}\)

\(=2x^2+\frac{x-3}{x-3}-\frac{5}{x-3}\)

\(=2x^2+1-\frac{5}{x-3}\)

M nguyên khi \(\frac{5}{x-3}\) nguyên

\(\Leftrightarrow5⋮\left(x-3\right)\Leftrightarrow x-3\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

\(\Leftrightarrow x\in\left\{-2;2;4;8\right\}\) thì M nguyên.

b) \(N=\frac{3x^2-x+3}{3x+2}\)

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

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

\(=\frac{\left(3x+2\right)\left(x-1\right)+5}{3x+2}\)

\(=x-1+\frac{5}{3x+2}\)

\(\Rightarrow\frac{5}{3x+2}\) phải là số nguyên và x nguyên

\(\Rightarrow3x+2\inƯ\left(5\right)=\left\{\pm1;\pm5\right\}\)

\(\Rightarrow x\in\left\{\pm1\right\}\)