Đáp án:
1a)\(-3x^{2}+1\)
2)\(4(x-2)\)
3)4(x-2)(x-1)
4a)\(x^{2}-2x+4\)
b)3
5)a)\(x\neq±2, x\neq0\)
b)\(\frac{x-2}{4}\)
Giải thích các bước giải:
Bài 1:
\(\frac{-6x^{3}+3x^{2}+2x-1}{2x-1}=\frac{-(6x^{3}-3x^{2})+(2x-1)}{2x-1}=\frac{-3x^{2}(2x-1)+(2x-1)}{2x-1}\)
=\(\frac{(-3x^{2}+1)(2x-1)}{2x-1}=-3x^{2}+1\)
Bài 2:
\(\frac{4x^{2}-16}{x^2+2x}=\frac{A}{x}\)
⇒A=\(\frac{x(4x^{2}-16)}{x^{2}+2x}=\frac{4x(x-4)}{x(x+2)}\)
=\(\frac{4x(x-2)(x+2)}{x(x+2)}=4(x-2)\)
Bài 3:
\(x^{2}-4+3(x- 2)^{2}=(x-2)(x+2)+3(x-2)^{2}=(x-2)(x+2+3(x-2))=4(x-2)(x-1)\)
Bài 4:
a)\(\frac{2x^{3}-5x^{2}+10x-4}{2x-1}=x^{2}-2x+4\)
b)Ta có:\( x^{2}-2x+4=(x^{2}-2x+1)+3=(x-1)^{2}+3≥3\)
⇒ GTNN là 3
Bài 5:
a) đk: \(x\neq±2, x\neq0\)
b)A=\((\frac{x+2}{2x-4}+\frac{2-3x}{x^{3}-4x}\cdot \frac{x^{2}-4}{x-2}):\frac{2}{x}\)
=\((\frac{x+2}{2(x-2)}+\frac{2-3x}{x(x-2)(x+2)}\cdot \frac{(x-2)(x+2)}{x-2})\cdot \frac{x}{2}\)
=\((\frac{x+2}{2(x-2)}+\frac{2-3x}{x(x-2)})\cdot \frac{x}{2}=\frac{x(x+2)+2(2-3x)}{2x(x-2)}\cdot \frac{x}{2}\)
=\((\frac{x^{2}-4x+4}{2x(x-2)})\cdot \frac{x}{2}\)
=\(\frac{(x-2)^{2}}{2x(x-2)}\cdot \frac{x}{2}=\frac{x-2}{4}\)