Bài 2: Tính chất cơ bản của phân thức

\(VT:\dfrac{\left(2-3x\right)\left(2+3x\right)}{2\left(2-3x\right)}=\dfrac{2+3x}{2}\)

\(\Rightarrow VT=VP=\dfrac{2+3x}{2}\)

Phân thức nhận được bằng phân thức đã cho

a) \(a,\dfrac{x-3}{x+3}=\dfrac{\left(x-3\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{\left(x-3\right)^2}{x^2-9}\\ b,\dfrac{x+1}{x-1}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x-1\right)}=\dfrac{x^2-1}{\left(x-1\right)^2}\\ c,\dfrac{2x+1}{3x-2}=\dfrac{-\left(2x+1\right)}{-\left(3x-2\right)}=\dfrac{-2x-1}{2-3x}\\ d,-\dfrac{x-3}{1-2x}=\dfrac{x-3}{-\left(1-2x\right)}=\dfrac{x-3}{2x-1}\)

a: \(=\dfrac{x^2}{x-1}\)

1.\(x^2+4x+4=\left(x+2\right)^2\)

2.\(x^2-8x+16=\left(x-4\right)^2\)

3.\(\left(x+5\right)\left(x-5\right)=x^2-25\)

4.\(x^3+12x+48x+64=\left(x+4\right)^3\)

5.\(x^3-6x^2+12x-8=\left(x-2\right)^3\)

6.\(\left(x+2\right)\left(x^2-2x+4\right)=x^3+8\)

7.\(\left(x-3\right)\left(x^2+3x+9\right)=x^3-27\)

8.\(4x^4-15=\left(2x^2\right)^2-\left(\sqrt{15}\right)^2=\left(2x^2-\sqrt{15}\right)\left(2x^2+\sqrt{15}\right)\)

Câu 11: A

Câu 10: D

Câu 2: 

a: \(\Leftrightarrow3x^2+2x-1=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)

b: \(\Leftrightarrow x^3-4x-x^3-8=4\)

hay x=-3

\(=7x\left(x+2y\right)-4\left(x+2y\right)=\left(7x-4\right)\left(x+2y\right)\)

= 6x ( 2x - 1 )