BÀi 3:
a: \(\left(2x-1\right)^2-\left(4x+1\right)\left(x-3\right)=0\)
=>\(4x^2-4x+1-\left(4x^2-12x+x-3\right)=0\)
=>\(4x^2-4x+1-\left(4x^2-11x-3\right)=0\)
=>\(4x^2-4x+1-4x^2+11x+3=0\)
=>7x+4=0
=>7x=-4
=>\(x=-\frac47\)
b: 5x(x-3)-2x+6=0
=>5x(x-3)-2(x-3)=0
=>(x-3)(5x-2)=0
=>\(\left[\begin{array}{l}x-3=0\\ 5x-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=3\\ x=\frac25\end{array}\right.\)
Bài 2:
a: \(2x^3-50x\)
\(=2x\cdot x^2-2x\cdot25\)
\(=2x\left(x^2-25\right)=2x\left(x-5\right)\left(x+5\right)\)
b: \(x^2-y^2+2x-2y\)
=(x-y)(x+y)+2(x-y)
=(x-y)(x+y+2)
c: \(5x^3-10x^2+5x\)
\(=5x\cdot x^2-5x\cdot2x+5x\cdot1\)
\(=5x\left(x^2-2x+1\right)=5x\left(x-1\right)^2\)
Bài 1:
a: \(\left(3x+2\right)\left(2-x\right)-\left(2x-3\right)^2+7x^2\)
\(=6x-3x^2+4-2x-\left(4x^2-12x+9\right)+7x^2\)
\(=4x^2+4x+4-4x^2+12x-9=16x-5\)
b: \(\left(15x^4y^5-30x^3y^4+5x^5y^4\right):\left(5x^3y^3\right)\)
\(=\frac{15x^4y^5}{5x^3y^3}-\frac{30x^3y^4}{5x^3y^3}+\frac{5x^5y^4}{5x^3y^3}=3xy^2-6y+x^2y\)
c: \(\frac{x+3}{x-3}-\frac{x-3}{x+3}+\frac{x^2-9x}{x^2-9}\)
\(=\frac{\left(x+3\right)^2-\left(x-3\right)^2+x^2-9x}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x^2+6x+9-\left(x^2-6x+9\right)+x^2-9x}{\left(x-3\right)\left(x+3\right)}=\frac{2x^2-3x+9-x^2+6x-9}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x^2+3x}{\left(x-3\right)\left(x+3\right)}=\frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{x}{x-3}\)
