Bài 2: 

a: =>4x(x+5)=0

=>x=0 hoặc x=-5

b: =>(x+3)(x-3)=0

=>x=-3 hoặc x=3

\(5-7x^2=\left(\sqrt{5}\right)^2-\left(x\sqrt{7}\right)^2\)

\(=\left(\sqrt{5}-x\sqrt{7}\right)\left(\sqrt{5}+x\sqrt{7}\right)\)

\(3+4x=\left(\sqrt{3}\right)^2-\left(2\sqrt{x}\right)^2\) ( do x<0 )

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

1/\(\left(x^2-25\right)^2-\left(x-5\right)^2\)

<=>\(\left[\left(x-5\right)\left(x+5\right)\right]^2-\left(x-5\right)^2\)

<=>\(\left(x-5\right)^2\left[\left(x+5\right)^2-1\right]\)

2/\(\left(4x^2-25\right)^2-9\left(2x-5\right)^2\)

<=>\(\left[\left(2x-5\right)\left(2x+5\right)\right]^2-9\left(2x-5\right)^2\)

<=>\(\left(2x-5\right)\left[\left(2x+5\right)^2-9\right]\)

 #hoctot<3# 

\(=\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)\)

\(=\left(2x-5\right)\left(2x+5-2x-7\right)\)

\(=-2\left(2x-5\right)\)

= ( 2x - 5 )(2x + 5) - ( 2x-  5 )(2x + 7 )

= (2x - 5 ) [ 2x+  5 - ( 2x+  7 ) ] 

= ( 2x- 5 ) ( 2x + 5 - 2x - 7 )

= -  2( 2x- 5 ) 

a: \(x^2+12x+36=0\) 

=>\(x^2+2\cdot x\cdot6+6^2=0\)

=>\(\left(x+6\right)^2=0\)

=>x+6=0

=>x=-6

b: \(4x^2-4x+1=0\)

=>\(\left(2x\right)^2-2\cdot2x\cdot1+1^2=0\)

=>\(\left(2x-1\right)^2=0\)

=>2x-1=0

=>2x=1

=>x=1/2

c: \(x^3+6x^2+12x+8=0\)

=>\(x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3=0\)

=>\(\left(x+2\right)^3=0\)

=>x+2=0

=>x=-2

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

\(9x^4-6x^2+1=\left(3x^2-1\right)^2\)

\(\dfrac{x^2}{9}-\dfrac{2}{3}x+1=\left(\dfrac{x}{3}+1\right)^2\)

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

\(x^2-4x+5y^2-10y+9=0\\ \Leftrightarrow\left(x^2-4x+4\right)+\left(5y^2-10y+5\right)=0\\ \Leftrightarrow\left(x-2\right)^2+5\left(y^2-2y+1\right)=0\\ \Leftrightarrow\left(x-2\right)^2+5\left(y-1\right)^2=0\)

Vì \(\left(x-2\right)^2\ge0;5\left(y-1\right)^2\ge0\) mà \(\left(x-2\right)^2+5\left(y-1\right)^2=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left(x-2\right)^2=0\\5\left(y-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)