Câu 14:
a: \(x^3+10x^2+25x\)
\(=x\left(x^2+10x+25\right)\)
\(=x\left(x+5\right)^2\)
b: \(x^2-9y^2-8x+16\)
\(=\left(x^2-8x+16\right)-9y^2\)
\(=\left(x-4\right)^2-9y^2\)
\(=\left(x-4-3y\right)\left(x-4+3y\right)\)
Câu 15:
a: \(2x\left(x-3\right)-x+3=0\)
=>2x(x-3)-(x-3)=0
=>(x-3)(2x-1)=0
=>\(\left[{}\begin{matrix}x-3=0\\2x-1=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)
b: \(\left(3x-1\right)\left(2x+1\right)-\left(x+1\right)^2=5x^2\)
=>\(6x^2+3x-2x-1-x^2-2x-1=5x^2\)
=>\(5x^2-x-2=5x^2\)
=>-x-2=0
=>x+2=0
=>x=-2
Câu 14
a)x3+10x2+25x
=x(x2+10x+25)
=x(x+5)2
b)x2-9y2-8x+16
=(x2-8x+16)-9y2
=(x-4)2-9y2
=[(x-4)-3y][(x-4)+3y]
=(x-4-3y)(x-4+3y)
a)2x(x-3)-x+3=0
2x(x-3)-1(x-3)=0
(x-3)(2x-1)=0
TH1:x-3=0 TH2:2x-1=0
x=3 2x=1
x=1/2
Vậy x=3 hoặc x=1/2
