Câu 3:
c: x(x+3)+2x+6=0
=>x(x+3)+2(x+3)=0
=>(x+3)(x+2)=0
=>\(\left[\begin{array}{l}x+3=0\\ x+2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=-3\\ x=-2\end{array}\right.\)
d: \(\left(x-3\right)^2-x\left(x-9\right)=0\)
=>\(x^2-6x+9-x^2+9x=0\)
=>3x+9=0
=>3x=-9
=>x=-3
Câu 2:
c; \(6x^3-12x^2-18x=6x\cdot x^2-6x\cdot2x-6x\cdot3\)
\(=6x\left(x^2-2x-3\right)\)
=6x(x-3)(x+1)
d: \(x^2+5x+xy+5y\)
=x(x+5)+y(x+5)
=(x+5)(x+y)
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