\(\dfrac{x-1}{x+3}=\dfrac{2}{10}\)

=>\(\dfrac{x-1}{x+3}=\dfrac{1}{5}\)

=>5(x-1)=x+3

=>5x-5=x+3

=>5x-x=3+5

=>4x=8

=>x=2

b) 15a² + 30ab + 15b² - 60c²

=15(a²+2ab+b²-4c²)

=15[(a²+2ab+b²)-4c²]

=15[(a+b)²-4c²]

=15(a+b-2c)(a+b+2c)

đặt chia như bt

các câu còn lại làm tương tự

a) x+5x²=0

=>x(1+5x)=0

=>x=0 hoặc 1+5x=0 =>x=\(\dfrac{1}{5}\)

điên ak

Do |2,5-x|\(\ge0\forall x\)

|x-3|\(\ge0\forall x\)

=>\(\left|2,5-x\right|+\left|x-3\right|=0\)

=>\(\left[{}\begin{matrix}\left|2,5-x\right|=0\\\left|x-3\right|=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}2,5-x=0\Rightarrow x=2,5\\x-3=0\Rightarrow x=3\end{matrix}\right.\)

vậy x=2,5 hoặc x=3

 \(8,5:0,5-7,6:0,5=\left(8,5-7,6\right)\times2=0,9\times2=1,8\)

2. x3+6x2+11x+6

=x³+x²+5x²+5x+6x+6

=(x³+x²)+(5x²+5x)+(6x+6)

=x²(x+1)+5x(x+1)+6(x+1)

=(x+1)(x²+5x+6)

=(x+1)(x²-x+6x+6)

=(x+1)(x²+2x+3x+6)

=(x+1)[(x²+2x)+(3x+6)]

=(x+1)[x(x+2)+3(x+2)

=(x+1)(x+2)(x+3)