a) 3x(x - 1) + 7x2(x - 1) = 0
<=> x(x - 1)(3 + 7x) = 0
<=> x = 0
hoặc : x - 1 = 0
hoặc 3 + 7x = 0
<=> x = 0
hoặc x = 1
hoặc x = -3/7
b) x2 - 2018x - 2019 = 0
<=> x2 - 2019x + x - 2019 = 0
<=> x(x - 2019) + (x - 2019) = 0
<=> (x + 1)(x - 2019) = 0
<=> \(\orbr{\begin{cases}x+1=0\\x-2019=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-1\\x=2019\end{cases}}\)
c) (x + 3)2 - x(x - 2) = 13
<=> x2 + 6x + 9 - x2 + 2x = 13
<=> 8x = 13 - 9
<=> 8x = 6
<=> x= 6/8 = 3/4
a/\(3x\left(x-1\right)+7x^2\left(x-1\right)=0.\)
\(\Leftrightarrow\left(x-1\right)\left(3x+7x^2\right)=0\)
\(\Leftrightarrow\left(x-1\right)x\left(3+7x\right)=0\)
Th1: x - 1 = 0
=> x = 1
Th2: x= 0
Th3: 3 + 7x = 0
=> x= -3/7
\(\Rightarrow x\in\left\{1;0;-\frac{3}{7}\right\}\)
b/ \(x^2-2018x-2019=0\)
\(\Leftrightarrow x^2+x-2019x-2019=0\)
\(\Leftrightarrow\left(x^2+x\right)-\left(2019x+2019\right)=0\)
\(\Leftrightarrow x\left(x+1\right)-2019\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-2019\right)\left(x+1\right)=0\)
Th1 : x -2019 = 0
=> x =2019
Th2: x + 1 =0
=> x = -1
\(\Rightarrow x\in\left\{2019;-1\right\}\)
c/ \(\left(x+3\right)^2-x\left(x-2\right)=13\)
\(\Leftrightarrow x^2+6x+9-x^2+2x=13\)
\(\Leftrightarrow8x=4\Rightarrow x=\frac{1}{2}\)
\(3x\left(x-1\right)+7x^2\left(x-1\right)=0\)
\(\Rightarrow\left(3x+7x^2\right)\left(x-1\right)=0\)
\(x\left(3+7x\right)\left(x-1\right)=0\)
\(\hept{\begin{cases}x=0\\3+7x=0\\x-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x=0\\x=\frac{-3}{7}\\x=1\end{cases}}\)
\(x\in\left\{0;\frac{-3}{7};1\right\}\)
b) \(x^2-2018x-2019=0\)
\(=x^2+x-2019x-2019=0\)
\(=x\left(x+1\right)-2019\left(x+1\right)=0\)
\(=\left(x-2019\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\x-2019=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=2019\end{cases}}\)
\(x\in\left\{-1;2019\right\}\)
c) \(\left(x+3\right)^2-x\left(x-2\right)=13\)
\(\Rightarrow x^2+6x+9-x^2+2x-13=0\)
\(\Rightarrow8x-4=0\)
\(\Rightarrow4\left(2x-1\right)=0\)
\(\Rightarrow2x-1=0\Rightarrow x=\frac{1}{2}\)
