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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/ \(\left(x+3\right)^2-x\left(x-2\right)=13\)

\(\Leftrightarrow x^2+6x+9-x^2+2x=13\)

\(\Leftrightarrow8x=4\Rightarrow x=\frac{1}{2}\)

|4x + 3| - x = 15

=> |4x + 3| = 15 + x

=> \(\orbr{\begin{cases}4x+3=x+15\\4x+3=-x-15\end{cases}}\)

=> \(\orbr{\begin{cases}3x=12\\5x=-12\end{cases}}\)

=> \(\orbr{\begin{cases}x=4\\x=-\frac{12}{5}\end{cases}}\)

\(x=-3,6\)