\(a,PT\Leftrightarrow3x^2+3x-2x^2-4x=-1-x\Leftrightarrow x^2=-1\left(\text{vô nghiệm}\right)\)

Vậy: ...

\(b,PT\Leftrightarrow4x\left(x-2019\right)-\left(x-2019\right)=0\Leftrightarrow\left(x-2019\right)\left(4x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2019\\x=\dfrac{1}{4}\end{matrix}\right.\)

Vậy: ...

\(c,PT\Leftrightarrow\left(x-4-6\right)\left(x-4+6\right)=0\Leftrightarrow\left(x-10\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=10\\x=-2\end{matrix}\right.\)

Vậy: ...

\(d,PT\Leftrightarrow\left(x+4\right)^2=0\Leftrightarrow x=-4\)

Vậy: ...

\(e,PT\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)

Vậy: ...

\(f,PT\Leftrightarrow\left(5x-3\right)\left(5x+3\right)=0\Leftrightarrow x=\pm\dfrac{3}{5}\)

Vậy: ...

a) \(\left(x+2\right)^2=4\left(2x-1\right)^2\)

\(\left(x+2\right)^2-4\left(2x-1\right)^2=0\)

\(\left(x+2\right)^2-\left[2\left(2x-1\right)\right]^2=0\)

\(\left(x+2\right)^2-\left(4x-2\right)^2=0\)

\(\left(x+2-4x+2\right)\left(x+2+4x-2\right)=0\)

\(6x\left(-3x+4\right)=0\)

\(\Rightarrow6x=0\) hoặc \(-3x+4=0\)

*) \(6x=0\)

\(x=0\)

*) \(-3x+4=0\)

\(3x=4\)

\(x=\dfrac{4}{3}\)

Vậy \(x=0;x=\dfrac{4}{3}\)

b) \(4x\left(x-2019\right)-x+2019=0\)

\(4x\left(x-2019\right)-\left(x-2019\right)=0\)

\(\left(x-2019\right)\left(4x-1\right)=0\)

\(\Rightarrow x-2019=0\) hoặc \(4x-1=0\)

*) \(x-2019=0\)

\(x=2019\)

*) \(4x-1=0\)

\(4x=1\)

\(x=\dfrac{1}{4}\)

Vậy \(x=\dfrac{1}{4};x=2019\)

a/ \(4x\left(x-2019\right)-x+2019=0\)

\(\Leftrightarrow4x\left(x-2019\right)-\left(x-2019\right)=0\)

\(\Leftrightarrow\left(x-2019\right)\left(4x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2019=0\\4x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2019\\x=\frac{1}{4}\end{matrix}\right.\)

Vậy..

b/ \(3x\left(2x-3\right)=6-4x\)

\(\Leftrightarrow3x\left(2x-3\right)-2\left(2x-3\right)=0\)

\(\Leftrightarrow\left(2x-3\right)\left(3x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\3x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=\frac{2}{3}\end{matrix}\right.\)

Vậy..

a, x²(x - 3) + 12 - 4x = 0

<=> x²(x - 3) + 4(3 - x) = 0

<=> x²(x - 3) - 4(x - 3) = 0

<=> (x - 3)(x² - 4) = 0

<=> x - 3 = 0    hoặc   x² - 4 = 0

<=> x = 3                    x² = 4

<=> x = 3                    x = 2 hoặc x = -2

b, 2(x + 5) - x² - 5x = 0

<=>  2(x + 5) - x(x + 5) = 0

<=> (x + 5)(2 - x) = 0

<=> x + 5 = 0   hoặc 2 - x = 0

<=> x = -5                  x = 2

c, 2x(x + 2019) - x - 2019 = 0

<=> 2x(x + 2019) - (x + 2019) = 0

<=> (x + 2019)(2x - 1) = 0

<=> x + 2019 = 0  hoặc  2x - 1 = 0

<=> x = -2019                 2x = 1

<=> x = -2019                  x = 1/2

Mình sẽ chia các bài ra nhìu lần viết nhé

A,|x| + x = 0

   |x|. =-x    (bài toán vô nghĩa )

B, |x| - x =0

    |X| = x

=>  x € Z

| x -1| +2019 = 0

|x-1| = 0- 2019

|x-1| = -2019 ( bài toán vô nghĩa)

C

\(Th1:x-2019>0\)

\(x-2019-x+2019=0\)

\(0x=0\)

Vậy \(|x-2019|-x+2019=0\)với tất cả giá trị x 

\(th2:x-2019< 0\)

\(-x+2019-x+2019=0\)

\(\Rightarrow2x=4038\)

\(\Rightarrow x=2019\)