Bài giải còn nhiều thiếu sót.Mong bạn thông cảm.

\(x^4+2x^3-4x^2-5x-6=0\)

\(\Leftrightarrow\left(x+3\right)\left(\frac{x^4+2x^3-4x^2-5x-6}{x+3}\right)=0\)

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

\(\Leftrightarrow\left(x+3\right)\left[\left(x-2\right)\left(\frac{x^3-x^2-x-2}{x-2}\right)\right]=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-2\right)\left(x^2+x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}}\) hoặc \(x^2+x+1=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=2\end{cases}}\) hoặc \(x^2+x+1=0\)

Ta sẽ c/m \(x^2+x+1=0\) vô nghiệm.Thật vậy:

\(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)

Mà \(\frac{3}{4}>0\Rightarrow x^2+x+1>0\Rightarrow\)vô nghiệm.

Vậy x = {-3;2}

\(\left(x^4+x^3-6x^2\right)+\left(x^3+x^2-6x\right)+\left(x^2+x-6\right)=0\)

\(\Leftrightarrow x^2\left(x^2+x-6\right)+x\left(x^2+x-6\right)+\left(x^2+x-6\right)=0\)

\(\Leftrightarrow\left(x^2+x+1\right)\left(x^2+x-6\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+3\right)\left(x^2+x+1\right)=0\)

Sửa đề :

(x - 2)² - 16 = 0

=> (x - 2)² = 16

=> (x - 2)² = (\(\pm\)4)²

=> \(\orbr{\begin{cases}x-2=4\\x-2=-4\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=-2\end{cases}}\)

(x - 2)² - x² + 4 = 0

=> x² - 4x + 4 - x² + 4 = 0

=> (x² - x²) - 4x + (4 + 4) = 0

=> -4x + 8 = 0

=> -4x = -8

=> x = 2

(2x + 3)² - (\(\frac{1}{3}-2x\))² = -2/3x + 5

=> \(\left(2x+3\right)\left(2x+3\right)-\left(\frac{1}{3}-2x\right)\left(\frac{1}{3}-2x\right)=-\frac{2}{3}x+5\)

=> \(2x\left(2x+3\right)+3\left(2x+3\right)-\frac{1}{3}\left(\frac{1}{3}-2x\right)+2x\left(\frac{1}{3}-2x\right)=-\frac{2}{3}x+5\)

=> \(4x^2+6x+6x+9-\frac{1}{9}+\frac{2}{3}x+\frac{2}{3}x-4x^2=-\frac{2}{3}x+5\)

=> \(\left(4x^2-4x^2\right)+\left(6x+6x+\frac{2}{3}x+\frac{2}{3}x\right)+\left(9-\frac{1}{9}\right)+\frac{2}{3}x-5=0\)

=> \(\frac{40}{3}x+\frac{80}{9}+\frac{2}{3}x-5=0\)

=> \(\frac{40}{3}x+\frac{2}{3}x+\frac{80}{9}-5=0\)

=> 14x + 80/9 - 5 = 0

=> x = -5/18

Giải tiêu biểu câu a nhé.

a/ \(5x\left(2x-7\right)+2x\left(8-5x\right)=5\)

\(\Leftrightarrow19x+5=0\)

\(\Leftrightarrow x=-\frac{5}{19}\)

cần câu mấy

Câu a:

(x + 2)(x -3) - (x -2)(x + 5) = 0

x^2 + 2x - 3x - 6 - x^2 - 5x + 2x + 10 = 0

(x^2 - x^2) + (2x - 3x - 5x + 2x) + (10 - 6) = 0

0 + (-x - 5x + 2x) + 4 = 0

-6x + 2x + 4 = 0

-4x + 4 = 0

4x = 4

x = 4 : 4

x = 1

Vậy x = 1

Câu b:

(2x + 3)(x - 4) + (x - 5)(x - 2) = (3x - 5)(x - 4)

2x^2 - 8x + 3x - 12+ x^2 - 2x - 5x + 10 = 3x^2 - 12x - 5x + 20

(2x^2 + x^2) - (8x+5x-3x +2x) - (12 - 10) = 3x^2 - (12x + 5x) + 20

3x^2 - (13x - 3x + 2x) - 2 = 3x^2 - 17x + 20

3x^2 - (10x + 2x) - 2 = 3x^2 - 17x + 20

3x^2 - 12x - 2 = 3x^2 - 17x + 20

3x^2 - 12x - 2 - 3x^2 + 17x - 20 = 0

(3x^2 - 3x^2) + (-12x + 17x) - (2 + 20) = 0

0 + 5x - 22 = 0

5x = 22

x = 22/5

Vậy x = 22/5

\(5X\left(X-2020\right)+X=2020\)

\(\Leftrightarrow5X^2-10100X+X=2020\)

\(\Leftrightarrow5X^2-10099X=2020\)

\(\Leftrightarrow5X^2-10099X-2020=0\)

\(\Leftrightarrow5X^2-10100X+x-2020=0\)

\(\Leftrightarrow5X\left(X-2020\right)+X-2020=0\)

\(\Leftrightarrow\left(X-2020\right)\left(5X+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=-\frac{1}{5}\end{cases}}\)

\(4\left(x-5\right)^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left[2\left(x-5\right)\right]^2-\left(2x+1\right)^2=0\)

\(\Leftrightarrow\left[2\left(x-5\right)-2x-1\right]\left[2\left(x-5\right)+2x+1\right]=0\)

\(\Leftrightarrow\left(2x-10-2x-1\right)\left(2x-10+2x+1\right)=0\)

\(\Leftrightarrow-11\left(4x-9\right)=0\)

\(\Leftrightarrow x=\frac{9}{4}\)