a: =>7-x=0

hay x=7

b: \(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\left(x+5\right)\left(3x-8\right)=0\)

hay \(x\in\left\{\sqrt{2};-\sqrt{2};-5;\dfrac{8}{3}\right\}\)

a: =>-x+7=0

hay x=7

b: \(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)\left(x+5\right)\left(3x-8\right)=0\)

hay \(x\in\left\{\sqrt{2};-\sqrt{2};-5;\dfrac{8}{3}\right\}\)

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

Do không có vế thứ 2 nên mình chỉ giải đc đến đây

ai nhanh mk tk cho

Suy ra 2x^2 - 6x = 1 

???????????????

3x² + 2x - 1 = 0

=> 3x² + 3x - x - 1 = 0

=> 3x(x + 1) - (x + 1) = 0

=> (3x - 1)(x + 1) = 0

=> \(\orbr{\begin{cases}3x-1=0\\x+1=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{1}{3}\\x=-1\end{cases}}\)

x² - 5x + 6 = 0

=> x² - 2x - 3x + 6 = 0

=> x(x - 2) - 3(x - 2) = 0

=> (x - 3)(x - 2) = 0

=> \(\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=3\\x=2\end{cases}}\)

3x² + 7x + 2 = 0

=> 3x² + 6x + x  + 2 = 0

=> 3x(x + 2) + (x + 2) = 0

=> (3x + 1)(x + 2) = 0

=> \(\orbr{\begin{cases}3x+1=0\\x+2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{1}{3}\\x=-2\end{cases}}\)

1, \(3x^2+2x-1=0\Leftrightarrow3x^2+3x-x-1=0\)

\(\Leftrightarrow3x\left(x+1\right)-\left(x+1\right)=0\)

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

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

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

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

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

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

3, \(3x^2+7x+2=0\Leftrightarrow3x^2+6x+x+2=0\)

\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)

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

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

x₁=1

x₂=\(\dfrac{-2}{7}\)

\(21x^3-15x^2-6x=0\\ \Leftrightarrow x\left(21x^2-15x-6\right)=0\\ \Leftrightarrow x\left[\left(21x^2-21x\right)+\left(6x-6\right)\right]=0\\ \Leftrightarrow x\left[21x\left(x-1\right)+6\left(x-1\right)\right]=0\\ \Leftrightarrow x\left(x-1\right)\left(21x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=\dfrac{-2}{7}\end{matrix}\right.\)

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

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

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

\(=\left(x+3\right)\left(2x+4\right)\)

\(=2\left(x+3\right)\left(x+2\right)\)

______

\(xy\left(x-y\right)-5x+5y\)

\(=xy\left(x-y\right)-\left(5x-5y\right)\)

\(=xy\left(x-y\right)-5\left(x-y\right)\)

\(=\left(x-y\right)\left(xy-5\right)\)

______

\(2x^2+3x-4xy-6y\)

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

\(=x\left(2x+3\right)-2y\left(2x+3\right)\)

\(=\left(x-2y\right)\left(2x+3\right)\)

2x² + 6x - 4(x + 3)

= (2x² + 6x) - 4(x + 3)

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

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

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

------------

xy(x - y) - 5x + 5y

= xy(x - y) - (5x - 5y)

= xy(x - y) - 5(x - y)

= (x - y)(xy - 5)

------------

2x² + 3x - 4xy - 6y

= (2x² - 4xy) + (3x - 6y)

= 2x(x - 2y) + 3(x - 2y)

= (x - 2y)(2x + 3)

ngắn gọn dễ hiểu nha

Ta có : 

\(x^2-6x-8=0\)

\(\Leftrightarrow\)\(\left(x^2-6x+9\right)-17=0\)

\(\Leftrightarrow\)\(\left(x^2-2.3x+3^2\right)-17=0\)

\(\Leftrightarrow\)\(\left(x-3\right)^2-17=0\)

\(\Leftrightarrow\)\(\left(x-3\right)^2=17\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-3=\sqrt{17}\\x-3=-\sqrt{17}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3+\sqrt{17}\\x=3-\sqrt{17}\end{cases}}}\)

Vậy \(x=3+\sqrt{17}\) hoặc \(x=3-\sqrt{17}\)

Chúc bạn học tốt ~ 

2x³ + 3x² + 6x + 5 = 0

⇔ 2x³ + 2x² + x² + x + 5x + 5 = 0

⇔ (2x³ + 2x²) + (x² + x) + (5x + 5) = 0

⇔ 2x²(x + 1) + x(x + 1) + 5(x + 1) = 0

⇔ (x + 1)(2x² + x + 5) = 0

⇔ (x + 1)[2(x² + 2.x.1/4 + 1/16) + 79/16] = 0

⇔ (x + 1)[(2(x + 1/4)² + 79/16] = 0

⇔ x + 1 = 0 (do 2(x + 1/4)² + 79/16 > 0 với mọi x)

⇔ x = -1

Vậy S = {-1}