Đề yêu cầu gì em?

\(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)

\(x^2-7x+12=\left(x-2\right)\left(x-5\right)\)

\(x^2+x-12=\left(x-5\right)\left(x+6\right)\)

\(x^2-9x+20=\left(x-4\right)\left(x-5\right)\)

\(2x^2-3x+2=2\left(x+\dfrac{1}{2}\right)\left(x-2\right)\)

\(4x^2-7x-2=4\left(x-2\right)\left(x+\dfrac{1}{4}\right)\)

\(4x^2+15x+9=4\left(x+\dfrac{3}{4}\right)\left(x+3\right)\)

`1)<=> -4x-3 + 5x+ 2 =0`

`<=> 5x-4x = -2+3`

`<=> x =1`

`2)<=> -5x +2-3x+6 =4`

`<=> -5x-3x = 4-6-2`

`<=> -8x=-4`

`<=> x=1/2`

`3) <=> -7x^2 +2 +7x^2 +14x =8`

`<=> 14x +2 =8`

`<=> 14x = 6`

`<=> x=3/7`

6x³ - 7x² + 5x - 2
= 6x³ - 4x² - 3x² + 2x + 3x - 2
= 6x²(x - 2/3) - 3x(x - 2/3) + 3(x - 2/3)
= (x - 2/3)(6x² - 3x + 3)
= 3(x - 2/3)(2x² - x + 1)

4x³ + 5x² + 10x - 12
= 4x³ - 3x² + 8x² - 6x + 16x - 12
= 4x²(x - 3/4) + 8x(x - 3/4) + 16(x - 3/4)
= (x - 3/4)(4x² + 8x + 16)
= 4(x - 3/4)(x² + 2x + 4)

4x³ - 7x² - x + 3
= 4x³ - 3x² - 4x² + 3x - 4x + 3
= 4x²(x - 3/4) - 4x(x - 3/4) - 4(x - 3/4)
= (x - 3/4)(4x² - 4x - 4)
= 4(x - 3/4)(x² - x - 1)

4x³ - 5x² + 6x + 9
= 4x³ + 3x² - 8x² - 6x + 12x + 9
= 4x²(x + 3/4) - 8x(x + 3/4) + 12(x + 3/4)
= (x + 3/4)(4x² - 8x + 12)
= 4(x + 3/4)(x² - 2x + 3)

3x³ - 5x² + 5x - 2
= 3x³ - 2x² - 3x² + 2x + 3x - 2
= 3x²(x - 2/3) - 3x(x - 2/3) + 3(x - 2/3)
= (x - 2/3)(3x² - 3x + 3)
= 3(x - 2/3)(x² - x + 1)

1) x² -7x + 10 = x² - 2x - 5x + 10 = x(x - 2) - 5(x - 2) = (x - 5)(x - 2)

2) x² + 3x + 2 = x² + 2x + x  + 2 = x(x + 2) + (x + 2) = (x + 1)(x + 2)

3) x² - 7x + 12 = x² - 3x - 4x + 12 = x(x - 3) - 4(x - 3) = (x - 3)(x - 4)

4) x² + 7x + 12 = x² + 3x + 4x + 12 = x(x + 3) + 4(x + 3) = (x + 3)(x + 4)

5) 16x - 5x² - 3 = 15x - 5x² + x - 3 = -5x(x - 3) + (x - 3) = (x - 3)(1 - 5x) 

6) 6x² + 7x - 3 = 6x² - 2x + 9x - 3 = 2x(3x - 1) + 3(3x - 1) = (2x + 3)(3x - 1)  

7) 3x² - 3x - 6 = 3x² - 6x + 3x - 6 = 3x(x - 2) + 3(x - 2) = (x - 2)(3x + 3) = 3(x - 2)(x + 1)

8) 3x² + 3x - 6 = 3x² - 3x + 6x - 6 = 3x(x - 1) + 6(x - 1) = (x - 1)(3x + 6) = 3(x - 1)(x + 2)

9) 6x² - 13x + 6 = 6x² - 9x -  4x + 6 = 3x(2x - 3) - 2(2x - 3) = (3x - 2)(2x - 3) 

10) 6x² + 15x  + 6 = 6x² + 12x + 3x + 6 = 6x(x + 2) + 3(x + 2) = (x + 2)(6x + 3) = 3(x + 2)(3x + 1)

11) 6x² - 20x + 6 = 6x² - 18x - 2x + 6 = 6x(x -3) - 2(x - 3) = (6x - 2)(x - 3) = 2(3x - 1)(x - 3)

12) 8x² + 5x - 3 = 8x² + 8x - 3x - 3 = 8x(x + 1) - 3(x + 1) = (x + 1)(8x - 3)

a) Ta có: \(x^2+3x-10=0\)

\(\Leftrightarrow x^2+5x-2x-10=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)

Vậy: S={-5;2}

b) Ta có: \(3x^2-7x+1=0\)

\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{1}{3}\right)=0\)

mà 3>0

nên \(x^2-\dfrac{7}{3}x+\dfrac{1}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}-\dfrac{37}{36}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=\dfrac{37}{36}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{7}{6}=\dfrac{\sqrt{37}}{6}\\x-\dfrac{7}{6}=-\dfrac{\sqrt{37}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{37}+7}{6}\\x=\dfrac{-\sqrt{37}+7}{6}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{\sqrt{37}+7}{6};\dfrac{-\sqrt{37}+7}{6}\right\}\)

c) Ta có: \(3x^2-7x+8=0\)

\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{8}{3}\right)=0\)

mà 3>0

nên \(x^2-\dfrac{7}{3}x+\dfrac{8}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}+\dfrac{47}{36}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=-\dfrac{47}{36}\)(vô lý)

Vậy: \(x\in\varnothing\)

ko bt

phân tích thành nhân tử ak?

\(a,2x^3+5x^2+5x+3\)

\(=2x^3+3x^2+2x^2+3x+2x+3\)

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

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

b) = 4x^3 - 3x^2 + 4x^2 - 3x + 4x - 3

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

= (4x - 3)(x^2 + x + 1)

c) = 5x^3 - 2x^2 - 10x^2 + 4x + 10x - 4

= x^2(5x - 2) - 2x(5x - 2) + 2(5x - 2)

= (5x - 2)(x^2 - 2x + 2)

d)= 6x^3 - 4x^2 - 3x^2 + 2x + 3x - 2

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

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

e) = 3x^3 - 2x^2 + 21x^2 - 14x + 18x - 12

= x^2( 3x - 2) + 7x(3x - 2) + 6(3x - 2)

= (3x - 2)(x^2 + 7x + 6)

= (3x - 2)(x+1)(x+6)

a) \(4x^2-7x+3=4x^2-4x-\left(3x-3\right)\)

\(=4x\left(x-1\right)-3\left(x-1\right)=\left(x-1\right)\left(4x-3\right)\)

Cho đa thức trên bằng 0 và tự tìm nghiệm:D

b)\(3x^2-7x+4=3x^2-3x-4x+4\)

\(=3x\left(x-1\right)-4\left(x-1\right)=\left(x-1\right)\left(3x-4\right)\)

Cho đa thức trên bằng 0 và tự tìm nghiệm:D

c) \(5x^2+7x+2=5x^2+5x+2x+2=5x\left(x+1\right)+2\left(x+1\right)=\left(x+1\right)\left(5x+2\right)\)

Cho đa thức trên bằng 0 và tự tìm nghiệm:D

d) \(6x^2-5x+1=6x^2-3x-2x+1=3x\left(2x-1\right)-\left(2x-1\right)=\left(2x-1\right)\left(3x-1\right)\)

Cho đa thức trên bằng 0 và tự tìm nghiệm:D

e) Tương tự

f)\(3x^2-6x-x+2=3x\left(x-2\right)-\left(x-2\right)=\left(3x-1\right)\left(x-2\right)\)

Cho đa thức trên bằng 0 và tự tìm nghiệm:D

a) \(4x^2-7x+3\)

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

\(=4x\left(x-1\right)-3\left(x-1\right)\)

\(=\left(4x-3\right)\left(x-1\right)\)

b) \(3x^2-7x+4\)

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

\(=3x\left(x-1\right)-4\left(x-1\right)\)

\(=\left(3x-4\right)\left(x-1\right)\)

c)\(5x^2+7x+2\)

\(=5x^2+5x+2x+2\)

\(=5x\left(x+1\right)+2\left(x+1\right)\)

\(=\left(5x+2\right)\left(x+1\right)\)

d) \(6x^2-5x+1\)

\(=6x^2-3x-2x+1\)

\(=3x\left(2x-1\right)-\left(2x-1\right)\)

\(=\left(3x-1\right)\left(2x-1\right)\)

e) \(12x^2-x-6\)

\(=12x^2-9x+8x-6\)

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

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

f) \(3x^2-7x+2\)

\(=3x^2-6x-x+2\)

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

\(=\left(3x-1\right)\left(x-2\right)\)