đề bài là j ?? ? và quế bênh phải là j ?? ?

a) \(2x^2-8x\Leftrightarrow2x\left(x-4\right)\)

b) \(2x^2-4x+2\Leftrightarrow2\left(x^2-2x+1\right)=2\left(x-1\right)^2\)

c) \(3x^3+12x^2+12x\Leftrightarrow3x\left(x^2+4x+4\right)=3x\left(x+2\right)^2\)

Đề bài là:phân tích các đa thực thành nhân tử

[9x³(x² - 1) - 6x²(x² - 1) + 12x(x² - 1)] : 3x(x² - 1)

= [9x³(x² - 1) : 3x(x² - 1)] - [6x²(x² - 1) : 3x(x² - 1) + [12x(x² - 1) : 3x(x² - 1)]

= 3x² - 2x + 4

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

\(\Rightarrow2\left(x-2\right)-x\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(2-x\right)=0\)

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

a)2

b)\(\frac{-11}{4}\)

a, <=> (x-1).(x-6) = 0

<=> x=1 hoặc x=6

b, <=> (x+1).(2x-5) = 0

<=> x=-1 hoặc x=5/2

c, <=> (2x-5).(2x-1) = 0

<=> x=5/2 hoặc x=1/2

d, <=> (x^2-x+1).(x^2+1) = 0

=> pt vô nghiệm vì x^2-x+1 và x^2+1 đều > 0

Tk mk nha

a) x² - 7x + 6 = 0

<=> x² - 6x - x + 6 = 0

<=>( x - 6 ) ( x - 1 ) = 0

<=> x - 6 = 0 hoặc x - 1 = 0

1. x - 6 = 0

<=> x = 6

2. x - 1 = 0

<=> x = 1

Vậy ......

b) 2x² - 3x - 5 = 0

<=> 2x² + 2x - 5x - 5 = 0

<=> ( x + 1 ) ( 2x - 5 ) = 0

<=> x + 1 = 0 hoặc 2x - 5 = 0

1. x + 1 = 0

<=> x = -1

2. 2x - 5 = 0

<=> x = 2.5

Vậy ............

c) 4x² - 12x + 5 = 0

<=> 4x² - 2x - 10x + 5 = 0

<=> 2x ( 2x - 1 ) - 5( 2x - 1 ) = 0

<=> ( 2x - 1 ) ( 2x - 5 ) = 0

<=> 2x - 1 = 0 hoặc 2x - 5 = 0

1. 2x - 1 = 0

<=> x = 0.5

2. 2x - 5 = 0

<=> x = 2.5

Vậy ....................

d) x⁴ - x³ + 2x² - x + 1 = 0

a) \(x^2-6x-x+6=0\)

\(x\left(x-6\right)-\left(x-6\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-6=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=6\end{cases}}\)

b) \(2x^2+2x-5x-5=0\)

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

\(\Rightarrow\orbr{\begin{cases}2x-5=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-1\end{cases}}\)

c) \(4x^2-12x+5=0\)

\(4x^2-2x-10x+5=0\)

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

\(\Rightarrow\orbr{\begin{cases}2x-5=0\\2x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{1}{2}\end{cases}}\)

\(\left(3x^2-2x+1\right)\left(x^2+2x+3\right)-4x\left(x^2-1\right)-3x^2\left(x^2+2\right)=3x^4+6x^3+9x^2-3x^3-4x^2-6x-4x^3+4x-3x^4-6x^2=0\)

a/ 4x²+x-4x-1

x(4x+1)-(4x+1)

(4x+1)(x-1)

b/(6-11)x²+3

-5x²+3

c/x²-3xy-4xy+12y²

x(x-3y)-4y(x-3y)

(x-3y)(x-4y)

d/(x-y)²+3(x-y)

(x-y+3)(x-y)

e/(2-12)x²+17x-2

-10x²+17x-2

g/x³+x²+2x²+2x+4x+4

x²(x+1)+2x(x+1)+4(x+1)

(x+1)(x²+2x+4)

h/x³+2x²+7x²+14x+12x+24

x²(x+2)+7x(x+2)+12(x+2)

(x+2)(x²+7x+12)

(x+2)(x²+4x+3x+12)

(x+2)(x+4)(x+3)

Giải:

a) 4x² - 3x - 1 = 4x² - 4x + x - 1 = 4x(x - 1) + (x -1) = (x - 1)(4x +1)

b) 6x² - 11x + 3 = 6x² - 2x - 9x + 3 = 2x(3x - 1) - 3(3x - 1) = (3x - 1)(2x - 3)

c) x² - 7xy + 12y² = x² - 6xy + 9y² - xy +3y² = (x - 3y)² - y(x - 3y) = (x - 3y)( x - 3y - y) = (x - 3y)(x - 4y)

d) x² - 2xy + y² + 3x - 3y = (x - y)² + 3(x - y) = (x - y)(x - y + 3)

e)Sửa đề: x² → x³
2x³ - 12x² + 17x - 2 = 2x³ - 4x² - 8x² + 16x + x - 2 = (2x²- 8x + 1)(x -2)

f) x³ - 3x + 2 = x³ - x - 2x + 2 = x(x + 1)(x - 1) - 2(x - 1) = (x - 1)(x² + x - 2) = (x - 1)²(x + 2)

g) x³ + 3x² + 6x + 4 = x³ + 3x² + 3x + 1 + 3x + 3 = (x +1)³ + (x + 1) = (x + 1)(x² + 2x + 4 )

h) x³ + 9x² + 26x + 24 = x³ + 4x² + 5x² + 20x + 6x + 24 = (x + 4)(x² + 5x + 6) = (x + 4)(x + 3)(x + 2)ư

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