Question

[Nguyễn Quốc Đại]

1.Phân tách các đa thức sau thành nhân tử:

a)x²-ax+bx-ab

b)x²+ay-y²-ax

c)x³-3x²-4x+12

2.Tìm x biết:

a)2x²-12x=-18

b)(4x²-4x+1)-x²=0

Answer 1

[headsot96]

1.a)\(x^2-ax+bx-ab=x\left(x-a\right)+b\left(x-a\right)=\left(x+b\right)\left(x-a\right)\)

b)\(x^2+ay-y^2-ax=\left(x-y\right)\left(x+y\right)-a\left(x-y\right)=\left(x+y-a\right)\left(x-y\right)\)

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

2.a)\(2x^2-12x=-18=>2x^2-12x+18=0=>x^2-6x+9=0=>\left(x-3\right)^2=0=>x-3=0=>x=3\)b)\(\left(4x^2-4x+1\right)-x^2=0=>3x^2-3x-x+1=3x\left(x-1\right)-\left(x-1\right)=\left(3x-1\right)\left(x-1\right)=0\)

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

Answer 2

[Lê Trung Hiếu]

a) 2x² - 12x = -18

<=> 2x² - 12x + 18 = 0

<=> 2(x² - 6x + 9) = 0

<=> 2(x² - 2.x.3 + 9) = 0

<=> 2(x - 3)² = 0

<=> x - 3 = 0

<=> x = 0 + 3

<=> x = 3

b) (4x² - 4x + 1) - x² = 0

<=> 4x² - 4x + 1 - x² = 0 

<=> 3x² - 4x + 1 = 0

<=> 3x² - x - 3x + 1 = 0

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

<=> \(\orbr{\begin{cases}\left(3x-1\right)=0\\\left(x-1\right)=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=\frac{1}{3}\\x=1\end{cases}}\)

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