Bài 6: Phân tích đa thức thành nhân tử bằng phương pháp đặt nhân tử chung

a. 2x²(x - 3) - 2x(3 - x) = 0

<=> 2x²(x - 3) + 2x(x - 3) = 0

<=> (2x² + 2x)(x - 3) = 0

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

<=> \(\left[{}\begin{matrix}2x=0\\x+1=0\\x-3=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=0\\x=-1\\x=3\end{matrix}\right.\)

b. 4x(x - 2) - x + 2 = 0

<=> 4x(x - 2) - (x - 2) = 0

<=> (4x - 1)(x - 2) = 0

<=> \(\left[{}\begin{matrix}4x-1=0\\x-2=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=2\end{matrix}\right.\)

giúp mình với ạ, cảm ơn rất nhiều<3

\(1,\\ a,=\left(x-2\right)\left(x-3\right)\\ b,=3\left(x^2+3x-10\right)=3\left(x-2\right)\left(x+5\right)\\ c,=\left(x-1\right)\left(x-2\right)\\ d,=\left(x-6\right)\left(x-3\right)\\ e,=\left(x^3-y^3\right)\left(x^3+y^3\right)=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\\ f,=\left(x+1\right)^3-27z^3=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\\ g,=\left(2x+1\right)^2-9y^2=\left(2x+1-3y\right)\left(2x+1+3y\right)\\ h,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\)

\(2,\\ a,\Rightarrow\left(x+2\right)\left(x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\\ b,\Rightarrow\left(x-8\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=8\end{matrix}\right.\\ c,\Rightarrow\left(x-3\right)\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\\ d,\Rightarrow\left(x+3\right)\left(2x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=-3\\x=-\dfrac{1}{2}\end{matrix}\right.\\ e,\Rightarrow4x^2-24x+36-4x^2+1=10\\ \Rightarrow-24x=-27\Rightarrow x=\dfrac{9}{8}\\ f,\Rightarrow25x^2+150x+225+1-25x^2=8\\ \Rightarrow150x=-218\Rightarrow x=-\dfrac{109}{75}\)

\(g,\Rightarrow9x^2+18x+9-9x^2+4=10\\ \Rightarrow18x=-3\Rightarrow x=-\dfrac{1}{6}\\ h,\Rightarrow-4x^2+8x-4+4x^2-1=3\\ \Rightarrow8x=8\Rightarrow x=1\\ 3,\\ a,=\left(xyz+xy\right)+\left(xz+yz\right)+\left(x+y\right)+\left(z+1\right)\\ =xy\left(z+1\right)+\left(z+1\right)+z\left(x+y\right)+\left(x+y\right)\\ =\left(xy+1\right)\left(z+1\right)+\left(x+y\right)\left(z+1\right)\\ =\left(xy+1+x+y\right)\left(z+1\right)\\ =\left[x\left(y+1\right)+\left(y+1\right)\right]\left(z+1\right)=\left(x+1\right)\left(y+1\right)\left(z+1\right)\)

\(b,ab\left(a+b\right)-bc\left(b+c\right)-ac\left(c-a\right)\\ =a^2b+ab^2-bc\left(b+c\right)-ac^2+a^2c\\ =\left(a^2b+a^2c\right)+\left(ab^2-ac^2\right)-bc\left(b+c\right)\\ =a^2\left(b+c\right)+a\left(b-c\right)\left(c+b\right)-bc\left(b+c\right)\\ =\left(a^2+ab-ac-bc\right)\left(b+c\right)\\ =\left(a+b\right)\left(a-c\right)\left(b+c\right)\)

\(3x^2-5x+2\\ \Leftrightarrow3x^2-2x-3x+2\\ \Leftrightarrow\left(3x^2-2x\right)-\left(3x-2\right)\\ \Leftrightarrow x\left(3x-2\right)-\left(3x-2\right)\\ \Leftrightarrow\left(3x-2\right)\left(x-1\right)\)

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

a,15.91,5 + 150.0,85

= 15.91,5 + 15.10.0,85

= 15.91,5 + 15.8,5

= 15.100

= 1500

b,x(x – 1) – y(1 – x)

= x(x – 1) – y[–(x – 1)]

= x(x – 1) + y(x – 1)

Tại x = 2001, y = 1999, giá trị biểu thức bằng:

(2001 – 1)(2001 + 1999) = 2000.4000 = 8000000

Trắng tinh tương luôn

Lỗi rồi kìa bn ơi

câu hỏi?

a)
(3x+1)^2 -9x (x-1) =46
9x²+6x+1-9x²+9x = 46
15x+1=46
15x=45
x=3
b)
5x (x-3) = 7 (3-x)
5x ( x - 3 ) - 7 ( 3 - x ) = 0
5x ( x - 3 ) + 7 ( x - 3 ) = 0
(5x + 7 ) ( x - 3 ) = 0
5x+7=0 hoặc x-3 = 0 
5x=-7    hoặc x=3
  x= -7 / 5 hoặc x=3

16a² - 1 + 2b - b²

= 16a² - (1 - 2b + b²)

= (4a)² - (1 - b)²

= (4a - 1 + b)(4a + 1 - b)

a) \(10x^2+10xy-x-y=10x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(10x-1\right)\)