Bài tập Phân tích đa thức thành nhân tử bằng cách đặt nhân tử chung (có lời giải chi tiết)

m(x + y + 1)

m(x + y + m)

m(x + y)

m(x + y – 1)

$y^5 – y^4 = y^4(y – 1)$

$y^5 – y^4 = y^3(y^2 – 1)$

$y^5 – y^4 = y^5(1 – y)$

$y^5 – y^4 = y^4(y + 1)$

$4x^3{y}^2 – 8x^2{y}^3 = 4x^{2}y(xy – 2y^2)$

$4x^3{y}^2 – 8x^2{y}^3 = 4x^2{y}^2(x – y)$

$4x^3{y}^2 – 8x^2{y}^3 = 4x^2{y}^2(x – 2y)$

$4x^3{y}^2 – 8x{{}^{2}y}^3 = 4x2y^2(x – 2y)$

${(x – 1)}^3 + 2{(x – 1)}^2 = {(x – 1)}^2(x + 1)$

$$\begin{gathered} {(x – 1)}^3 + 2(x – 1) \\ = (x – 1)[{(x – 1)}^2 + 2] \end{gathered}$$

$(x – 1{)}^3 + 2{(x – 1)}^2 = (x – 1\left)\right[{(x – 1)}^2 + 2x – 2]$

${(x – 1)}^3 + 2{(x – 1)}^2 = (x – 1)(x + 3)$

${(x – 2)}^2 – {(2 – x)}^3 = {(x – 2)}^2(x – 1)$

${(x – 2)}^2 – (2 – x) = (x – 2)(x – 1)$

${(x – 2)}^3 + {(2 – x)}^2 = {(x – 2)}^2(3 – x)$

${(x – 2)}^2 + x – 2 = (x – 2)(x – 1)$

$3{(x – 3y)}^2$

(x – 3y)(3x + 9y)

(x – 3y) + (3 – 9y)

(x – 3y) + (3x – 9y)

5x(x – y) – (y – x) = (x – y)(5x + 1)

5x(x – y) – (y – x) = 5x(x – y)

5x(x – y) – (y – x) = (x – y)(5x – 1)

5x(x – y) – (y – x) = (x + y)(5x – 1)

$3a^2 – b$

$3a^2+ 4b$

$3a^2 – 4b$

$3a^2 + b$

2a + b

1 + b

$a^2$ + ab

a + b

5 – 2x

5 + 2x

4x – 10

4x + 10

x + 2

3(x – 2)

${(x – 2)}^2$

${(x + 2)}^2$

$x=2; x=\frac{-1}{3}$

$x=-2; x=\frac{1}{3}$

x = 2; x = 3

$x=2; x=\frac{1}{3}$

$x=3; x=-\frac{1}{2}$

$x=-3; x=-\frac{1}{2}$

$x=3; x=\frac{1}{2}$

$x=-3; x=\frac{1}{2}$

1

2

3

0

1

2

3

0

$\frac{1}{2}$

-3

$\frac{-5}{2}$

$\frac{-7}{2}$

-4

4

6

-6

$x_0$ < 2

$x_0$ < 0

$x_0$ > 3

1 < $x_0$ < 5

$x_0$ < 1

$x_0$ = 0

$x_0$ > 3

1 < $x_0$ < 2

7xy + (xy – 3yz + z + 2)

7xy(xy – 21yz + z + 14)

7xy(xy – 3$y^2$z + z + 2)

7xy(xy – 3yz + z + 2)