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

A. m(x + y + 1)

B. m(x + y + m)

C. m(x + y)

D. m(x + y – 1)

A. $y^5 – y^4 = y^4(y – 1)$

B. $y^5 – y^4 = y^3(y^2 – 1)$

C. $y^5 – y^4 = y^5(1 – y)$

D. $y^5 – y^4 = y^4(y + 1)$

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

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

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

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

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

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

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

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

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

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

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

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

A. $3{(x – 3y)}^2$

B. (x – 3y)(3x + 9y)

C. (x – 3y) + (3 – 9y)

D. (x – 3y) + (3x – 9y)

A. 5x(x – y) – (y – x) = (x – y)(5x + 1)

B. 5x(x – y) – (y – x) = 5x(x – y)

C. 5x(x – y) – (y – x) = (x – y)(5x – 1)

D. 5x(x – y) – (y – x) = (x + y)(5x – 1)

A. $3a^2 – b$

B. $3a^2+ 4b$

C. $3a^2 – 4b$

D. $3a^2 + b$

A. 2a + b

B. 1 + b

C. $a^2$ + ab

D. a + b

A. 5 – 2x

B. 5 + 2x

C. 4x – 10

D. 4x + 10

A. x + 2

B. 3(x – 2)

C. ${(x – 2)}^2$

D. ${(x + 2)}^2$

A.  $x=2; x=\frac{-1}{3}$

B. $x=-2; x=\frac{1}{3}$

C. x = 2; x = 3

D. $x=2; x=\frac{1}{3}$

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

B. $x=-3; x=-\frac{1}{2}$

C.  $x=3; x=\frac{1}{2}$

D. $x=-3; x=\frac{1}{2}$

A. 1

B. 2

C. 3

D. 0

A. 1

B. 2

C. 3

D. 0

A. $\frac{1}{2}$

B. -3 

C. $\frac{-5}{2}$

D. $\frac{-7}{2}$

A. -4

B. 4

C. 6

D. -6

A. $x_0$ < 2

B. $x_0$ < 0

C. $x_0$ > 3

D. 1 < $x_0$ < 5

A. $x_0$ < 1

B. $x_0$ = 0

C. $x_0$ > 3

D. 1 < $x_0$ < 2

A. 7xy + (xy – 3yz + z + 2)

B. 7xy(xy – 21yz + z + 14)

C. 7xy(xy – 3$y^2$z + z + 2)

D. 7xy(xy – 3yz + z + 2)