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

A. $x^2$ (x + 12)

B. x( $x^2$ + 12)

C. x( $x^2$ – 12)

D. $x^2$ (x – 12)

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

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)