a. x2 + 5x2 + (-3x2) = (1 + 5 – 3)x2 = 3x2

b. 5xy2 + 1/2 xy2 + 1/4 xy2 + (-1/2 )xy2 = (5 + 1/2 + 1/4 - 1/2 )xy2 = 21/4 xy2

c. 3x2y2z2 + x2y2z2 = (3 + 1) x2y2z2 = 4 x2y2z2

a. x2 + 5x2 + (-3x2) = (1 + 5 – 3)x2 = 3x2

b. 5xy2 + 1/2 xy2 + 1/4 xy2 + (-1/2 )xy2 = (5 + 1/2 + 1/4 - 1/2 )xy2 = 21/4 xy2

c. 3x2y2z2 + x2y2z2 = (3 + 1) x2y2z2 = 4 x2y2z2

Bài 4: 

b: \(=x^2z\left(-1+3-7\right)=-5x^2z=-5\cdot\left(-1\right)^2\cdot\left(-2\right)=10\)

c: \(=xy^2\left(5+0.5-3\right)=2.5xy^2=2.5\cdot2\cdot1^2=5\)

\(\left(-\dfrac{1}{4}x^2y^3\right)\left(\dfrac{6}{5}xy^2\right)\\ =\left(-\dfrac{1}{4}.\dfrac{6}{5}\right)\left(x^2.x\right)\left(y^3.y^2\right)\\ =-\dfrac{3}{10}x^3y^5\)

Hệ số: \(-\dfrac{3}{10}\)

Biến:x³y⁵

Bậc:8

5xy2 + 1/2 xy2 + 1/4 xy2 + (-1/2 )xy2 = (5 + 1/2 + 1/4 - 1/2 )xy2 = 21/4 xy2

thay x=1; y=-2 vào bt b ta có:

\(\Rightarrow3.1+5.1.\left(-2\right)^2=23\)

thay x=1; y=-2 vào bt b ta có:

Thay \(x=1;y=-2:3.1+5.1.\left(-2\right)^2=3+20=23.\)

\(\dfrac{1}{2}x^2y\left(2x^3-\dfrac{2}{5}xy^2-1\right)\)

\(=\dfrac{1}{2}x^2y\cdot2x^3-\dfrac{1}{2}x^2y\cdot\dfrac{2}{5}xy^2-\dfrac{1}{2}x^2y\)

\(=x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\)

\(\dfrac{1}{2}x^2y\left(2x^3-\dfrac{2}{5}xy^2-1\right)\)

\(=\dfrac{1}{2}x^2y\cdot2x^3-\dfrac{1}{2}x^2y\cdot\dfrac{2}{5}xy^2-\dfrac{1}{2}x^2y\cdot1\)

\(=x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\)

5 x y 2 + 9 x y - x 2 y 2 : - x y = 5 x y 2 - - x y + 9 x y : - x y + - x 2 y 2 : - x y = - 5 y - 9 + x y

(5x2y – 5xy2 + xy) + (xy – x2y2 + 5xy2)

= 5x2y – 5xy2 + xy + xy – x2y2 + 5xy2

= 5x2y + (5xy2 – 5xy2) + (xy + xy) – x2y2

= 5x2y + 2xy – x2y2

Ta có: P(x) - Q(x)

= (xy2z + 3x2y - 5xy2)-(x2y + 9xy2z - 5xy2 - 3)

= xy2z + 3x2y - 5xy2 - x2y-9xy2z + 5xy2 + 3

= -8xy2z + 2x2y + 3

Chọn D