a, \(A=\dfrac{1}{4}x^4y^3z^2\)

b, bậc 9 

\(a,A=\left(\dfrac{-3}{8}x^2y\right)\left(\dfrac{2}{3}xy^2z^2\right)\left(\dfrac{4}{5}x^3y\right)\\ =\left(\dfrac{-3}{8}.\dfrac{2}{3}.\dfrac{4}{5}\right)\left(x^2.x.x^3\right)\left(y.y^2.y\right).z^2\\ =\dfrac{-1}{5}x^6y^4z^2\)

b, Hệ số: \(-\dfrac{1}{5}\)

Biến: \(x^6y^4z^2\)

c, Bậc: 12

d,Thay x=-1, y=-2, z=3 vào A ta có:
\(A=\dfrac{-1}{5}x^6y^4z^2=\dfrac{-1}{5}.\left(-1\right)^2.\left(-2\right)^4.3^2=\dfrac{-1}{5}.1.16.9=\dfrac{-144}{5}\)

a: A=-3/8x^2z*2/3xy^2z^2*4/5x^3y=-1/5x^6y^3z^3

b: Khi x=-1;y=-2;z=-3 thì -3/8x^2z=-3/8*(-1)^2*(-3)=9/8

2/3xy^2z^2=2/3*(-1)*(2*3)^2=-2/3*36=-24

4/5x^3y=4/5*(-1)^3*(-3)=12/5

A=-1/5*(-1)^6*(-2)^3*(-3)^3=-216/5

a) \(\left(-\dfrac{3}{8}x^2z\right).\left(\dfrac{2}{3}xy^2z^2\right).\dfrac{4}{5}x^3y=-\dfrac{1}{5}x^6y^3z^3\)

b) Gía trị đơn thức :

\(-\dfrac{1}{5}.\left(-1\right)^6\left(-2\right)^3.3^3=-\dfrac{1}{5}.1.\left(8\right).27=\dfrac{216}{5}\)

Bài tập `17`

`a,` ` @` Tớ nghĩ là tính tích ba đơn thức chứ nhỉ ?

\(-\dfrac{3}{8}x^2z.\dfrac{2}{3}xy^2z^2.\dfrac{4}{5}x^3y\\ =\left(-\dfrac{3}{8}.\dfrac{2}{3}.\dfrac{4}{5}\right)\left(x^2.x.x^3\right)\left(y^2.y\right)\left(z.z^2\right)\\ =-\dfrac{1}{5}x^6y^3z^3\)

`b,` Tại `x=-1 ; y=-2;z=-3`

Thì \(-\dfrac{3}{8}x^2z=-\dfrac{3}{8}.\left(-1\right)^2.\left(-3\right)=-\dfrac{3}{8}.1.\left(-3\right)=\dfrac{9}{8}\\ \dfrac{2}{3}xy^2z^2=\dfrac{2}{3}.\left(-1\right)\left(-2\right)^2\left(-3\right)^2=\dfrac{2}{3}.\left(-1\right).4.9=-24\\ \dfrac{4}{5}x^3y=\dfrac{4}{5}.\left(-1\right)^3.\left(-2\right)=\dfrac{4}{5}.\left(-1\right).\left(-2\right)=\dfrac{8}{5}\)

a: F=9/25x^2y^4*20/27x^3y=4/15x^5y^5

Bậc: 10

b: y=-x/3 và x+y=2

=>x+y=2 và -1/3x-y=0

=>x=3 và y=-1

Khi x=3 và y=-1 thì F=4/15*(-3)^5=-324/5

a: C=1/3*36x^4y^4*1/2x^3y=6x^7y^5

b: Khi x=1 và y=-1 thì C=-6

1.\(A=-\dfrac{3}{4}x^2yz;B=\dfrac{1}{3}xy^2;C=-\dfrac{8}{7}xy^2\)

\(A.\left(B+C\right)=-\dfrac{3}{4}x^2yz\left[\dfrac{1}{3}xy^2+\left(-\dfrac{8}{7}xy^2\right)\right]\)

\(=-\dfrac{3}{4}x^2yz\left(\dfrac{1}{3}xy^2-\dfrac{8}{7}xy^2\right)\)

\(=\left(-\dfrac{3}{4}x^2yz\right)\dfrac{1}{3}xy^2-\left(-\dfrac{3}{4}x^2yz\right)\dfrac{8}{7}xy^2\)

\(=-\dfrac{1}{4}x^3y^3z+\dfrac{6}{7}x^3y^3z\)

1. Ta có: \(-\dfrac{3}{4}x^2yz;B=\dfrac{1}{3}xy^2;C=-\dfrac{8}{7}xy^2\)

\(B+C=\dfrac{1}{3}xy^2-\dfrac{8}{7}xy^2=-\dfrac{17}{21}xy^2\)

\(A.\left(B+C\right)=\left(-\dfrac{3}{4}x^2yz\right).\left(-\dfrac{17}{21}xy^2\right)\)

\(\Rightarrow A.\left(B+C\right)=\dfrac{17}{28}x^3y^3z\)

2. a) Ta có: \(3x^3y^2=\dfrac{3}{5}x^3y^2+a.x^3y^2\)

\(\Rightarrow3=\dfrac{3}{5}+a\)

\(\Rightarrow a=3-\dfrac{3}{5}=\dfrac{12}{5}\)

Vậy: \(3x^3y^2=\dfrac{3}{5}x^3y^2+\dfrac{12}{5}x^3y^2\)

b) \(3x^3y^2=bx^3y^2-\left(-\dfrac{1}{4}x^3y^2\right)\)

\(\Rightarrow3=b+\dfrac{1}{4}\)

\(\Rightarrow b=3-\dfrac{1}{4}=\dfrac{11}{4}\)

Vậy: \(3x^3y^2=\dfrac{11}{4}x^3y^2-\left(-\dfrac{1}{4}x^3y^2\right)\)

c) Ta có: \(3x^3y^2=\left(-\dfrac{5}{7}x^2y\right).F\)

\(\Rightarrow F=\left(3x^3y^2\right):\left(-\dfrac{5}{7}x^2y\right)\)

\(\Rightarrow F=\left(3:\left(-\dfrac{5}{7}\right)\right)\dfrac{x^3y^2}{x^2y}\)

\(\Rightarrow F=-\dfrac{21}{5}xy\)

Vậy: \(3x^3y^2=\left(-\dfrac{5}{7}x^2y\right).\left(\dfrac{-21}{5}xy\right)\)

\(a,B=\left(-\dfrac{1}{2}xy^3\right)\left(2x^3y\right)^2\)

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

\(=-2x^7y^6\)

\(b,x=2,y=\dfrac{1}{2}\Rightarrow B=-2.2^7.\dfrac{1}{2}^6=-4\)

\(a,B=\left(-\dfrac{1}{2}xy^3\right).\left(2x^3y\right)^2\\ =\left(-\dfrac{1}{2}\right).4.x.x^6.y^3.y^2\\ =-2x^7y^5\)

b, Thay \(x=2;y=\dfrac{1}{2}\) vào B

\(B=\left(-2\right).2^7.\left(\dfrac{1}{2}\right)^5=-8\)

\(A=\dfrac{2}{5}x^7y^3\)

Hệ số: \(\dfrac{2}{5}\)

Bậc: 10

Bài tâpj `18`

`a, -3/2 x^3 y^2 z . (-6xy^3 z^5)`

\(\left[-\dfrac{3}{2}.\left(-6\right)\right]\left(x^3.x\right)\left(y^2.y^3\right)\left(z.z^5\right)\\ =9x^4y^5z^6\)

`b,` Hệ số : `9`

Phần biến : `x^4 y^5z^6`

Bậc : `15`