a) Ta có: \(A=\left(-\dfrac{3}{4}x^2y^5z^3\right)\cdot\left(\dfrac{5}{3}x^3y^4z^2\right)\)

\(=\left(\dfrac{-3}{4}\cdot\dfrac{5}{3}\right)\cdot\left(x^2\cdot x^3\right)\cdot\left(y^5\cdot y^4\right)\cdot\left(z^3\cdot z^2\right)\)

\(=\dfrac{-5}{4}x^5y^9z^5\)

thay x = -1 , y = -1 , z = -1 vào N ta có

N = 1 + (-1) + 1 + ... + 1 + (-1)

= [1 + (-1)] + [1 + (-1) ] + ... + [1 + (-1)]

= 0 + 0 + ... + 0

= 0

thay x = -1 , y = -1 , z = -1 và N ta có

N = 1 + (-1) + 1 + ... + 1 + (-1)

= [1 + (-1)] + [1 + (-1) ] + ... + [1 + (-1)]

= 0 + 0 + ... + 0

= 0

=(-1).(-1)².(-1)^3+(-1)^2.(-1)^3.(-1)^4+(-1)^3.(-1)^4.(-1)^5+...+(-1)^2014.(-1)^2015.(-1)^2016

=(-1).1.(-1)+1.(-1).1+(-1).1.(-1)+...+1.(-1).1

=1+(-1)+1+...+(-1)

=0+0+..+0= 0

`#3107.101117`

a)

`x \div y \div z = 4 \div 3 \div 9`

`=> x/4 = y/3 = z/9`

`=> x/4 = (3y)/9 = (4z)/36`

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

`x/4 = (3y)/9 = (2z)/8 = (x - 3y + 4z)/(4 - 9 + 36) = 62/31 = 2`

`=> x/4 = y/3 = z/9 = 2`

`=> x = 4*2 = 8` $\\$ `y = 3*2 = 6` $\\$ `z = 9*2 = 18`

Vậy, `x = 8; y = 6; z = 18`

c)

\(x \div y \div z = 1 \div 2 \div 3\)

`=> x/1 = y/2 = z/3`

`=> (4x)/4 = (3y)/6 = (2z)/6`

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

`(4x)/4 = (3y)/6 = (2z)/6 = (4x - 3y + 2z)/(4 - 6 + 6) = 36/4 = 9`

`=> x/1 = y/2 = z/3 = 9`

`=> x = 1*9=9` $\\$ `y = 2*9 = 18` $\\$ `z = 3*9 = 27`

Vậy, `x = 9; y = 18; z = 27`

Các câu còn lại cậu làm tương tự nhé.

Đặt x/3=y/4=z/5=k

=>x=3k; y=4k; z=5k

\(\dfrac{4x-3y}{2016}=\dfrac{4\cdot3k-3\cdot4k}{2016}=0\)

\(\dfrac{5y-4z}{2017}=\dfrac{5\cdot4k-4\cdot5k}{2017}=0\)

\(\dfrac{3z-5x}{2018}=\dfrac{3\cdot5k-5\cdot3k}{2018}=0\)

=>\(\dfrac{4x-3y}{2016}=\dfrac{5y-4z}{2017}=\dfrac{3z-5x}{2018}\)

3x²y²z² = x³y³ y³z³ z³x³ 
(3x²y²z²) / (x³y³ y³z³ z³x³) = 1
3.[(x²y²z²) / (x³y³ y³z³ z³x³)] = 1
(x²y²z²) / (x³y³ y³z³ z³x³) = 1/3
(x²y²z²) / (x³y³) (x²y²z²) / (y³z³) (x²y²z²) / (z³x³) = 1/3
z²/(xy) x/(yz) y²/(zx) = 1/3
Vậy x²/(yz) y²/(xz) z²/(xy) = 1/3

TH1: a+b+c  khác 0

\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)

\(\Rightarrow2+\frac{a+b-c}{c}=2+\frac{b+c-a}{a}=2+\frac{c+a-b}{b}\)

\(\Rightarrow\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)

\(\Rightarrow a=b=c\)

thay a=b=c vào B ta có:

\(B=\left(1+\frac{a}{a}\right)\cdot\left(1+\frac{a}{a}\right)\cdot\left(1+\frac{a}{a}\right)=2\cdot2\cdot2=8\)

TH2: a+b+c=0

=> c=-a-b

=>a=-b-c

=>b=-a-c

thay a,b,c vào B ta có:

\(B=\left(1+\frac{-\left(a+c\right)}{a}\right)\cdot\left(1+\frac{-\left(b+c\right)}{c}\right)\cdot\left(1+\frac{-\left(a+b\right)}{b}\right)\)

\(B=\left(-\frac{c}{a}\right)\cdot\left(-\frac{b}{c}\right)\cdot\left(-\frac{a}{b}\right)=-1\)

p/s: th2 ko chắc nhá