Ta có :\(15x=10y=6z\Rightarrow\hept{\begin{cases}15x=10y\\10y=6z\end{cases}}\Rightarrow\hept{\begin{cases}3x=2y\\5y=3z\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{y}{3}=\frac{z}{5}\end{cases}}\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)

Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\Rightarrow\hept{\begin{cases}x=2k\\y=3k\\z=5k\end{cases}}\)

Khi đó 5x³ + 2y³ - z³ = 31

=> 5(2k)³ + 2(3k)³ - (5k)³ = 31

=> 40k³ + 54k³ - 125k³ = 31

=> -31k³ = 31

=> k³ = -1

=> k = -1

=> x = -2 ; y = -3 ; z = -5

b) Ta có 7x = 14y = 6z =>  \(\hept{\begin{cases}7x=14y\\14y=6z\end{cases}}\Rightarrow\hept{\begin{cases}x=2y\\7y=3z\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{2}=\frac{y}{1}\\\frac{y}{3}=\frac{z}{7}\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{6}=\frac{y}{3}\\\frac{y}{3}=\frac{z}{7}\end{cases}}\Rightarrow\frac{x}{6}=\frac{y}{3}=\frac{z}{7}\)

Đặt \(\frac{x}{6}=\frac{y}{3}=\frac{z}{7}=k\Rightarrow\hept{\begin{cases}x=6k\\y=3k\\z=7k\end{cases}}\)

Khi đó 2x² - 3y² = 5

<=> 2.(6k)² - 3.(3k)² = 5

=> 72k² - 27k² = 5

=> 45k² = 5

=> k² = 1/9

=> k = \(\pm\frac{1}{3}\)

Nếu k = 1/3 => x = 2 ; y = 1 ; z = 7/3

Nếu k = -1/3 => x = -2 ; y = - 1 ; z = -7/3

Vậy các cặp (x;y;z) thỏa mãn là : (2;1;7/3) ; (-2 ; - 1; -7/3)

c) Ta có : \(3x=8y=5z\Rightarrow\frac{3x}{120}=\frac{8y}{120}=\frac{5z}{120}\Rightarrow\frac{x}{40}=\frac{y}{15}=\frac{z}{24}\)

Đặt \(\frac{x}{40}=\frac{y}{15}=\frac{z}{24}=k\Rightarrow\hept{\begin{cases}x=40k\\y=15k\\z=24k\end{cases}}\)

Khi đó |x - 2y| = 5

<=> |40k - 2.15k| = 5

=>  |10k| = 5

=> \(\orbr{\begin{cases}10k=5\\10k=-5\end{cases}}\Rightarrow\orbr{\begin{cases}k=\frac{1}{2}\\k=-\frac{1}{2}\end{cases}}\)

Nếu k = 5 => x = 20 ; y = 7,5 ; z = 12

Nếu k = -5 => x = -20 ; y =-7,5 ; z = -12

d) 4x = 5y = 6z => \(\frac{4x}{60}=\frac{5y}{60}=\frac{6z}{60}\Rightarrow\frac{x}{15}=\frac{y}{12}=\frac{z}{10}\)

Đặt \(\frac{x}{15}=\frac{y}{12}=\frac{z}{10}=k\Rightarrow\hept{\begin{cases}x=15k\\y=12k\\z=10k\end{cases}}\)

Khi đó (3x - 2y)² = 16

<=> (3.15k - 2.12k)² = 16

=> (45k -24k)² = 16

=> (21k)² = 16

=> \(\orbr{\begin{cases}21k=4\\21k=-4\end{cases}}\Rightarrow\orbr{\begin{cases}k=\frac{4}{21}\\k=-\frac{4}{21}\end{cases}}\)

Nếu k = 4/21 => x = 20/7 ; y = 16/7 ; z = 40/21

Nếu k = -4/21 => x = -20/7 ; y = -16/7 ; z = -40/21

Ai có cách làm khác không 

a,A=3x^2y^4+5x^3+xy-3x^2y^4

   A=5x³ +xy

=> bậc của A là 3

b,B=7x^3y.(-4x^2y^2)+17x^2y^3-4x^2y+28x^2y^4

  => bậc của B là 8

c,C=5x^4y^2-7x^3y^2.(-2xy^2)-5x^4y^2+x^3-14x^4y^4

   C = 5x⁴y² -7x³y² (-2xy²) - 5x⁴y² +x³ -14x⁴y⁴ 

   C =  5x⁴y² + 14x⁴y⁴ -5x⁴y² +x³ -14x⁴y⁴ 

   C = x³ 

=> Bậc của C là 3

cám ơn

a: Ta có: \(x^2-4x\left(3x-4\right)+7x-5\)

\(=x^2-12x^2+16x+7x-5\)

\(=-11x^2+23x-5\)

b: Ta có: \(7x\left(x^2-5\right)-3x^2y\left(xy-6y^2\right)\)

\(=7x^3-35x-3x^3y^2+18x^2y^3\)

c: Ta có: \(\left(5x+4\right)\left(2x-7\right)\)

\(=10x^2-35x+8x-28\)

\(=10x^2-27x-28\)

a) \(\left(3x+y-z\right)-\left(4x-2y+6z\right)\)

\(=3x+y-z-4x+2y-6z\)

\(=-x+3y-7z\)

b) \(\left(x^3+6x^2+5y^3\right)-\left(2x^3-5x+7y^3\right)\)

\(=x^3+6x^2+5y^3-2x^3+5x-7y^3\)

\(=-x^3+6x^2+5x-2y^3\)

c) \(\left(5,7x^{2y}-3,1xy+8y^3\right)-\left(6,9xy-2,3x^{2y}-8y^3\right)\)

\(=5,7x^{2y}-3,1xy+8y^3-6,9xy+2,3x^{2y}+8y^3\)

\(=8x^{2y}-10xy+16y^3\)

6: =x^2-7xy+5xy-35y^2

=x(x-7y)+5y(x-7y)

=(x-7y)(x+5y)

7: =x^2-2xy-8xy+16y^2

=x(x-2y)-8y(x-2y)

=(x-2y)(x-8y)

8: =3x^2-6xy-4xy+8y^2

=3x(x-2y)-4y(x-2y)

=(x-2y)(3x-4y)

9: =4x^2+4xy+y^2-16y^2

=(2x+y)^2-16y^2

=(2x+y-4y)(2x+y+4y)

=(2x-3y)*(2x+5y)

10: =2(x^2+5xy+4y^2)

=2(x+y)(x+4y)

11: =5x(x+2y+y^2)