\(\left(-4x+2y\right)\left(-4x-2y\right)+\left(x-5y\right)^2-\left(3x+2y\right)^2-7x\left(x-3y\right)\)

\(=\left(16x^2-4y^2\right)+\left(x^2-10xy+25y^2\right)-\left(9x^2+12xy+4y^2\right)-7x^2+21xy\)

\(=16x^2-4y^2+x^2-10xy+25y^2-9x^2-12xy-4y^2-7x^2+21xy\)

\(=x^2+17y^2-xy\)

\(=\left(4x-2y\right)\left(4x+2y\right)+x^2-10xy+25y^2-9x^2-12xy-4y^2-7x^2+21xy\)

\(=16x^2-4y^2-15x^2-xy+21y^2\)

\(=x^2-xy+17y^2\)

a) Xem lại đề

b) x³ - 4x²y + 4xy² - 9x

= x(x² - 4xy + 4y² - 9)

= x[(x² - 4xy + 4y² - 3²]

= x[(x - 2y)² - 3²]

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

c) x³ - y³ + x - y

= (x³ - y³) + (x - y)

= (x - y)(x² + xy + y²) + (x - y)

= (x - y)(x² + xy + y² + 1)

d) 4x² - 4xy + 2x - y + y²

= (4x² - 4xy + y²) + (2x - y)

= (2x - y)² + (2x - y)

= (2x - y)(2x - y + 1)

e) 9x² - 3x + 2y - 4y²

= (9x² - 4y²) - (3x - 2y)

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

= (3x - 2y)(3x + 2y - 1)

f) 3x² - 6xy + 3y² - 5x + 5y

= (3x² - 6xy + 3y²) - (5x - 5y)

= 3(x² - 2xy + y²) - 5(x - y)

= 3(x - y)² - 5(x - y)

= (x - y)[(3(x - y) - 5]

= (x - y)(3x - 3y - 5)

a: =>A-B=3x^2y-4xy^2+x^2y-2xy^2=4x^2y-6xy^2

b: =>B-A=-7xy^2+8x^2y-5xy^2+6x^2y=-12xy^2+14x^2y

=>A-B=12xy^2-14x^2y

c: =>B-A=8x^2y^3-4x^3y-3x^2y^3+5x^3y^2=5x^2y^3+x^3y^2

=>A-B=-5x^2y^3-x^3y^2

d: =>A-B=2x^2y^3-7x^3y+6x^2y^3+3x^3y^2=8x^2y^3-7x^3y+3x^3y^2

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