Vậy x + y =27.

Chọn đáp án D.

2x² - 9y² = 14

thay 2x = 9y va ngược lại 

ta có

9xy - 2xy = 14

7xy = 14

xy = 2

=> x = 2/y

2x = 9y

=> 4/y = 9y

=>(3y)² = 4 = 2²

=> y = 2/3

x= 3

1) (x+3)(x²- 3x + 9) = x³ + 27
2) (x² + 2y)² = x⁴ + 4xy + 4y² 
3) (2x-3)(2x+3) = 4x² - 9
4) (x + 3y)³ = x³ + 9x²y + 9xy² + y³
5) (2x²- y)³ = 8x⁶ - 6x⁴y + 6x²y² - y³
6) (x-3y)(x² + 3xy +9y²)= x³- 27y³
7) (2x + 3y)(4x² - 6xy +9y²)= 8x³ + 27y³
8) (3x - y²)²= 9x² - 6xy² + y⁴

Sửa đề bài: \(2^x=8^{y+1}\)và  \(9^y=3^{x-9}\)

Có: \(2^x=8^{y+1}\)

\(\Leftrightarrow2^x=\left(2^3\right)^{y+1}\)

\(\Leftrightarrow2^x=2^{3y+3}\)

\(\Leftrightarrow x=3y+3\)   (1)

Lại có: \(9^y=3^{x-9}\)

\(\Leftrightarrow\left(3^2\right)^y=3^{x-9}\)

\(\Leftrightarrow3^{2y}=3^{x-9}\)

\(\Leftrightarrow2y=x-9\)    (2)

Thay (1) vào (2), ta có:

=> 2y = 3y + 3  - 9

=> 2y = 3y - 6

=> 2y - 3y = -6

=> -1y = -6

=> y = 6 \(\left(y\in N\right)\)

Từ x = 3y + 3 (theo điều 1)

=> x = 3.6 + 3 = 21 \(\left(x\in N\right)\)

Vậy x + y = 21 + 6 = 27

Bạn huy sai rồi::::2x chứ ko phải 2^x

1) \(3x\left(x-1\right)+5\left(x-1\right)\)

\(=\left(x-1\right)\left(3x+5\right)\)

2) \(4x(x-2y)-8y(2y-x)\)

\(=4x\left(x-2y\right)+8y\left(x-2y\right)\)

\(=\left(4x+8y\right)\left(x-2y\right)\)

\(=4\left(x+2y\right)\left(x-2y\right)\)

3) \(a^2\left(x-1\right)+b^2\left(1-x\right)\)

\(=a^2\left(x-1\right)-b^2\left(x-1\right)\)

\(=\left(a^2-b^2\right)\left(x-1\right)\)

\(=\left(a-b\right)\left(a+b\right)\left(x-1\right)\)

4) \(3x\left(x-a\right)+4a\left(a-x\right)\)

\(=3x\left(x-a\right)-4a\left(x-a\right)\)

\(=\left(x-a\right)\left(3x-4a\right)\)

5) \(5x\left(x-y\right)^2+10y^2\left(y-x\right)^2\)

\(=5x\left(x-y\right)^2+10y^2\left(x-y\right)^2\)

\(=\left(5x+10y^2\right)\left(x-y\right)^2\)

\(=5\left(x+2y^2\right)\left(x-y\right)^2\)

6) \(3x\left(x-3\right)^2+9\left(3-x\right)^2\)

\(=3x\left(x-3\right)^2+9\left(x-3\right)^2\)

\(=\left(3x+9\right)\left(x-3\right)^2\)

\(=3\left(x+3\right)\left(x-3\right)^2\)

7) \(x\left(m-a\right)^2-y\left(a-m\right)^2\)

\(=x\left(a-m\right)^2-y\left(a-m\right)^2\)

\(=\left(x-y\right)\left(a-m\right)^2\)

8) \(6y^2\left(x-1\right)^2+9y\left(1-x\right)^2\)

\(=6y^2\left(x-1\right)^2+9y\left(x-1\right)^2\)

\(=\left(6y^2+9x\right)\left(x-1\right)^2\)

\(=3\left(2y^2+3x\right)\left(x-1\right)^2\)

#Ayumu

khiếp,ít ít thôi, t giải phụ chứ nhìn lóa mắt quá