Ta có: \(\left(2y+2x+1\right)\left(2y-2x-1\right)=51\) 

<=> \(4y^2-\left(2x+1\right)^2=51\) 

<=> \(4y^2-\left(2x+1\right)^2=100-49\) 

=> \(\hept{\begin{cases}4y^2=100\\\left(2x+1\right)^2=49\end{cases}}\) => \(\hept{\begin{cases}2y=\pm10\\2x+1=\pm7\end{cases}}\) 

Đến đây bn tự giải tiếp nhé

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

a: M=3/4xy^2-2x^2y+2y^3-1/3x^2+1/2x^2y-5xy^2+x^3-y^3

=y^3-1/3x^2+x^3-17/4xy^2-3/2x^2y

1) ADTCDTSBN, ta có:

 \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)= \(\frac{2x^2+2y^2-3z^2}{18+32-75}=\frac{-100}{-25}\)= 4

* \(\frac{x}{3}=4\)=> x = 3 . 4 = 12

- \(\frac{y}{4}=4\)=> y = 4 . 4 = 16

* \(\frac{z}{5}=4\)=> z = 5 . 4 = 20

Vậy x = 12

       y = 16

       z = 20

x=12

y=16

z=20

      2x² + 2y² -2xy+2x+2y+2=0

<=>x²-2xy+y²+x²+2x+1+y²+2y+1=0

<=>(x-y)²+(x+1)²+(y+1)²=0

<=>x=-1;y=-1

còn x²⁰¹⁶+a chia hết cho x-1 khi a =-1.đúng chuẩn

sao lại bằng -1

Bài 1:
a) (2x - y) + (2x - y) + (2x - y) + 3y
= 3(2x - y) + 3y
= 3(2x - y + 3y)
= 3(2x + 2y)
= 3.2(x + y)
= 6(x + y)

b) (x + 2y) + (x - 2y) + (8x - 3y)
= x + 2y + x - 2y + 8x - 3y
= 9x - 3y
= 3(3x - y)

c) (x + 2y) - 2(x - 2y) - (2x - 3y)
= x + 2y - 2x + 4y - 2x + 3y
= 9y - 3x
= 3(3y - x)

Bài 2:
M + 2(x² - 4y²) + Q = 6x² - 4xy + 5y² + P
M + 2x² - 8y²   -3x² + 7xy - 2y² = 6x² - 4xy + 5y² + 9x² - 6xy + 3y²
M + 2x² - 3x² - 6x² - 9x² - 8y² - 2y² - 5y² - 3y² + 7xy + 4xy + 6xy = 0
M - 16x² - 18y² + 17xy = 0
M = 16x² + 18y² - 17xy