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

\(=4x^2-9y^2-4x^2+4x-1+9y^2-6y+1=4x-6y\)

Thay x = 1 ; y = -1 ta được : 

\(4+6=10\)

(2x+3y)²+2(2x+3y)+1

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

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

(2x + 3y)\(^2\) + 2(2x + 3y) + 1

= (2x + 3y + 1)\(^2\)

AD HĐT : (a + b)\(^2\) = a\(^2\) + 2ab + b\(^2\)

\(\left(x+3y\right)^2-\left(2x-3y\right)^2-2x^2+12y^2\)

\(=x^2+2\cdot x\cdot3y+\left(3y\right)^2-\left[\left(2x\right)^2-2\cdot2x\cdot3y+\left(3y\right)^2\right]-2x^2+12y^2\)

\(=x^2+6xy+9y^2-4x^2+12xy-9y^2-2x^2+12y^2\)

\(=-5x^2+18xy+12y^2\)  

2x+\(\dfrac{1}{5}\) = 3y - \(\dfrac{2}{7}\) = 2x+3y -\(\dfrac{1}{6x}\) và 2x + 3y - z =50

có phải đề như này ko

bn viết rõ đề đi ạ:)

a) \(\left(2x-3\right)^2=4x^2-12x+9\)

\(b.\left(x-3y\right)^2=x^2-6xy+9y^2\)

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

\(=\left(4x^2-9y^2\right)-\left(4x^2+4xy+y^2\right)\)

\(=-10y^2-4xy\)

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

d) \(\left(x+3y^2\right)^2\)

\(=x^2+6xy^2+9y^4\)

Điều kiện \(x\ne\pm3;y\ne-2\):

 \(P=\frac{2x+3y}{xy+2x-3y-6}-\frac{6-xy}{xy+2x+3y+6}-\frac{x^2+9}{x^2-9}.\)

=> \(P=\frac{2x+3y}{\left(y+2\right)\left(x-3\right)}-\frac{6-xy}{\left(y+2\right)\left(x+3\right)}-\frac{x^2+9}{\left(x-3\right)\left(x+3\right)}\)

\(P=\frac{\left(2x+3y\right)\left(x+3\right)-\left(6-xy\right)\left(x-3\right)-\left(x^2+9\right)\left(y+2\right)}{\left(y+2\right)\left(x-3\right)\left(x+3\right)}\)

\(P=\frac{2x^2+3xy+6x+9y-6x+x^2y+18-3xy-x^2y-9y-2x^2-18}{\left(y+2\right)\left(x-3\right)\left(x+3\right)}\)

\(P=\frac{0}{\left(y+2\right)\left(x-3\right)\left(x+3\right)}=0\)

=> P=0 (với mọi x khác 3, -3 và y khác -2)

Câu 1: D

Câu 2: D