2x(3y-2)+(3y-2)
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\(\left(2x+3y\right)\left(2x-3y\right)-\left(2x-1\right)^2+\left(3y-1\right)^2\)
\(=4x^2-9y^2-4x^2+4x-1+9y^2-6y+1=4x-6y\)
Thay x = 1 ; y = -1 ta được :
\(4+6=10\)
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(2x+3y)2+2(2x+3y)+1
= (2x+3y)(2x+3y+2+1)
= (2x+3y)(2x+3y+3)
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\(\left(x+3y\right)^2-\left(2x-3y\right)^2-2x^2+12y^2\)
\(=x^2+2\cdot x\cdot3y+\left(3y\right)^2-\left[\left(2x\right)^2-2\cdot2x\cdot3y+\left(3y\right)^2\right]-2x^2+12y^2\)
\(=x^2+6xy+9y^2-4x^2+12xy-9y^2-2x^2+12y^2\)
\(=-5x^2+18xy+12y^2\)
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2x+\(\dfrac{1}{5}\) = 3y - \(\dfrac{2}{7}\) = 2x+3y -\(\dfrac{1}{6x}\) và 2x + 3y - z =50
có phải đề như này ko
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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) \(\left(x+3y^2\right)^2\)
\(=x^2+6xy^2+9y^4\)
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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)