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dễ mà , chỉ cần đổi trật tự một chút thì bạn có thể nhận ra đây là HĐT bình phương của 1 hiệu
\(A=\left(x+y\right)^2+\left(x-y\right)^2-2x-y\times x+y\)
\(A=\left(x+y\right)^2-2x-y\times x+y+\left(x-y\right)^2\)
\(A=\left(x+y-x-y\right)^2\)
\(2\left(x+y\right)\left(x-y\right)-\left(x-y\right)^2+\left(x+y\right)^2-4y^2\)
\(=\left[\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]-4y^2\)
\(=\left(x+y-x+y\right)^2-\left(2y\right)^2\)
\(=\left(2y\right)^2-\left(2y\right)^2=0\)
Sửa:
\(2\left(x+y\right)\left(x-y\right)-\left(x-y\right)^2+\left(x+y\right)^2-4y^2\)
\(=2\left(x^2-y^2\right)-\left(x^2-2xy+y^2\right)+\left(x^2+2xy+y^2\right)-4y^2\)
\(=2x^2-2y^2-x^2+2xy-y^2+x^2+2xy+y^2-4y^2\)
\(=2x^2-6y^2+4xy\)
\(=2\left(x^2-3y^2+2xy\right)\)
\(=2\left(x^2-3y^2+3xy-xy\right)\)
\(=2\left[x\left(x-y\right)+3y\left(x-y\right)\right]\)
\(=2\left(x-y\right)\left(3y+x\right)\)
x2(y - z) + y2(z - x) + z2(x - y)
= z2(x - y) + x2 y - x2 z + y2 z - y2 x
= z2(x - y) + (x2 y - y2 x) + (- x2 z + y2 z)
= (x - y)(z2 + xy - zx - zy)
= (x - y)[(z2 - zx) + (xy - zy)]
= (x - y)(z - x)(z -y)
\(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)
\(=x^2\left(y-z\right)+y^2\left(z-y+y-x\right)\)
\(=x^2\left(y-z\right)-y^2\left(y-z\right)-y^2\left(x-y\right)+z^2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(y-z\right)+\left(z-y\right)\left(y+z\right)\left(x-y\right)\)
\(=\left(x-y\right)\left(y-x\right)\left(x-z\right)\)
y(x−2z)2+8xyz+x(y−2z)2−2z(x+y)2
=y(x2−4xz+4z2)+8xyz+x(y2−4yz+4z2)−2z(x2+y2+2xy)
=(yx2+xy2)+(4yz2+4xz2)−(2zx2+2zy2+4xyz)
=xy(x+y)+4z2(x+y)−2z(x2+y2+2xy)
=xy(x+y)+4z2(x+y)−2z(x+y)2
=(x+y)(xy+4z2−2xz−2yz)
=(x+y)[y(x−2z)−2z(x−2z)]
=(x+y)(y−2z)(x−2z)
x^2-x-y^2-y
=(x^2-y^2)-(x+y)
=(x-y)(x+y)-(x+y)
=(x+y)(x-y-1)
tick nha
a) \(xy-y^2-x+y\)
\(=y\left(x-y\right)-\left(x-y\right)\)
\(=\left(x-y\right)\left(y-1\right)\)
b) \(xy-x^2-x+y\)
\(=x\left(y-x\right)+\left(y-x\right)\)
\(=\left(y-x\right)\left(x+1\right)\)
\(x^2-x-y^2-y=\left(x^2-y^2\right)-\left(x+y\right)=\left(x+y\right)\left(x-y\right)-\left(x+y\right)\)\(=\left(x+y\right)\left(x-y-1\right).\)
a) \(x^2-x-y^2-y\)
\(=\left(x-y\right).1\)
b) \(x^2-2xy+y^2-z^2\)
\(=\left(x-y\right)^2-x^2\)
\(=\left(x-y-z\right)\left(x-y+z\right)\)
Mik tl nhanh nhất đấy
\(-x-y^2+x^2-y\)
\(=\left(x^2-y^2\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)