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(x+y+z)2 - x2-y2-z2 = x2+y2+z2+2xy+2yz+2zx -x2-y2-z2= 2(xy+yz+zx)
a) xy(x + y) + yz(y + z) + xz(z + x) + 3xyz
= xy(X + y + z) + yz(x + y + z) + xz(X + y + z)
= (x + y +z)(xy + yz+ xz)
b) xy(x + y) - yz(y + z) - xz(z - x)
= x2y + xy2 - y2z - yz2 - xz2 + x2z
= x2(y + z) - yz(y + z) + x(y2 - z2)
= x2(y + z) - yz(y + z) + x(y + z)(y - z)
= (y + z)(x2 - yz + xy - xz)
= (y + z)[x(x + y) - z(x + y)]
= (y + z)(x + y)(x - z)
c) x(y2 - z2) + y(z2 - x2) + z(x2 - y2)
= x(y - z)(y + z) + yz2 - yx2 + x2z - y2z
= x(y - z)(y + z) - yz(y - z) - x2(y - z)
= (y - z)((xy + xz - yz - x2)
= (y - z)[x(y - x) - z(y - x)]
= (y - z)(x - z)(y -x)
\(\left(x+y\right)\left(x^2-y^2\right)+\left(y+z\right)\left(y^2-z^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)
\(=-y^3-xy^2+x^2y+x^3-z^3-yz^2+y^2z+y^3-x^3-zx^2+z^2x+z^3\)
\(=-xy^2+x^2y-yz^2+y^2z-zx^2+z^2x\)
\(=\left(x-y\right)\left(z-x\right)\left(z-y\right)\)
\(x\left(y^2-z^2\right)+y\left(z^2-x^2\right)+z\left(x^2-y^2\right)\)
\(=xy^2-xz^2+yz^2-ỹx^2+zx^2-zy^2\)
\(=\left(xy^2-yx^2\right)+\left(-xz^2+yz^2\right)+\left(zx^2-zy^2\right)\)
\(=-xy\left(x-y\right)-z^2\left(x-y\right)+z\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(-xy-z^2+zx+zy\right)\)
\(=\left(x-y\right)\left[\left(-xy+zx\right)-\left(z^2-zy\right)\right]\)
\(=\left(x-y\right)\left[-x\left(y-z\right)+z\left(y-z\right)\right]\)
\(=\left(x-y\right)\left(y-z\right)\left(z-x\right)\)
Làm như vầy là sai hướng rồi.
Tham khảo :
\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y+z\right)-x\right]\left[\left(x+y+z\right)^2+x^2+x\left(x+y+z\right)\right]-\left(y+z\right)\left(y^2+z^2-yz\right)\)
\(=\left(y+z\right)\left[x^2+y^2+z^2+2\left(xy+yz+xz\right)+x^2+x^2+xy+yz+xz\right]-\left(y+z\right)\left(y^2+z^2-yz\right)\)
\(=\Rightarrow\left(y+z\right)\left[x^2+y^2+z^2+2\left(xy+yz+xz\right)+x^2+x^2+xy+yz+xz-y^2-z^2+yz\right]\)
\(=\left(y+z\right)\left[3x^2+3xy+3yz+3xz\right]\)
\(=3\left(y+z\right)\left[\left(x^2+xy\right)+\left(yz+xz\right)\right]\)
\(=3\left(y+z\right)\left[x\left(x+y\right)+z\left(x+y\right)\right]\)
\(=3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
\(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)
\(=x^2\left(y-z\right)+y^2z-xy^2+xz^2-yz^2\)
\(=x^2\left(y-z\right)+yz\left(y-z\right)-x\left(y^2-z^2\right)\)
\(=\left(y-z\right)\left(x^2+yz-xy-xz\right)\)
\(=\left(y-z\right)\left[x\left(x-y\right)-z\left(x-y\right)\right]\)
\(=\left(y-z\right)\left(x-y\right)\left(x-z\right)\)
\(x^2\left\{\left(y-z-x\right)+x\right\}+y^2\left\{\left(z-x-y\right)+y\right\}+z^2\left\{\left(x-y-z\right)+z\right\}.\)
đặt ( x-y-z) = p ta được
\(x^2p-x^3+y^2p-y^3+z^2p-z^3.\)
\(P\left(x^2+y^2+z^2\right)-\left(x^3+y^3+z^3\right)\)
\(P\left(x^2+y^2+z^2\right)-P.\frac{1}{p}\left(x^3+y^3+z^3\right)\)
\(P\left(x^2+y^2+z^2+\frac{1}{p}\left(x^3+y^3+z^3\right)\right)\)
p/s đúng 100% :))
khi những con gà chép mạng :)
éo C/M được bài của t sai nhưng vẫn tích sai :)
best :)
(bí mật làm nên sự quyến rũ của phụ nữ)
Của Giang ko sai nhưng mk chỉ cho cách khác nha!!
\(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)=x^2\left(y-z\right)+y^2\left(z-x+y-y\right)+z^2\left(x-y\right)\)
\(=x^2\left(y-z\right)-y^2\left(y-z\right)-y^2\left(x-y\right)+z^2\left(x-y\right)\)
\(=\left(y-z\right)\left(x^2-y^2\right)-\left(x-y\right)\left(y^2-z^2\right)\)
\(=\left(y-z\right)\left(x-y\right)\left(x+y\right)-\left(x-y\right)\left(y-z\right)\left(y+z\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
cảm ơn bn nha, NGUYỄN XUÂN ANH và ĐƯỜNG QUỲNH GIANG
còn thằng Pain Thiên Đạo ko biết làm thì cúttttttttttt