Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)
\(=\left(x-y+4+2x+3y-1\right)\left(x-y+4-2x-3y+1\right)\)
\(=\left(3x+2y+3\right)\left(-x-4y+5\right)\)
\(49\left(y-4\right)^2-9y^2-36y-36\)
\(=49\left(y-4\right)^2-\left(9y^2+36y+36\right)\)
\(=49\left(y-4\right)^2-\left(3y+6\right)^2\)
\(=[7\left(y-4\right)]^2-\left(3y+6\right)^2\)
\(=\left(7y-28\right)^2-\left(3y+6\right)^2\)
\(=\left(7y-28+3y+6\right)\left(7y-28-3y-6\right)\)
\(=\left(10y-22\right)\left(4y-34\right)\)
Ta có :
\(1)\left(x^2+y^2-5\right)-4x^2y^2-16xy-16\)
\(=\left(x^2+y^2-5\right)^2-[\left(2xy\right)^2+2.2xy.4+4^2]\)
\(=\left(x^2+y^2-5\right)^2-\left(2xy+4\right)^2\)
\(=\left(x^2+y^2-5-2xy-4\right)\left(x^2+y^2-5+2xy+4\right)\)
\(=\left(x^2+y^2-2xy-9\right)\left(x^2+y^2+2xy-1\right)\)
\(=\left[\left(x-y\right)^2-3^2\right]\left[\left(x+y\right)^2-1\right]\)
\(=\left(x-y+3\right)\left(x-y-3\right)\left(x+y-1\right)\left(x+y+1\right)\)
\(2)x^2y^2\left(y-x\right)+y^2z^2\left(z-y\right)-z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(y-z+z-x\right)+y^2z^2\left(z-y\right)-z^2x^2\left(z-x\right)\)
\(=x^2y^2\left(y-z\right)+x^2y^2\left(z-x\right)+y^2z^2\left(z-y\right)-z^2x^2\left(z-x\right)\)
\(=\left(y-z\right)\left(x^2y^2-y^2z^2\right)+\left(z-x\right)\left(x^2y^2-z^2x^2\right)\)
\(=\left(y-z\right)\left(xy-yz\right)\left(xy+yz\right)+\left(z-x\right)\left(xy-zx\right)\left(xy+xz\right)\)
\(=y^2\left(y-z\right)\left(x-z\right)\left(x+z\right)+x^2\left(z-x\right)\left(y-z\right)\left(y+z\right)\)
\(=\left(y-z\right)\left(x-z\right)[y^2\left(x+z\right)-x^2\left(y+z\right)]\)
\(=\left(y-z\right)\left(x-z\right)(y^2x+y^2z-x^2y-x^2z)\)
\(=\left(y-z\right)\left(x-z\right)[(y^2x-x^2y)+(y^2z-x^2z)]\)
\(=\left(y-z\right)\left(x-z\right)[xy(y-x)+z(y^2-x^2)]\)
\(=\left(y-z\right)\left(x-z\right)[xy(y-x)+z(y-x)\left(x+y\right)]\)
\(=\left(y-z\right)\left(x-z\right)(y-x)\left(xy+xz+yz\right)\)
a, \(x^3+y^3+z^3=3xyz\Rightarrow x^3+y^3+z^3-3xyz=0\)( 1 )
Nhận xét : \(\left(x+y\right)^3=x^3+y^3+3x^2y+3xy^2\Rightarrow x^3+y^3=\left(x+y\right)^3-3x^2-3xy^2\)
Thay vào ( 1 ) ta có :
\(\left(x+y\right)^3+c^3-3x^2y-3xy^2-3xyz\)
\(=\left(z+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(z+y+z\right)\left(z^2+2xy+y^2-xz-yz+z^2\right)-3xyz\left(z+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(z^2+x^2+y^2-xy-yz-xz\right)\)
Vì theo đầu bài ta có: \(x+y+z=0\)nên ta có ( DPCM ) ..... học cho tốt nhé!
x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz
=x^2y+xy^2+xyz+x^2z+xz^2+xyz+y^2z+yz^2
=xy(x+y+z)+zx(x+y+z)+yz(y+z)
=x(y+z)(x+y+z)+yz(y+z)
=(y+z)(x^2+xy+zx+yz)
=(x+y)(y+z)(z+x)
t i c k mk nha!!! 565464556756768768787669789789776575656767676945645645654
a, \(xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+2xyz\)\(=x^2y+xy^2+y^2z+yz^2+x^2z+xz^2+2xyz\)
\(=\left(x^2y+xy^2+xyz\right)+\left(x^2z+xz^2+xyz\right)+\left(y^2z+yz^2\right)\)
\(=xy\left(x+y+z\right)+xz\left(x+z+y\right)+yz\left(y+z\right)\)
\(=x\left(x+y+z\right)\left(y+z\right)+yz\left(y+z\right)\)
\(=\left(y+z\right)\left(x^2+xy+xz+yz\right)\)
\(=\left(y+z\right)\left[x\left(x+z\right)+y\left(x+z\right)\right]\)
\(=\left(y+z\right)\left(x+z\right)\left(x+y\right)\)
b, \(2x^2+2y^2-x^2z+z-y^2z-2\)
\(=\left(2x^2-x^2z\right)+\left(2y^2-y^2z\right)-\left(2-z\right)\)
\(=x^2\left(2-z\right)+y^2\left(2-z\right)-\left(2-z\right)\)
\(=\left(2-z\right)\left(x^2+y^2-1\right)\)
Vụ này khoai à nha !
\(b,9x^2+90x+225-\left(x-y\right)^2\)
\(=\left(3x+15\right)^2-\left(x-y\right)^2\)
\(=\left(3x+15-x+y\right)\left(3x+15+x-y\right)\)
\(=\left(2x+y+15\right)\left(4x-y+15\right)\)