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\(yz\left(y+z\right)+zx\left(z-x\right)-xy\left(x+y\right)\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left[\left(y+z\right)-\left(z-x\right)\right]\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left(y+z\right)+xy\left(z-x\right)\)
\(=y\left(y+z\right)\left(z-x\right)+x\left(z-x\right)\left(z-y\right)\)
\(=\left(z-x\right)\left(yz-xy+xz-xy\right)\)
tuổi con HN là :
50 : ( 1 + 4 ) = 10 ( tuổi )
tuổi bố HN là :
50 - 10 = 40 ( tuổi )
hiệu của hai bố con ko thay đổi nên hiệu vẫn là 30 tuổi
ta có sơ đồ : bố : |----|----|----|
con : |----| hiệu 30 tuổi
tuổi con khi đó là :
30 : ( 3 - 1 ) = 15 ( tuổi )
số năm mà bố gấp 3 tuổi con là :
15 - 10 = 5 ( năm )
ĐS : 5 năm
mình nha
\(2xyz+x^2y+xy^2+x^2z+xz^2+y^2z+yz^2\)
\(=x^2\left(y+z\right)+yz\left(y+z\right)+x\left(y^2+z^3\right)+2xyz\)
\(=\left(y+z\right)\left(x^2+yz\right)+x\left(y^2+z^2+2yz\right)\)
\(=\left(y+z\right)\left(x^2+yz\right)+x\left(y+z\right)^2\)
\(=\left(y+z\right)\left(x^2+yz\right)+xy+xz\)
\(=\left(y+z\right)\left[x\left(x+2\right)+y\left(x+2\right)\right]\)
\(=\left(y+z\right)\left(x+y\right)\left(x+2\right)\)
\(b,x^2\left(y-z\right)+y^2\left(z-y\right)+z^2\left(x-y\right)\)
\(=x^2\left(y-z\right)+y^2z-y^2x+z^2x-z^2y\)
\(=x^2\left(y-z\right)+yz\left(y-z\right)-x\left(y^2-z^2\right)\)
\(=\left(y-z\right)\left[x^2+yz-x\left(y+z\right)\right]\)
\(=\left(y-z\right)\left[x\left(x-y\right)-z\left(x-y\right)\right]\)
\(=\left(y-z\right)\left[\left(x-z\right)\left(x-y\right)\right]\)
b \(x^8y^8+x^4y^4+1=x^8y^8+2x^4y^4+1-x^4y^4=\left(x^4y^4\right)^2+2x^4y^4+1-\left(x^2y^2\right)^2\)
\(=\left(x^4y^4+1\right)^2-\left(x^2y^2\right)^2=\left(x^4y^4-x^2y^2+1\right)\left(x^4y^4+x^2y^2+1\right)\)
c \(x^2y+xy^2+xz^2+x^2z+y^2z+yz^2+2xyz=\left(x^2y+x^2z+xyz+xy^2\right)+\left(xz^2+yz^2+xyz+y^2z\right)\)
\(=x\left(xy+xz+yz+y^2\right)+z\left(xz+yz+xy+y^2\right)=\left(x+z\right)\left(xy+xz+yz+y^2\right)\)
\(=\left(x+z\right)\left(x\left(y+z\right)+y\left(y+z\right)\right)=\left(x+z\right)\left(x+y\right)\left(y+z\right)\)
a \(3xyz+x\left(y^2+z^2\right)+y\left(x^2+z^2\right)+z\left(x^2+y^2\right)=3xyz+xy^2+xz^2+x^2y+yz^2+x^2z+y^2z\)
\(=\left(x^2y+x^2z+xyz\right)+\left(xy^2+xyz+y^2z\right)+\left(xyz+xz^2+yz^2\right)\)
\(=x\left(xy+xz+yz\right)+y\left(xy+xz+yz\right)+z\left(xy+xz+yz\right)=\left(x+y+z\right)\left(xy+xz+yz\right)\)
a) \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz\)
\(=x^2y+xy^2+xyz+x^2z+xz^2+xyz+y^2z+yz^2\)
\(=xy\left(x+y+z\right)+xz\left(x+z+y\right)+yz\left(y+z\right)\)
\(=\left(x+y+z\right)\left(xy+xz\right)+yz\left(y+z\right)\)
\(=x\left(x+y+z\right)\left(y+z\right)+yz\left(y+z\right)\)
\(=\left(y+z\right)\left(x^2+xy+xz+yz\right)\)
\(=\left(y+z\right)\left[x\left(x+y\right)+z\left(x+y\right)\right]=\left(y+z\right)\left(x+y\right)\left(x+z\right)\)
b) \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+3xyz\)
\(=\left(x^2y+xy^2+xyz\right)+\left(x^2z+xz^2+xyz\right)+\left(y^2z+yz^2+xyz\right)\)
\(=xy\left(x+y+z\right)+xz\left(x+z+y\right)+yz\left(y+z+x\right)\)
\(=\left(x+y+z\right)\left(xy+xz+yz\right)\)
P/s: Sai sót xin bỏ qua.
\(1\hept{\begin{cases}6x^2-8x+3x-4\\2x\left(3x-4\right)+\left(3x-4\right)\\\left(3x-4\right)\left(2x+1\right)\end{cases}}\)
\(2\hept{\begin{cases}7x^2-7xy-5x+5y+6xy\\7x\left(x-y\right)-5\left(x-y\right)+\frac{6xy\left(x-y\right)}{\left(x-y\right)}\\\left(x-y\right)\left(7x-5+\frac{6xy}{\left(x-y\right)}\right)\end{cases}}\)
\(3\hept{\begin{cases}5x\left(x-y\right)-15\left(x-y\right)\\\left(x-y\right)\left(5x-15\right)\end{cases}}\)
\(4,,2x^2+x=x\left(2x+1\right)\)
\(5\hept{\begin{cases}x^3-4x-3x^2+12\\x\left(x^2-4\right)-3\left(x^2-4\right)\\\left(x+2\right)\left(x-2\right)\left(x-3\right)\end{cases}}\)
\(6\hept{\begin{cases}2x+2y+x^2-y^2\\2\left(x+y\right)+\left(x+y\right)\left(x-y\right)\\\left(x+y\right)\left(2+x-y\right)\end{cases}}\)
\(7\hept{\begin{cases}\left(x^2y-2xy\right)-\left(xy-2y\right)+\left(xy-y\right)\\xy\left(x-2\right)-y\left(x-2\right)+y\left(x-1\right)\\y\left(X-2\right)\left(x-1\right)+y\left(x-1\right)\end{cases}}\Leftrightarrow y\left(x-1\right)\left(x-2+1\right)\)
\(8\hept{\begin{cases}x\left(2-y\right)+z\left(2-y\right)\\\left(2-y\right)\left(x+1\right)\end{cases}}\)
a) \(2\left(x-y\right)+x^2-y^2\\ =2\left(x-y\right)+\left(x^2-y^2\right)\\ =2\left(x-y\right)+\left(x+y\right)\left(x-y\right)\\ =\left(x-y\right)\left(2+x-y\right)\)
b) \(x^3-4x^2-9x+36\\ =x^2\cdot x-4x^2-9x+36\\ =x^2\left(x-4\right)-9\left(x-4\right)\\=\left(x-4\right)\left(x^2-9\right)\\ =\left(x-4\right)\left(x-3\right)\left(x+3\right)\)
c) \(2x^2+2y^2-x^2z+2-y^2z-2\\ =2\left(x^2+y^2\right)-z\left(x^2+y^2\right)+\left(2-2\right)\\ =\left(x^2+y^2\right)\left(2-z\right)\)
d) \(x^3+y^3+2x^2-2xy+2y^2\\ =\left(x+y\right)\left(x^2-xy+y^2\right)+2\left(x^2-xy+y^2\right)\\ =\left(x^2-xy+y^2\right)\left(x+y+2\right)\)
e) \(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+2xyz\\ =x^2y+xy^2+xyz+x^2z+xz^2+xyz+y^2z+yz^2\\ =xy\left(x+y+z\right)+xz\left(x+y+z\right)+yz\left(y+z\right)\\ =\left(y+z\right)\left(x^2+xy+xz+yz\right)\\ =\left(y+z\right)\left(z+x\right)\left(x+y\right)\)
Câu a kq lầ (x-y)(2+x+y) chứ