Phân tích đa thức thành nhân tử:
(x2+y2+z2)(x+y+z)2+(xy+yz+zx)2
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\(2\left(xy+yz+zx\right)-x^2-y^2-z^2\)
\(2xy+2yz+2zx-x^2-y^2-z^2\)
\(-\left(x^2+y^2+z^2-2xy-2yz-2xz\right)\)
\(-\left(x+y+z\right)^2\)
A ) xy(z+y)+yz(y+z)+zx(z+x)
=y.[x(z+y)+z(y+z)]+zx(z+x)
=y.(xz+xy+zy+z2)+zx(z+x)
=y.(xz+z2+xy+zy)+zx(z+x)
=y.[z.(z+x)+y.(z+x)]+zx(z+x)
=y.(z+x)(z+y)+zx(z+x)
=(z+x)[y(z+y)+zx]
=(z+x)(yz+y2+zx)
B )xy(x+y)-yz(y+z)-zx(z-x)
=y.[x(x+y)-z(y+z)]-zx(z-x)
=y.(x2+xy-zy-z2)-zx(z-x)
=y.(x2-z2+xy-zy)-zx(z-x)
=y.[(x+z)(x-z)+y.(x-z)]-zx(z-x)
=y.(x-z)(x+z+y)+zx(x-z)
=(x-z)[y(x+z+y)+zx]
=(x-z)(yx+yz+y2+zx)
=(x-z)(yx+zx+yz+y2)
=(x-z)[x.(y+z)+y.(y+z)]
=(x-z)(y+z)(x+y)
b. \(\text{ xy(x+y)-yz(y+z)-xz(z-x) =xy(x+y+z-z)+yz(y+z)+xz(x-z) =xy(x-z)+xy(y+z)+yz(y+z)+xz(x-z) =(x+y)(y+z)(x-z) }\)
\(1,\\ a,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ b,=a^2\left(a-x\right)-y\left(a-x\right)=\left(a^2-y\right)\left(a-x\right)\\ c,=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\\ d,=x\left(x-2y\right)+t\left(x-2y\right)=\left(x+t\right)\left(x-2y\right)\\ 2,\\ \Rightarrow x^2-4x+4-x^2+9=6\\ \Rightarrow-4x=-7\Rightarrow x=\dfrac{7}{4}\\ 3,\\ a,x^2+2x+2=\left(x+1\right)^2+1\ge1>0\\ b,-x^2+4x-5=-\left(x-2\right)^2-1\le-1< 0\)
x 2 y + x y 2 + x 2 z + x z 2 + y 2 z + y z 2 + 3xyz.
= ( x 2 y + x 2 z + xyz) + (x y 2 + y 2 z + xyz) + (x z 2 + y z 2 + xyz)
= x(xy + xz + yz) + y(xy + yz + xz) + z(xz + yz + xy)
= (x + y + z)(xy + xz + yz).
\(x^2y+xy^2+x^2z+xz^2+y^2z+yz^2+3xyz\)
\(=\left(x^2y+x^2z+xyz\right)+\left(xz^2+yz^2+xyz\right)+\left(xy^2+y^2z+xyz\right)\)
\(=x\left(xy+xz+yz\right)+z\left(xz+yz+xy\right)+y\left(xy+yz+xz\right)\)
\(=\left(x+y+z\right)\left(xy+yz+xz\right)\)
a/ \(\left(x-y\right)\left(z-x\right)\left(z-y\right)\)
b/ \(\left(1-y\right)\left(y-x\right)\)
a. \(\left(x-y\right)\left(z-x\right)\left(z-y\right)\)
b. \(\left(1-y\right)\left(y-x\right)\)
xy(x+y)-yz(y+z)-zx(z-x)
=y.[x.(x+y)-z.(y+z)]-zx.(z-x)
=y.(x2+xy-zy-z2)-zx.(z-x)
=y.[(x-z)(x+z)-y.(z-x)]-zx.(z-x)
=y.[-(z-x)(x+z)-y.(z-x)]-zx.(z-x)
=y.(z-x)(-x-z-y)-zx.(z-x)
=(z-x)(-xy-zy-y2-zx)
=(z-x)[-x.(y+z)-y.(y+z)]
=(z-x)(y+z)(-x-y)
=-(z-x)(y+z)(x+y)
\(xy\left(x+y\right)+yz\left(y-z\right)-xz\left(x+z\right)\)
\(=xy\left(x+y\right)+z\left[y\left(y-z\right)-x\left(x+z\right)\right]\)
\(=xy\left(x+y\right)+z\left(y^2-yz-x^2-xz\right)\)
\(=xy\left(x+y\right)+z\left[\left(y^2-x^2\right)-\left(yz+xz\right)\right]\)
\(=xy\left(x+y\right)+z\left[\left(y-x\right)\left(y+x\right)-z\left(x+y\right)\right]\)
\(=xy\left(x+y\right)+z\left(x+y\right)\left(y-z-x\right)\)
\(=\left(x+y\right)\left[xy+z\left(y-x-z\right)\right]\)
\(=\left(x+y\right)\left(xy-yz-xz-z^2\right)\)
Bài này dùng cách đặt ẩn phụ. Nhiều bài lớp 8 phải làm vậy. Mong bạn hiểu được cách giải.
Đặt x^2 +y^2 +z^2 =a , xy+yz+zx =b
Ta có: (x^2 +y^2 +z^2)(x+y+z)^2 +(xy+yz+zx)^2
= a (x^2 +y^2 +z^2 +2xy +2yz +2xz) +b^2
= a (a+2b)+ b^2
= a^2 + 2ab+ b^2
= (a+b)^2
= (x^2 +y^2 +z^2 +xy+yz+zx)^2
Chúc bạn học tốt.