phân tích đa thức xyz − xy − yz − zx + x + y + z − 1
Cho xy - yz - zx = 0 và xyz khác 0. Tính giá trị biểu thức B = yz/x^2 - zx/y^2 - xy/z^2 .
Lời giải:
\(yz-xz-xy=0\Rightarrow yz-xz=xy\)
\(B=\frac{yz}{x^2}-\frac{zx}{y^2}-\frac{xy}{z^2}\)\(=\frac{(yz)^3-(xz)^3-(xy)^3}{x^2y^2z^2}\)
Xét: \((yz)^3-(xz)^3-(xy)^3=(yz-xz)^3+3yz.xz(yz-xz)-(xy)^3\)
\(=(xy)^3+3yz.xz.xy-(xy)^3=3x^2y^2z^2\)
\(\Rightarrow B=\frac{(yz)^3-(xz)^3-(xy)^3}{x^2y^2z^2}=\frac{3x^2y^2z^2}{x^2y^2z^2}=3\)
\(\dfrac{xyz-xy-yz-zx+x+y+z-1}{xyz+xy+yz-zx-x+y-z-1}\) với x = 5001;y=5002;z=5003
\(=\dfrac{xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(z-1\right)}{xy\left(z+1\right)+y\left(z+1\right)-x\left(z+1\right)-\left(z+1\right)}\\ =\dfrac{\left(z-1\right)\left(xy-y-x+1\right)}{\left(z+1\right)\left(xy+y-x-1\right)}=\dfrac{\left(z-1\right)\left(x-1\right)\left(y-1\right)}{\left(z+1\right)\left(x+1\right)\left(y-1\right)}=\dfrac{\left(z-1\right)\left(x-1\right)}{\left(z+1\right)\left(x+1\right)}\\ =\dfrac{\left(5003-1\right)\left(5001-1\right)}{\left(5003+1\right)\left(5001+1\right)}=\dfrac{5002\cdot5000}{5004\cdot5002}=\dfrac{5000}{5004}=\dfrac{1250}{1251}\)
MỌI NGƯỜI GIÚP MK VỚI!!!
Phân tích đa thức sau thành nhân tử:
A=xyz+(x+y+z)-1-( xy+yz+zx)
B=x2y+y2z+z2x+xy2+yz2+zx2+3xyz
C=yz(y+z)+zx(z-x)-xy(x+y)
D=(x+2)(x+3)(x+4)(x+5)-24
phân tích đa thức thành nhân tử
(xy+yz+zx)(x+y+z)-xyz
xy(x+y+z) -xyz +(yz+zx)(x+y+z)
=xy(x+y+z -z) + z(x+y)(x+y+z)
=xy(x+y) +z(x+y)(x+y+z)
=(x+y)(xy+z(x+y+z))
=(x+y)(xy+zx+z(y+z))
=(x+y)(x(y+z)+z(y+z))
=(x+y)(y+z)(x+z)
bài 1: Phân tích đa thức thành nhân tử
a, (xy-1)2+ (x+y)2
b, a2+2a2+2a+1
c, (1+2a).(1-2a)-a.(a+2).(a-2)
d, a2+b2-a2b2+ab-a-b
e, xy.(x+y)-yz.(y+z)+xz(x-z)
f, xyz-(xy+yz+zx)+(x+y+z)-1
giúp em với ạ ! em đang cần gấp
\(a,=\left(xy-1-x-y\right)\left(xy-1+x+y\right)\\ b,Sửa:a^3+2a^2+2a+1\\ =a^3+a^2+a^2+a+a+1=\left(a+1\right)\left(a^2+a+1\right)\\ c,=1-4a^2-a\left(a^2-4\right)=1-4a^2-a^3+4a\\ =\left(1-a\right)\left(1+a+a^2\right)+4a\left(1-a\right)\\ =\left(1-a\right)\left(1+5a+a^2\right)\\ d,=\left(a^2-a^2b^2\right)+\left(b^2-b\right)+\left(ab-a\right)\\ =a^2\left(1-b\right)\left(1+b\right)+b\left(b-1\right)+a\left(b-1\right)\\ =\left(b-1\right)\left(-a^2-ab+b+a\right)\\ =\left(b-1\right)\left(b-1\right)\left(a+b\right)\left(1-a\right)\)
\(e,=x^2y+xy^2-yz\left(y+z\right)+x^2z-xz^2\\ =\left(x^2y+x^2z\right)+\left(xy^2-xz^2\right)-yz\left(y+z\right)\\ =x^2\left(y+z\right)+x\left(y-z\right)\left(y+z\right)-yz\left(y+z\right)\\ =\left(y+z\right)\left(x^2+xy-xz-yz\right)\\ =\left(y+z\right)\left(x+y\right)\left(x-z\right)\)
\(f,=xyz-xy-yz-xz+x+y+z-1\\ =xy\left(z-1\right)-y\left(z-1\right)-x\left(z-1\right)+\left(x-1\right)\\ =\left(z-1\right)\left(xy-y-x+1\right)=\left(z-1\right)\left(x-1\right)\left(y-1\right)\)
11. xyz - xy - yz - zx + x + y + z - 1
12. xy(x + y) + yz(y + z) + zx(z + x) + 2xyz
13. xy(x + y) + yz(y + z) + zx(z + x) + 3xyz
giúp mik vs mik đang cần gấp =(((
13:
xy(x+y)+yz(y+z)+xz(x+z)+2xyz
= xy(x + y) + yz(y + z) + xyz + xz(x + z) + xyz
= xy(x + y) + yz(y + z + x) + xz(x + z + y)
= xy(x + y) + z(x + y + z)(y + x)
= (x + y)(xy + zx + zy + z²)
= (x + y)[x(y + z) + z(y + z)]
= (x + y)(y + z)(z + x)
Phân tích thành nhân tử:
a/ xy(x+y) + yz(y+z) +zx(z+x) + 3xyz (tách 3xyz=xyz+xyz+xyz)
xy( x+ y) + yz(y+z) + xz(x+z) + 3xyz
= xy(x+y) + xyz + yz(y+z) + xyz + xz(x+z) + xyz
= zy(x+y+z) + yz(x + y + z) + xz ( x+y+z)
= ( x+ y +z )( xy + yz + zx)
Cho \(x+y+z=xyz\) và \(xy+yz+zx\ne-3\)
Chứng minh: \(\dfrac{x.\left(y^2+z^2\right)+y.\left(z^2+x^2\right)+z.\left(x^2+y^2\right)}{xy+yz+zx-3}=xyz\)
Phân tích đa thức thành nhân tử
A ) xy(z+y)+yz(y+z)+zx(z+x)
B )xy(x+y)-yz(y+z)-zx(z-x)
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. \(\text{ xy(x+y)-yz(y+z)-xz(z-x) =xy(x+y+z-z)+yz(y+z)+xz(x-z) =xy(x-z)+xy(y+z)+yz(y+z)+xz(x-z) =(x+y)(y+z)(x-z) }\)