phân tích đa thức thành nhân tử
a) \(x^4+x^3+2x^2+x+1\)
b) \(a^3+b^3+c^3-3abc\)
c) \(\left(x-y\right)^3+\left(y-z\right)^3+\left(z-x\right)^3\)
d) \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
e) \(8\left(x+y+z\right)^3+\left(x+y\right)^3-\left(y+z\right)^3-\left(z+x\right)^3\)
f) \(x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\)
a) x4+x3+2x2+x+1=(x4+x3+x2)+(x2+x+1)=x2(x2+x+1)+(x2+x+1)=(x2+x+1)(x2+1)
b)a3+b3+c3-3abc=a3+3ab(a+b)+b3+c3 -(3ab(a+b)+3abc)=(a+b)3+c3-3ab(a+b+c)
=(a+b+c)((a+b)2-(a+b)c+c2)-3ab(a+b+c)=(a+b+c)(a2+2ab+b2-ac-ab+c2-3ab)=(a+b+c)(a2+b2+c2-ab-ac-bc)
c)Đặt x-y=a;y-z=b;z-x=c
a+b+c=x-y-z+z-x=o
đưa về như bài b
d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung
e)x2(y-z)+y2(z-x)+z2(x-y)=x2(y-z)-y2((y-z)+(x-y))+z2(x-y)
=x2(y-z)-y2(y-z)-y2(x-y)+z2(x-y)=(y-z)(x2-y2)-(x-y)(y2-z2)=(y-z)(x2-2y2+xy+xz+yz)
a. =\(\left(x^2+1\right).\left(x^2+x+1\right)\)
b = \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
c=\(\left(3z+3y\right)\left(x+y\right)\left(z-x\right)\)
d= \(3\left(ab+bc+ca\right)\)
2 câu còn lại mình ko biết
xin lỗi mình chỉ viết đc đáp án vì nó dài quá
phân tích đa thức thành nhân tử :
a. \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
b.\(8\left(x+y+z\right)^3-\left(x+y\right)^3-\left(y+z\right)^3-\left(x+z\right)^3\)
a: \(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=\left[\left(a+b+c\right)^3-a^3\right]-\left(b^3+c^3\right)\)
\(=\left(a+b+c-a\right)\left[\left(a+b+c\right)^2+a\left(a+b+c\right)+a^2\right]-\left(b+c\right)\left(b^2-bc+c^2\right)\)
\(=\left(b+c\right)\left[a^2+b^2+c^2+a^2+a^2+2ab+2bc+2ac+ab+ac-b^2+bc-c^2\right]\)
\(=\left(b+c\right)\left(3a^2+3ab+3bc+3ac\right)\)
\(=3\left(b+c\right)\left(a+b\right)\left(a+c\right)\)
b: \(=\left(2x+2y+2z\right)^3-\left(x+y\right)^3-\left[\left(y+z\right)^3+\left(x+z\right)^3\right]\)
\(=\left(x+y+2z\right)\left[\left(2x+2y+2z\right)^2+2\left(x+y+z\right)\left(x+y\right)+\left(x+y\right)^2\right]-\left(x+y+2z\right)\left[\left(y+z\right)^2-\left(y+z\right)\left(x+z\right)+\left(x+z\right)^2\right]\)
\(=3\left(x+y+2z\right)\left(x+z+2y\right)\left(y+z+2x\right)\)
phân tích đa thức thành nhân tử:
a.\(x^3\left(y-z\right)+y^3\left(z-x\right)+z^3\left(x-y\right)\)
b.\(x^3\left(z-y^2\right)+y^3\left(x-z^2\right)+z^3\left(y-z^2\right)+xyz\left(xyz-1\right)\)
Phân tích đa thức thành nhân tử
a) \(\left(x+y-2z\right)^3+\left(y+z-2x\right)^3+\left(z+x-2y\right)^3\)
b) \(a\left(c^2+b^2+bc\right)+b\left(c^2+a^2+ca\right)+c\left(a^2+b^2+bc\right)\)
c) (a+b+c)(ab+ac+bc)-abc
d) \(c\left(a+2b\right)^3-b\left(2a+b\right)^3\)
e) xy(x+y)-yz(y+z)+xz(x-z)
Phân tích đa thức sau thành nhân tử :
\(i,\left(b-c\right)^3+\left(c-a\right)^3+\left(a-b^3\right)\)
\(k,\left(x^2+y^2\right)^3+\left(z^2-x^2\right)^3-\left(y^2+z^2\right)^3\)
Phân tích đa thức sau thành nhân tử :
\(i,\left(b-c\right)^3+\left(c-a\right)^3+\left(a-b^3\right)\)
\(k,\left(x^2+y^2\right)^3+\left(z^2-x^2\right)^3-\left(y^2+z^2\right)^3\)
Phân tích đa thức thành nhân tử :
a) \(\left(x^2+y^2\right)^3+\left(z^2-x^2\right)^3-\left(y^2+z^2\right)^3\)
b)\(\left(x+y\right)^3-x^3-y^3\)
c)\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
a) \(\left(x^2+y^2\right)^3+\left(z^2-x^2\right)^3-\left(y^2+z^2\right)^3\)
\(=\left[\left(x^2+y^2\right)^3+\left(z^2-x^2\right)^3\right]-\left(y^2+z^2\right)^3\)
\(=\left(x^2+y^2+z^2-x^2\right)\left[\left(x^2+y^2\right)^2-\left(x^2+y^2\right)\left(z^2-x^2\right)+\left(z^2-x^2\right)^2\right]-\left(y^2+z^2\right)^3\)
\(=\left(y^2+z^2\right)\left(x^4+2x^2y^2+y^4-x^2z^2+x^4-y^2z^2+x^2y^2+z^4-2z^2x^2+x^4\right)-\left(y^2+z^2\right)^3\)
\(=\left(y^2+z^2\right)\left[x^4+2x^2y^2+y^4-x^2z^2+x^4-y^2z^2+x^2y^2+z^4-2z^2x^2+x^4-\left(y^2+z^2\right)^2\right]\)
\(=\left(y^2+z^2\right)\left(x^4+2x^2y^2+y^4-x^2z^2+x^4-y^2z^2+x^2y^2+z^4-2z^2x^2+x^4-y^4-2y^2z^2-z^4\right)\)
\(=\left(y^2+z^2\right)\left(3x^4+3x^2y^2-3x^2z^2-3y^2z^2\right)\)
= 3(y2+z2)(x4+x2y2-x2z2-y2z2)
= 3(y2+z2)[x2(x2+y2)-z2(x2+y2)]
= 3(y2+z2)(x2-z2)(x2+y2)
= 3(y2+z2)(x-z)(x+z)(x2+y2)
b) \(\left(x+y\right)^3-x^3-y^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3-y^3\)
\(=3x^2y+3xy^2=3xy\left(x+y\right)\)
c) \(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+3\left(x+y\right)^2.z+3\left(x+y\right).z^2+z^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+3\left(x+y\right)^2.z+3\left(x+y\right).z^2-\left(x^3+y^3\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2+3\left(x+y\right).z+3z^2\right]-\left(x+y\right)\left(x^2-xy+y^2\right)\)
\(=\left(x+y\right)\left(x^2+2xy+y^2+3xz+3yz+3z^2-x^2+xy-y^2\right)\)
= (x+y)[3xy+3xz+3yz+3z2 ]
= 3(x+y)(xy+xz+yz+z2)
= 3(x+y)[x(y+z)+z(y+z)]
= 3(x+y)(x+z)(y+z)
a) \(\left(x^2+y^2\right)^3+\left(z^2-x^3\right)-\left(y^2+z^2\right)^3\)
\(=x^6+3x^4y^2+3x^4y^2+y^6+z^2+-x^2+-y^6+-3y^4z+-3y^2z^4+-z^6\)
\(=x^6+3x^4y^2+3x^2y^4+-3y^4z^4+-z^6+-x^2+z^2\)
b) \(\left(x+y\right)^3-x^3-y^3\)
\(=x^3+3x^2y+3xy^2+y^3+-x^3+-y^3\)
\(=\left(x^3+-x^3\right)+\left(3x^2y\right)+\left(3xy^2\right)+\left(y^3+-y^3\right)\)
\(=3x^2y+3xy^2\)
c) \(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=x^3+3x^2y+3x^2z+3xy^2+6xyz+3xz^2+y^3+3y^2z+3yz^2+z^2-x^3-y^3-z^3\)
\(=3x^2y+3x^2z+3xy^2+3xy^2+6xyz+3xz^2+3y^2z+3yz^2\)
P/s: Ko chắc
Rút gọn phân thức
1/\(\frac{x^{3^{ }}-y^{3^{ }}+z^{3^{ }}+3xyz}{\left(x+y\right)^{2^{ }}+\left(y+z\right)^2+\left(z-x\right)^2}\)
2/\(\frac{x^{3^{ }}+y^{3^{ }}+z^3-3xyz}{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}\)
3/\(\frac{a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)}{a^4\left(b^2-c^2\right)+b^4\left(c^2-a^3\right)+c^4\left(a^2-b^2\right)}\)
Phân tích đa thưc thành nhân tử:
\(\left(x+y+x\right)^3-x^3-y^3-z^3\)
\(xy\left(x+y\right)+yx\left(y-z\right)-zx\left(z+x\right)\)
\(x^3+y^3+z^3-3xyz\)
\(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(\left(a+b-c\right)^2-\left(a-c\right)^2-2ab+2bc\)
mn giúp đc câu nào cứ giúp nhá!
Sửa đề chút :
\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+3\left(x+y\right)^2z+3\left(x+y\right)z^2+z^3-x^3-y^3-z^3\)
\(=x^3+3x^2y+3xy^2+y^3+3\left(x+y\right)^2z+3\left(x+y\right)z^2-x^3-y^3\)
\(=3x^2y+3xy^2+3\left(x+y\right)^2z+3\left(x+y\right)z^2\)
\(=3xy\left(x+y\right)+3\left(x+y\right)^2z+3\left(x+y\right)z^2\)
\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=3\left(x+y\right)\left[x\left(y+z\right)+z\left(y+z\right)\right]\)
\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
c) x3 + y3 + z3 - 3xyz
= x3 + 3x2y + 3xy2 + y3 + z3 - 3xyz - 3x2y - 3xy2
= (x+y)3 + z3 - 3xy.( z+x+y)
= (x+y+z).[(x+y)2 - (x+y).z + z2 ] - 3xy.(x+y+z)
= (x+y+z). ( x2 + 2xy + y2 - xz - yz + z2 - 3xy)
= (x+y+z) .(x2 + y2 + z2 - xy - xz -yz)
e) (a+b-c)2 - (a-c)2 - 2ab + 2bc
= (a+b-c - a+c).(a+b+c+a-c) - 2b.(a-c)
= b.(2a+b) - 2b.(a-c)
= b.(2a+b - a +c)
= b.( a+b+c)
xl bn nha! mk chỉ nghĩ đk 2 câu thoy, 1 câu bn kia làm r! 2 câu còn lại bn đợi người tiếp theo làm nhé
