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1 , \(x^5+x^4+1=\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\)
= \(x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)=\(\left(x^2+x+1\right)\left(x^3-x+1\right)\)
2 , \(x\left(x+4\right)\left(x+6\right)\left(x+10\right)+128=\left(x^2+10x\right)\left(x^2+10x+24\right)+128\)(*)
Đặt x2 + 10 = a , a>0 (1)
=> (*) <=> a(a+24)+128=a2 + 24a+128=(a+8)(a+16) (**)
Thay (1) vào (**) ta được :
(*) <=> \(\left(x^2+10+8\right)\left(x^2+10+16\right)\)
a) \(\left(x+y\right)^5-x-y=\left(x+y\right)^5-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^4-1\right]\)
= \(\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\) #áp dụng hàng đẳng thức#
c) \(x^9-x^7-x^6-x^5+x^4+x^3+x^2+1\)nhóm vào là đc
b) \(\left(x^2+y^2\right)^3+\left(z^2-x^2\right)^3+\left(y^2+z^2\right)^3\)
=\(\left(y^2+x^2\right)\left[\left(x^2+y^2\right)^2-\left(x^2+y^2\right)\left(z^2-x^2\right)+\left(z^2-x^2\right)^2\right]+\left(y^2+z^2\right)^3\)
= \(\left(y^2+z^2\right)\left[x^4+y^4+2x^2y^2-x^2z^2+x^4-y^2z^2+x^2y^2+z^4+x^4-2x^2z^2+y^4+z^4+2y^2z^2\right]\)
=\(=\left(y^2+z^2\right)\left(2x^4+2y^4+2z^4+3x^2y^2-3x^2z^2+y^2z^2\right)\)
\(yz\left(y+z\right)+zx\left(z-x\right)-xy\left(x+y\right)\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left[\left(y+z\right)-\left(z-x\right)\right]\)
\(=yz\left(y+z\right)+zx\left(z-x\right)-xy\left(y+z\right)+xy\left(z-x\right)\)
\(=y\left(y+z\right)\left(z-x\right)+x\left(z-x\right)\left(z-y\right)\)
\(=\left(z-x\right)\left(yz-xy+xz-xy\right)\)
1.
\(3x^2-16x+5\\ =3x^2-x-15x+5\\ =x\left(3x-1\right)-5\left(3x-1\right)\\ =\left(x-5\right)\left(3x-1\right)\)
2.
\(3x^3-14x^2+4x+3\\ =\left(3x^3+x^2\right)-\left(15x^2+5x\right)+\left(9x+3\right)\\ =x^2\left(3x+1\right)-5x\left(3x+1\right)+3\left(3x+1\right)\\ =\left(x^2-5x+3\right)\left(3x+1\right)\)
3. \(x^8+x^7+1\\ =\left(x^8-x^2\right)+\left(x^7-x\right)+\left(x^2+x+1\right)\\ =x^2\left(x^6-1\right)+x\left(x^6-1\right)+\left(x^2+x+1\right)\\ =x^2\left(x^3+1\right)\left(x^3-1\right)+x\left(x^3+1\right)\left(x^3-1\right)+\left(x^2+x+1\right)\\ =x^2\left(x^3+1\right)\left(x-1\right)\left(x^2+x+1\right)+x\left(x^3+1\right)\left(x+1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\\ =\left(x^2+x+1\right)[x^2\left(x^3+1\right)\left(x-1\right)+x\left(x^3+1\right)\left(x-1\right)+1]\\ =\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+x^5-x^4+x^2-x+1\right)\\ =\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)4.
\(64x^4+y^4\\ =\left(64x^4+16x^2y^2+y^4\right)-16x^2y^2\\ =\left(8x^2+y^2\right)^2-16x^2y^2\\ =\left(8x^2+y^2-4xy\right)\left(8x^2+y+4xy\right)\)
5.
\(\left(x+a\right)\left(x+2a\right)\left(x+3a\right)\left(x+4a\right)+a^4\\ =\left(x+a\right)\left(x+4a\right)\left(x+2a\right)\left(x+3a\right)+a^4\\ =\left(x^2+5ax+4a^2\right)\left(x^2+5ax+6a^2\right)+a^4\\=\left(x^2+5ax+4a^2\right)\left(x^2+5ax+4a^2+2a^2\right)+a^4\\=\left(x^2+5ax+4a^2\right)+2a^2\left(x^2+5ax+4a^2\right)+a^4\\ =\left(x^2+5ax+5a^2\right)^2\)
a/ x3 + x2 z + y2 z - xyz + y3
= (x + y)(x2 - xy + y2) + z(x2 - xy + y2)
= (x2 - xy + y2)(x + y + z)
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)
2) Để sau đi (em chưa nghĩ ra)
3) \(A=\left(x+y\right)\left(x^2-y^2\right)+\left(y+z\right)\left(y^2-z^2\right)+\left(z+x\right)\left(z^2-x^2\right)\)
\(=\left(x+y\right)^2\left(x-y\right)+\left(y+z\right)^2\left(y-z\right)+\left(z+x\right)^2\left(z-x\right)\)
Đặt x - y = a; y - z = b => z - x = -(a+b)
\(A=\left(x+y\right)^2a+\left(y+z\right)^2b-\left(z+x\right)^2a-\left(z+x\right)^2b\)
\(=a\left[\left(x+y\right)^2-\left(z+x\right)^2\right]+b\left[\left(y+z\right)^2-\left(z+x\right)^2\right]\)
\(=\left(x-y\right)\left(x+y-z-x\right)\left(x+y+z+x\right)+\left(y-z\right)\left(y+z-z-x\right)\left(y+z+z+x\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(2x+y+z\right)-\left(y-z\right)\left(x-y\right)\left(2z+x+y\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
Em tính sai sót chỗ nào thì thông cảm cho em ạ :>
1) a4+b4+c4−2a2b2−2a2c2−2b2c2
=2(a4+b4+c4-4a2b2-4a2c2-4b2c2)
=2a4+2b4+2c4-4a2b2-4a2c2-4b2c2
=(a4-2a2b2+b4)+(a4-2a2c2+c4)+(b4-2b2c2+c4
a/ Nó là cái gì mà không phải nhân tử b
b/ \(\left(x^2+x+1\right)\left(x^5-x^4+x^3-x+1\right)\)
c/ \(3\left(2x+y+z\right)\left(x+2y+z\right)\left(x+y+2z\right)\)