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A= x4 + 64
A= (x2)2 + 2.x2.8 +82 - (2.x2 .8)
A=(x2+8)2 -16x2
A =(x2+8+4x).(x2+8-4x)
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G=(x2+y2+z2)2 (có sẵn hdt rồi mak_)
a) \([(x-y)3 + (y-z)3]+ (z-x)3\)=\(\left(x-y+y-z\right)\left[\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2\right]-\left(x-z\right)^3\)
\(=\left(x-z\right)\left[\left(\left(x-y\right)^2-\left(x-y\right)\left(y-z\right)+\left(y-z\right)^2-\left(x-z\right)^2\right)\right]\)
\(=\left(x-z\right)\left[\left(x-y\right)\left(x-y-y+z\right)+\left(y-z-x+z\right)\left(y-z+x-z\right)\right]=\left(x-z\right)\left[\left(x-2y+z\right)\left(x+z\right)-\left(x-y\right)\left(x+y-2z\right)\right]\)
\(=\left(x-z\right)\left(x-y\right)\left(x-2y+z-x-y+2z\right)=\left(x-z\right)\left(x-y\right)\left(z-y\right)3\)
b) \(=y^2\left(x^2y-x^3+z^3-z^2y\right)-z^2x^2\left(z-x\right)=y^2\left[-y\left(z^2-x^2\right)-\left(z^3-x^3\right)\right]-z^2x^2\left(z-x\right)\)
\(=y^2\left(z-x\right)\left(-yz-xy-z^2-zx-x^2\right)-z^2x^2\left(z-x\right)=\left(z-x\right)\left(-y^3z-xy^2-z^2y^2-xyz-x^2y^2-z^2x^2\right)\)
đến đây coi như là thành nhân tử rồi nha. em muốn gọn thì ráng ngồi nghĩ rồi tách nha. chỉ cần nhóm mấy cái có ngoặc giống nhau là đc. k khó đâu. chịu khó nghĩ để rèn luyện nha
c) \(x^8+2x^4+1-x^4=\left(x^4+1\right)^2-x^4=\left(x^4+1-x^2\right)\left(x^4+1+x^2\right)\)
\(\left(9a^3-6a^2\right)+\left(6a^2-4a\right)+\left(-9a+6\right)=3a^2\left(3a-2\right)+2a\left(3a-2\right)-3\left(3a-2\right)=\left(3a-2\right)\left(3a^2+2a-3\right)\)
d) em sửa đề đi. đề sai rồi. đồng nhất hệ số phải có dấu bằng nha.
có gì liên hệ chị. đúng nha ;)
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
Áp dụng tính chất \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\) ta đc
\(x^3+y^3+z^3-3xyz=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y\right)^3+z^3-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)^3-3\left(x+y\right)z\left(x+y+z\right)-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)^3-\left(x+y+z\right)\left(3xz-3yz-3xy\right)\)
\(=\left(x+y+z\right)\left[\left(x+y+z\right)^2-3xz-3yz-3xy\right]\)
\(=\left(x+y+z\right)\left(x^2+y^2+x^2+2xy+2yz+2xz-3xz-3yz-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-xz\right)\)
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
\(x^3+y^3+z^3-3xyz\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
Ta có :
\(x^3+y^3+z^3-3xyz\)
\(\Rightarrow\left(x+y\right)^3-3xy\left(x+y\right)+z^3-3xyz\)
\(\Rightarrow\left(x+y+z\right)\left[\left(x+y^2\right)-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)
\(\Rightarrow\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
P/s tham khảo nha
hok tốt
1) \(x^6+1\)
\(=x^6+x^4-x^4+x^2-x^2+1\)
\(=\left(x^6-x^4+x^2\right)+\left(x^4-x^2+1\right)\)
\(=x^2\left(x^4-x^2+1\right)+\left(x^4-x^2+1\right)\)
\(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
2) \(x^6-y^6\)
\(=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)\left(x-y\right)\left(x^2+xy+y^2\right)\)
6,
=a4 [-(a-b)-(c-a)] + [b4(c-a)+c4(a-b)]
=rồi nhóm hạng tử chung lại
=và sau đó tách ra bằng hằng đẳng thức
kết quả =(a-b)(c-a)(c-b)(a2+b2+c2+ab+bc+ca)
Bài này khá dài nên mk nhác viết , bn cố gắng làm bài nhé !
a)x4+5x3+10x-4
=x4+2x2+5x3+10x-2x2-4
=x2(x2+2)+5x(x2+2)-2(x2+2)
=(x2+5x-2)(x2+2)
b)x3+y3+z3-3xyz
= (x+y)³ - 3xy(x-y) + z³ - 3xyz
= [(x+y)³ + z³] - 3xy(x+y+z)
= (x+y+z)³ - 3z(x+y)(x+y+z) - 3xy(x-y-z)
= (x+y+z)[(x+y+z)² - 3z(x+y) - 3xy]
= (x+y+z)(x² + y² + z² + 2xy + 2xz + 2yz - 3xz - 3yz - 3xy)
= (x+y+z)(x² + y² + z² - xy - xz - yz).
c)x7+x2+1
=x7-x+x2+x+1
=x(x6-1)+x2+x+1
=x(x3-1)(x3+1)+x2+x+1
=x(x-1)(x2+x+1)(x3+1)+x2+x+1
=(x2+x+1)(x5-x4+x2-x)
e)x5+x4+1
=x5-x2+x4-x+x2+x+1
=x2(x3-1)+x(x3-1)+x2+x+1
=(x2+x)(x3-1)+x2+x+1
=(x2+x)(x-1)(x2+x+1)+x2+x+1
=(x2+x+1)(x3-x+1)
g)x10+x5+1
=x10-x+x5-x2+x2+x+1
=x(x9-1)+x2(x3-1)+x2+x+1
=x(x3-1)(x6+x3+1)+x2(x-1)(x2+x+1)+x2+x+1
=(x7+x4+x)(x-1)(x2+x+1)+x2(x-1)(x2+x+1)+x2+x+1
=(x2+x+1)(x8-x7+x5-x4+2x2-x+1)
d)x8+x+1
=x8-x2+x2+x+1
=x2(x6-1)+x2+x+1
=x2(x3-1)(x3+1)+x2+x+1
=(x5+x2)(x-1)(x2+x+1)+x2+x+1
=(x2+x+1)(x6-x5+x3-x2+1)
a) x4 + 5x3 + 10x - 4
= (x4 - 4) + (5x3 + 10x)
= (x2 + 2)(x2 - 2) + 5x(x2 + 2)
= (x2 + 2)(x2 - 2 + 5x)
b) x3 + y3 + z3 - 3xyz
= x3 + y3 + z3 + 3x2y + 3xy2 - 3x2y - 3xy2 - 3xyz
= (x3 + 3x2y + 3xy2 + y3) + z3 - (3x2y + 3xy2 + 3xyz)
= (x + y)3 + z3 - 3xy(x + y + z)
= (x + y + z)[(x + y)2 - (x + y)z + z2] - 3xy(x + y + z)
= (x + y + z)[(x + y)2 - (x + y)z + z2 - 3xy]
= (x + y + z)(x2 + 2xy + y2 - xz - yz + z2 - 3xy)
= (x + y + z)(x2 + y2 + z2 - xz - yz - xy)
c) x7 + x2 + 1
= x7 + x2 + 1 - x + x
= (x7 - x) + (x2 + x + 1)
= x(x6 - 1) + (x2 + x + 1)
= x(x3 - 1)(x3 + 1) + (x2 + x + 1)
= x(x3 + 1)(x - 1)(x2 + x + 1) + (x2 + x + 1)
= (x2 + x + 1)[x(x3 + 1)(x - 1) + 1]
= (x2 + x + 1)[(x4 + x)(x - 1) + 1]
= (x2 + x + 1)(x5 - x4 + x2 - x + 1)
d) x8 + x + 1
= x8 + x + 1 + x2 - x2
= (x8 - x2) + (x2 + x + 1)
= x2(x6 - 1) + (x2 + x + 1)
= x2(x3 + 1)(x3 - 1) + (x2 + x + 1)
= x2(x3 + 1)(x -1)(x2 + x + 1) + (x2 + x + 1)
= (x2 + x + 1)[x2(x3 + 1)(x - 1) + 1]
= (x2 + x + 1)[(x5 + x2)(x - 1) + 1]
= (x2 + x + 1)(x6 - x5 + x3 - x2 + 1)
e) x5 + x4 + 1
= x5 + x4 + x3 - x3 + x2 - x2 + x - x +1
= (x5 + x4 + x3) - (x3 + x2 + x) + (x2 + x + 1)
= x3(x2 + x + 1) - x(x2 + x + 1) + (x2 + x + 1)
= (x3 - x + 1)(x2 + x + 1)