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Bạn hội con bò gì đó ơi cho mk tham gia đc không vì là hội học hành nên .....
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)
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 ;)
\(x^8+x+1\)
\(=x^8+x^7+x^6-x^7-x^6-x^5+x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)
\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x^2\left(x^2+x+1\right)+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x^2+1\right)\)
a, \(x^3-x^2-4\)
\(=x^3-2x^2+x^2-2x+2x-4\)
\(=x^2\left(x-2\right)+x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+x+2\right)\)
a) \(x^3-x^2-4\)
\(=x^3-2x^2+x^2-2x+2x-4\)
\(=x^2\left(x-2\right)+x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+x+2\right)\)
b) \(x^8-98x^4+1\)
\(=\left(x^4\right)^2+2\cdot x^4\cdot1+1^2-100x^4\)
\(=\left(x^4+1\right)^2-\left(10x^2\right)^2\)
\(=\left(x^4-10x^2+1\right)\left(x^4+10x^2+1\right)\)
\(A=x^3+4x^2-8x-8=\left(x^3-8\right)+4x\left(x-2\right)=\left(x^3-2^3\right)+4x\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+2x+4\right)+4x\left(x-2\right)=\left(x-2\right)\left(x^2+2x+4+4x\right)=\left(x-2\right)\left(x^2+6x+4\right)\)
\(B=a^2+b^2-a^2b^2+ab-a-b=\left(ab-a\right)-\left(a^2b^2-a^2\right)+\left(b^2-b\right)\)
\(=a\left(b-1\right)-a^2\left(b^2-1\right)+b\left(b-1\right)=a\left(b-1\right)-a^2\left(b-1\right)\left(b+1\right)+b\left(b-1\right)\)
\(=\left(b-1\right)\left(a-a^2b-a^2+b\right)\)
\(C=x^4-x^3-x+1=x^3\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)\)
Đoàn Thị Huyền Đoan: Hình như câu A bạn chép xuống bị sai đề rồi!
a) \(2x\left(x-3\right)^2+5x\left(3-x\right)\)
\(=2x\left(x-3\right)^2-5x\left(x-3\right)\)
\(=\left(x-3\right)\left[2x\left(x-3\right)-5x\right]\)
\(=\left(x-3\right)\left(2x^2-6x-5x\right)\)
\(=\left(x-3\right)\left(2x^2-11x\right)\)
\(=x\left(x-3\right)\left(2x-11\right)\)
b) \(\left(x+3\right)^2-4\left(y^2-2y+1\right)\)
\(=\left(x+3\right)^2-2^2\left(y-1\right)^2\)
\(=\left(x+3\right)^2-\left[2\left(y-1\right)\right]^2\)
\(=\left[\left(x+3\right)-2\left(y-1\right)\right]\left[\left(x+3\right)+2\left(y-1\right)\right]\)
\(=\left(x+3-2y+2\right)\left(x+3+2y-2\right)\)
\(=\left(x-2y+5\right)\left(x+2y+1\right)\)
a) \(2x.\left(x-3\right)^2+5x.\left(-x+3\right)=2x.\left(x-3\right)^2-5x.\left(x-3\right)\)
\(=\left(x-3\right).\left(2x^2-11x\right)=\left(x-3\right).x.\left(2x-11\right)\)
b) \(\left(x+3\right)^2-4.\left(y^2-2y+1\right)=\left(x+3\right)^2-2^2.\left(y-1\right)^2\)
\(=\left(x+3\right)^2-\left[2.\left(y-1\right)\right]^2=\left(x-2y+1\right).\left(x+2y+5\right)\)
a)\(x^4+x^3+x+1=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)=\left(x+1\right)^2\left(x^2-x+1\right)\)
b)\(x^4-x^3-x^2+1=\left(x^4-x^3\right)-\left(x^2-1\right)=x^3\left(x-1\right)-\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^3-x-1\right)\)
c)\(x^2y+xy^2-x-y=xy\left(x+y\right)-\left(x+y\right)=\left(xy-1\right)\left(x+y\right)\)
câu này hình như mình thấy rồi , trong câu hỏi tương tự đó
(x^2+x)^2+3(x^2+x)+2
đặt x^2+x=t
suy ra t^2+3t+2
=t^2+t+2t+2
=t(T+1)+2(T+1)
=(T+1)(t+2)
=(x^2+x+1)(x^2+x+2)
(x+1)(x+2)(x+3)(x+4)-8
=(x+1)(x+4)(x+2)(x+3)-8
=(x^2+4x+x+4)(x^2+3x+2x+6)-8
=(x^2+5x+4)(x^2+5x+6)-8
đặt x^2+5x=t
suy ra (T+4)(t+6)-8
=t^2+6t+4t+25-8
=t^2+10t+16
=t^2+2t+8t+16
=t(T+2)+8(t+2)
=(T+2)(t+8)
=(x^2+5x+2)(x^2+5x+8)