Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, x^4 - 5x^2 + 4
= x^4 - 4x^2- x+ 4
= x^2 . (x^2 - 4) - (x^2 - 4)
= (x^2 - 4) . (x^2 - 1)
= (x - 2) . (x + 2) . (x - 1) . (x + 1)
\(\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+3\left(x+y\right)z\left(x+y+z\right)+z^3-x^3-y^3-z^3\)
\(=x^3+y^3+z^3+3xy\left(x+y\right)+3\left(x+y\right)z\left(x+y+z\right)-x^3-y^3-z^3\)
\(=3\left(x+y\right)\left(xy+xz+yz+z^2\right)\)
\(=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)
a, \(3x^2-9x-2x+6=3x\left(x-3\right)-2\left(x-3\right)=\left(x-3\right)\left(3x-2\right)\)
b. \(8x^2-2x+12x-3=2x\left(4x-1\right)+3\left(4x-1\right)=\left(4x-1\right)\left(2x+3\right)\)
c. đề kiểu gì vậy? -2x-x để thành -3x à? xem lại đi nha
d. \(\left(x^2+10x+25\right)-\left(y^2+6y+9\right)=\left(x+5\right)^2-\left(y+3\right)^2=\left(x+5-y-3\right)\left(x+5+y+3\right)=\left(x-y+2\right)\left(x+y+8\right)\)
e. \(=x^4+2x^2y^2+y^4-x^2y^2=\left(x^2+y^2\right)^2-x^2y^2=\left(x^2+y^2-xy\right)\left(x^2+y^2+xy\right)\)
nhớ L I K E
x6+3x4y2-8x3y3+3x2y4+y6= x6+3x4y2+3x2y4+y6-8x3y3=(x2+y2)3-(2xy)3
= (x2+y2-2xy)[(x2+y2)2+2xy(x2+y2)+(2xy)2]= (x-y)2(x4+6x2y2+y4+2x3y+2xy3)
(x2+y2-5)2-4x2y2-16xy-16=(x2+y2-5)2-(4x2y2+16xy+16)=(x2+y2-5)2-(2xy+4)2
=(x2+y2-5+2xy+4)(x2+y2-5-2xy-4)=(x2+2xy+y2-1)(x2-2xy+y2-9)=[(x+y)2-1][(x-y)2-32]=(x+y-1)(x+y+1)(x-y-3)(x-y+3)
x4+324=x4+36x2+324-36x2=(x2+18)2-(6x)2=(x2+18-6x)(x2+18+6x)
a) \(x^3-3x^2-3x+1\)
\(=\left(x^3+1\right)-\left(3x^2+3x\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1-3x\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
b) \(4x^2+4x+1-y^2-16y-64\)
\(=\left(2x+1\right)^2-\left(y+8\right)^2\)
\(=\left(2x+1-y-8\right)\left(2x+1+y+8\right)\)
\(=\left(2x-7-y\right)\left(2x+9+y\right)\)
c) \(x^3+3x^2+3x+1-27z^3\)
\(=\left(x+1\right)^3-\left(3z\right)^3\)
\(=\left(x+1-3z\right)\left[\left(x+1\right)^2+3z\left(x+1\right)+9z^2\right]\)
\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)
d) \(\left(x^2+y^2-5\right)^2-4\left(x^2y^2+4xy+4\right)\)
\(=\left(x^2+y^2-4-1\right)^2-4\left(xy+2\right)^2\)
\(=\left(x^2+y^2-5\right)^2-4\left(xy+2\right)^2\)
\(=\left(x^2+y^2-5\right)^2-\left(2xy+4\right)^2\)
\(=\left(x^2+y^2-5-2xy-4\right)\left(x^2+y^2-5+2xy+4\right)\)
\(=\left[\left(x-y\right)^2-9\right]\left[\left(x+y\right)^2-1\right]\)
\(=\left(x-y-3\right)\left(x-y+3\right)\left(x+y-1\right)\left(x+y+1\right)\)
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)
a.\(3x^2-11x+6\)
= \(3x^2-9x-2x+6\)
=\(3x\left(x-3\right)-2\left(x-3\right)\)
=\(\left(x-3\right)\left(3x-2\right)\)
b\(8x^2+10x-3\)
=.\(8x^2-2x+12x-3\)
=\(2x\left(4x-1\right)+3\left(4x-1\right)\)
=\(\left(4x-1\right)\left(2x+3\right)\)
d.\(x^2-y^2+10x-6y+16\)
=\(\left(x^2+10x+25\right)-\left(y^2+6y+9\right)\)
=\(\left(x+5\right)^2-\left(y+3\right)^2\)
=\(\left(x+5-y-3\right)\left(x+5+y+3\right)\)
=\(\left(x-y+2\right)\left(x+y+8\right)\)
e.\(x^4+x^2y^2+y^4\)
=\(x^4+2x^2y^2+y^4-x^2+y^2\)
=\(\left(x^2+y^2\right)^2-x^2y^2\)
=\(\left(x^2+y^2-xy\right)\left(x^2+y^2+xy\right)\)
a)
\(=3x^2-9x-2x+6=3x\left(x-3\right)-2\left(x-3\right)=\left(x-3\right)\left(3x-2\right)\)
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 ;)
\(g,x^4-16=\left(x^2-4\right)\left(x^2+4\right)=\left(x-2\right)\left(x+2\right)\left(x^2+4\right)\\ i,-x^2+10x-25=-\left(x-5\right)^2\\ k,x^3+3x^2+3x+1-27z^3\\ =\left(x+1\right)^3-27z^3\\ =\left(x+1-3z\right)\left[\left(x+1\right)^2+3z\left(x+1\right)+9z^2\right]\\ =\left(x-3z+1\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\\ m,\left(x+y\right)^2-25\left(x+y\right)+24=\left(x+y-5\right)^2-1=\left(x+y-4\right)\left(x+y-6\right)\)
g. x4 - 16
<=> x4 - 42
<=> (x2)2 - 42
<=> (x2 - 4)(x2 + 4)
i. -x2 + 10x - 25
<=> -(x2 - 10x + 25)
<=> -(x2 -10x + 52)
<=> -(x - 5)2