Phân tích đa thức thành nhân tử
(4x2 - 25)2 - 9(4x2 - 9)2
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a/ \(4x^2-9\)
\(=\left(2x-3\right)\left(2x+3\right)\)
b/ \(3x\left(3x-2\right)+1\)
\(=9x^2-6x+1\)
\(=\left(3x-1\right)^2\)
\(a,=\left(2x-3\right)\left(2x+3\right)\)
\(b,=9x^2-6x+1=\left(3x-1\right)^2\)
b) 4x2 – 9 + (2x + 3)2
= (4x2 - 9) + (2x + 3)2
= (2x + 3)(2x - 3) + (2x + 3)2
= (2x + 3)(2x - 3 + 2x + 3)
= 4x(2x + 3)
a) \(x^2-9+2\left(x+3\right)=\left(x-3\right)\left(x+3\right)+2\left(x+3\right)=\left(x+3\right)\left(x-3+2\right)=\left(x+3\right)\left(x-1\right)\)
b) \(x^2-10x+25-3\left(x-5\right)=\left(x-5\right)^2-3\left(x-5\right)=\left(x-5\right)\left(x-5-3\right)=\left(x-5\right)\left(x-8\right)\)
c) \(x^3-4x^2+3x=x\left(x^2-4x+3\right)=x\left(x-1\right)\left(x-3\right)\)
a) $4x^2+4x+1$
$=(2x)^2+2\cdot2x\cdot1+1^2$
$=(2x+1)^2$
b) $x^2+6x-y^2+9$
$=(x^2+6x+9)-y^2$
$=(x^2+2\cdot x\cdot3+3^2)-y^2$
$=(x+3)^2-y^2$
$=(x+3-y)(x+3+y)$
$\text{#}Toru$
a: \(4x^2+4x+1\)
\(=\left(2x\right)^2+2\cdot2x\cdot1+1^2\)
\(=\left(2x+1\right)^2\)
b: \(x^2+6x-y^2+9\)
\(=\left(x^2+6x+9\right)-y^2\)
\(=\left(x+3\right)^2-y^2\)
\(=\left(x+3+y\right)\left(x+3-y\right)\)
b) 4x2 - 25 + (2x + 5)2
= (2x + 5)(2x - 5) + (2x + 5)2
= (2x + 5)(2x - 5 + 2x + 5)
= 4x(2x + 5)
phân tích các đa thức thành nhân tử
a) ( x2 + 4)2- 16x2
b) ( x+3)2- 8x3
c) (4x2-3x -18)2- ( 4x2 + 3x)2
\(a,\left(x^2+4\right)^2-16x^2=\left(x^2+4\right)-\left(4x\right)^2=\left(x^2+4-4x\right).\left(x^2+4+4x\right)=\left(x-2\right)^2.\left(x+2\right)^2\)
\(b,\left(x+3\right)^3-8x^3=\left(x+3\right)^3-\left(2x\right)^3=\left(x+3-2x\right).\left[x^2+\left(x+3\right).2x+\left(2x\right)^2\right]=\left(3-x\right).\left(x^2+2x^2+6x+4x^2\right)\)
\(c,\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2=\left(4x^2-3x-18-4x^2-3x\right).\left(4x^2-3x-18+4x^2+3x\right)=\left(-6x-18\right).\left(8x^2-18\right)\)
\(4x^2-7x-2\\ =4x^2-8x+x-2\\ =4x\left(x-2\right)+\left(x-2\right)\\ =\left(x-2\right)\left(4x+1\right)\)
Ta có:
\(\left(x^4+2x^3-x-2\right)+\left(4x^2+4x+4\right)\)
\(=\left[\left(x^4+2x^3\right)-\left(x+2\right)\right]+4\left(x^2+x+1\right)\)
\(=\left[x^3\left(x+2\right)-\left(x-2\right)\right]+4\left(x^2+x+1\right)\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+1\right)+4\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[\left(x-1\right)\left(x+2\right)+4\right]\)
\(=\left(x^2+x+1\right)\left(x^2+x+2\right)\)
1b.=2((x+y)+(x+y)(x-y)+(x-y))=2(x2-y2+x+y+x-y)=2(x2-y2+2x)=2x2-2y2+4x
2a.=4xy+4xy+2y=8xy+2y=2y(4x+1)
b.=(3x)2+2.3x.y+y2-(2z)2=(3x+y)2-(2z)2=(3x+y-2z)(3x+y+2z)
c.=x2-x-7x+7=x(x-1)-7(x-1)=(x-1)(x-7)
\(\left(x+y\right)^2+2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x+y+x-y\right)^2\)
\(=\left(2x\right)^2\)
\(=4x^2\)
hk tốt
^^
\(\left(4x^2-25\right)^2-9\left(4x^2-9\right)^2\)
\(=\left(4x^2-25\right)^2-\left(12x^2-27\right)^2\)
\(=\left(4x^2-25+12x^2-27\right)\left(4x^2-25-12x^2+27\right)\)
\(=\left(16x^2-52\right)\left(2-8x^2\right)\)
\(=8\left(4x^2-13\right)\left(1-4x^2\right)\)
\(=8\left(4x^2-13\right)\left(1+2x\right)\left(1-2x\right)\)