Phân tích đa thức thành nhân tử:
a/ 4x4+81
b/ 64x4+y4
c/ (2x2-4)2+9
K
Khách
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24 tháng 8 2019
a. = \(\left(x^3+x^2\right)+\left(7x^2+7x\right)+\left(10x+10\right)\)
= \(x^2\left(x+1\right)+7x\left(x+1\right)+10x\left(x+1\right)\)
= \(\left(x+1\right)\left(x^2+7x+10x\right)\)
= \(\left(x+1\right)\left(x+2\right)\left(x+5\right)\)
9 tháng 7 2023
64x^4+y^4
=64x^4+16x^2y^2+y^4-16x^2y^2
=(8x^2+y^2)^2-(4xy)^2
=(8x^2-4xy+y^2)(8x^2+4xy+y^2)
KN
4 tháng 8 2019
a) \(6x^2-11x+3\)
\(=6x^2-9x-2x+3\)
\(=3x\left(2x-3\right)-\left(2x-3\right)\)
\(=\left(3x-1\right)\left(2x-3\right)\)
KN
4 tháng 8 2019
b) \(2x^2+3x-27\)
\(=2x^2-6x+9x-27\)
\(=2x\left(x-3\right)+9\left(x-3\right)\)
\(=\left(2x+9\right)\left(x-3\right)\)
CM
31 tháng 1 2018
a) x6 – x4 + 2x3 + 2x2
= x2(x4 – x2 + 2x + 2)
= x2[x2(x2 – 1) + 2(x + 1)]
= x2. [x2.(x -1).(x + 1) + 2(x+ 1)]
= x2 (x+ 1).[x2(x- 1)+ 2]
= x2(x + 1)(x3 – x2 + 2)
= x2(x + 1)[(x3 + 1) – (x2 – 1)]
= x2(x + 1).[(x + 1).(x2 – x + 1) - (x - 1).(x + 1)]
= x2(x + 1)(x + 1)( x2 – x + 1 – x + 1)
= x2(x + 1)2(x2 – 2x + 2).
b) 4x4 + y4 = 4x4 + 4x2y2 + y4 - 4x2y2
= (2x2 + y2)2 - (2xy)2
= (2x2 + y2 + 2xy)(2x2 + y2 - 2xy)
AH
Akai Haruma
Giáo viên
4 tháng 9 2021
Lời giải:
a.
$=(x^2)^2+(\frac{1}{2}y^4)^2+2.x^2.\frac{1}{2}y^4-x^2y^4$
$=(x^2+\frac{1}{2}y^4)^2-(xy^2)^2$
$=(x^2+\frac{1}{2}y^4-xy^2)(x^2+\frac{1}{2}y^4+xy^2)$
b.
$=(\frac{1}{2}x^2)^2+(y^4)^2+2.\frac{1}{2}x^2.y^4-x^2y^4$
$=(\frac{1}{2}x^2+y^4)^2-(xy^2)^2$
$=(\frac{1}{2}x^2+y^4-xy^2)(\frac{1}{2}x^2+y^4+xy^2)$
c.
$=(8x^2)^2+(y^2)^2+2.8x^2.y^2-16x^2y^2$
$=(8x^2+y^2)^2-(4xy)^2=(8x^2+y^2-4xy)(8x^2+y^2+4xy)$
d.
$=\frac{64x^4+y^4}{64}=\frac{1}{64}(8x^2+y^2-4xy)(8x^2+y^2+4xy)$
4 tháng 9 2021
c: \(64x^4+y^4\)
\(=64x^4+16x^2y^2+y^4-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)
\(=\left(8x^2+y^2-4xy\right)\left(8x^2+y^2+4xy\right)\)
KV
17 tháng 7 2023
a) 6x² + 7xy + 2y²
= 6x² + 4xy + 3xy + 2y²
= (6x² + 4xy) + (3xy + 2y²)
= 2x(3x + 2y) + y(3x + 2y)
= (3x + 2y)(2x + y)
b) x² - y² + 10x - 6y + 16
= x² + 10x + 25 - y² - 6y - 9
= (x² + 10x + 25) - (y² + 6y + 9)
= (x + 5)² - (y + 3)²
= (x + 5 - y - 3)(x + 5 + y + 3)
= (x - y + 2)(x + y + 8)
c) 4x⁴ + y⁴
= 4x⁴ + 4x²y² + y⁴ - 4x²y²
= (2x² + y²)² - (2xy)²
= (2x² + y² - 2xy)(2x² + y² + 2xy)
6 tháng 6 2017
1) \(x^2-7x+6=x^3+1-7x-7=\left(x^3+1\right)-7\left(x+1\right)=\left(x+1\right)\left(x^2-x-6\right)\)
2) \(x^3-9x^2+6x+16\)
\(\left(x^3+1\right)-\left[\left(9x^2-6x+1\right)-16\right]\)
\(=\left(x^3+1\right)-\left[\left(3x-1\right)^2-16\right]=\left(x^3+1\right)-\left(3x-1+4\right)\left(3x-1-4\right)\)\(=\left(x^3+1\right)-3\left(3x-5\right)\left(x+1\right)\)\(=\left(x+1\right)\left[x^2-x+1-9x+15\right]=\left(x+1\right)\left(x^2-10x+16\right)\)
\(=\left(x+1\right)\left[x\left(x-2\right)-8\left(x-2\right)\right]\)\(\left(x+1\right)\left(x-2\right)\left(x-8\right)\)
3) \(x^3-6x^2-x+30\)
\(=x^3-5x^2-x^2+5x-6x+30\)
\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-x-1\right)\)
4) \(2x^3-x^2+5x+3=\left(2x^3+x^2\right)-\left(2x^2+x\right)+\left(6x+3\right)\)
\(=x^2\left(2x+1\right)-x\left(2x+1\right)+3\left(2x+1\right)\)
\(=\left(2x+1\right)\left(x^2-x+3\right)\)
5) \(27x^3-27x^2+18x-4=\left(27x^3-1\right)-\left(27x^2-18x+3\right)\)
\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(9x^2-6x+1\right)\)
\(=\left(3x-1\right)\left(9x^2+3x+1\right)-3\left(3x-1\right)^2\)
\(=\left(3x-1\right)\left(9x^2+3x+1-9x+3\right)=\left(3x-1\right)\left(9x^2-6x+4\right)\)
gửi phần này trước còn lại làm sau !!! tk mk nka !!!
26 tháng 12 2021
a: =(x+6)(x-1)
n: \(=4x^4+36x^2+81-36x^2\)
\(=\left(2x^2+9-6x\right)\left(2x^2+9+6x\right)\)
16 tháng 11 2021
\(1,\\ a,=6x^4-15x^3-12x^2\\ b,=x^2+2x+1+x^2+x-3-4x=2x^2-x-2\\ c,=2x^2-3xy+4y^2\\ 2,\\ a,=7x\left(x+2y\right)\\ b,=3\left(x+4\right)-x\left(x+4\right)=\left(3-x\right)\left(x+4\right)\\ c,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ d,=x^2-5x+3x-15=\left(x-5\right)\left(x+3\right)\\ 3,\\ a,\Leftrightarrow3x\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
16 tháng 11 2021
Câu 1
a)\(3x^2\left(2x^2-5x-4\right)=6x^4-15x^3-12x^2\)
b)\(\left(x+1\right)^2+\left(x-2\right)\left(x+3\right)-4x=x^2+2x+1+x^2+3x-2x-6-4x=2x^2-x-5\)
a) Ta có: \(4x^4+81\)
\(=\left(2x^2\right)^2+9^2+2\cdot2x^2\cdot9-36x^2\)
\(=\left(2x^2+9\right)^2-\left(6x\right)^2\)
\(=\left(2x^2-6x+9\right)\left(2x^2+6x+9\right)\)
b) Ta có: \(64x^4+y^4\)
\(=\left(8x^2\right)^2+\left(y^2\right)^2+2\cdot8x^2\cdot y^2-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)
\(=\left(8x^2-4xy+y^2\right)\left(8x^2+4xy+y^2\right)\)
c) Ta có: \(\left(2x^2-4\right)^2+9\)
\(=4x^4-16x^2+16+9\)
\(=4x^4-16x^2+25\)
\(=4x^4+20x^2+25-36x^2\)
\(=\left(2x^2+5\right)^2-\left(6x\right)^2\)
\(=\left(2x^2-6x+5\right)\left(2x^2+6x+5\right)\)
Lời giải:
a) $4x^4+81=(2x^2)^2+9^2=(2x^2)^2+9^2+2.2x^2.9-36x^2$
$=(2x^2+9)^2-(6x)^2=(2x^2+9-6x)(2x^2+9+6x)$
b)
$64x^4+y^4=(8x^2)^2+(y^2)^2+2.8x^2.y^2-16x^2y^2$
$=(8x^2+y^2)^2-(4xy)^2=(8x^2+y^2-4xy)(8x^2+y^2+4xy)$
c) Biểu thức không phân tích được thành nhân tử.