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TJ
1
23 tháng 5 2022
1: \(x^4-4+2x^3-4x\)
\(=\left(x^2-2\right)\left(x^2+2\right)+2x\left(x^2-2\right)\)
\(=\left(x^2-2\right)\left(x^2+2x+2\right)\)
4: \(-6x^3+18x^2+60x\)
\(=-6x\left(x^2-3x-10\right)\)
\(=-6x\left(x-5\right)\left(x+2\right)\)
6: \(x^4+x^3-5x^2-5x\)
\(=x\left(x^3+x^2-5x-5\right)\)
\(=x\left(x+1\right)\left(x^2-5\right)\)
KT
16 tháng 7 2018
1) \(x^6-x^4-9x^3+9x^2\)
\(=x^2\left(x^4-x^2-9x+9\right)\)
\(=x^2\left[x^2\left(x^2-1\right)-9\left(x-1\right)\right]\)
\(=x^2\left(x-1\right)\left[x^2\left(x+1\right)-9\right]\)
\(=x^2\left(x-1\right)\left(x^3+x^2-9\right)\)
2) \(x^4-4x^3+8x^2-16x+16\)
\(=x^2\left(x^2+4\right)-4x\left(x^2+4\right)+4\left(x^2+4\right)\)
\(=\left(x^2+4\right)\left(x-2\right)^2\)
18 tháng 7 2018
3) \(x^4-25x^2+20x-4=x^4+5x^3-2x^2-5x^3-25x^2+10x+2x^2+10x-4\)
\(=x^2\left(x^2+5x-2\right)-5x\left(x^2+5x-2\right)+2\left(x^2+5x-2\right)\)
\(=\left(x^2+5x-2\right)\left(x^2-5x+2\right)\)
4) \(5x\left(x-2y\right)+2\left(2y-x\right)^2\)\(=5x\left(x-2y\right)+2\left(x-2y\right)^2=\left(x-2y\right)\left(5x+2x-4y\right)=\left(x-2y\right)\left(7x-4y\right)\)
5) \(x^2\left(x^2-6\right)-x^2+9=x^4-7x^2+9\)
\(=x^4+x^3-3x^2-x^3-x^2+3x-3x^2-3x+9\)
\(=x^2\left(x^2+x-3\right)-x\left(x^2+x-3\right)-3\left(x^2+x-3\right)\)
\(=\left(x^2+x-3\right)\left(x^2-x-3\right)\)
6) \(7x\left(y-4\right)^2-\left(4-y\right)^3=7x\left(y-4\right)^2+\left(y-4\right)^3=\left(y-4\right)^2\left(7x+y-4\right)\)
7) \(x^3+2x^2-6x-27=x^3-3x^2+5x^2-15x+9x-27\)
\(=x^2\left(x-3\right)+5x\left(x-3\right)+9\left(x-3\right)=\left(x-3\right)\left(x^2+5x+9\right)\)
CN
24 tháng 12 2017
a) (x3 + 8y3) : (2y + x)
= (x + 2y)(x2 - 2xy + 4y2) : (2y + x)
= x2 - 2xy + 4y2
b) (x3 + 3x2y + 3xy2 + y3) : (2x + 2y)
= (x + y)3 : 2(x + y)
= \(\dfrac{\left(x+y\right)^2}{2}\)
c) (6x5y2 - 9x4y3 + 15x3y4) : 3x3y2
= 3x3y2(2x2 - 3xy + 5y2) : 3x3y2
= 2x2 - 3xy + 5y2
28 tháng 1 2020
Bài 1:
a)\(5x^2y^3-25x^3y^4+10x^3y^3=5x^2y^3\left(1-5xy+2x\right)\)
b)\(x^3-2xy-x^2y+2y^2=\left(x^3-x^2y\right)-\left(2xy-2y^2\right)=x^2\left(x-y\right)-2y\left(x-y\right)=\left(x-y\right)\left(x^2-2y\right)\)
c)Đề sai hoàn toàn
d) \(2x^2+4xy+2y^2-8z^2=2\left(x^2+2xy+y^2-4z^2\right)=2\left[\left(x+y\right)^2-\left(2z\right)^2\right]=2\left(x+y-2z\right)\left(x+y+2z\right)\)e) \(3x-3a+yx-ya=3\left(x-a\right)+y\left(x-a\right)=\left(x-a\right)\left(3+y\right)\)
f)\(\left(x^2+y^2\right)^2-4x^2y^2=\left(x-y\right)^2\left(x+y\right)^2\)
g)\(2x^2-5x+2=2x^2-x-4x+2=x\left(2x-1\right)-2\left(2x-1\right)=\left(2x-1\right)\left(x-2\right)\)
i)\(14x\left(x-y\right)-21y\left(y-x\right)+28z\left(x-y\right)=14x\left(x-y\right)+21y\left(x-y\right)+28z\left(x-y\right)=7\left(x-y\right)\left(2x+3y+4z\right)\)
15 tháng 11 2019
6) c) x3 - x2 + x = 1
<=> x3 - x2 + x - 1 = 0
<=> (x3 - x2) + (x - 1) = 0
<=> x2 (x - 1) + (x - 1) = 0
<=> (x - 1) (x2 + 1) = 0
=> x - 1 = 0 hoặc x2 + 1 = 0
* x - 1 = 0 => x = 1
* x2 + 1 = 0 => x2 = -1 => x = -1
Vậy x = 1 hoặc x = -1
KN
15 tháng 11 2019
Bài 5:
a) Đặt \(A=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=3^{32}-1\)
\(\Rightarrow A=\frac{3^{32}-1}{8}\)
b) (7x+6)2 + (5-6x)2 - (10-12x)(7x+6)
=(7x+6)2 + (5-6x)2 - 2(5-6x)(7x+6)
\(=\left(7x+6-5+6x\right)^2\)
\(=\left(13x+1\right)^2\)
23 tháng 2 2020
Bài 1 : Khai triển :
a, \(\left(x+5\right)^2=x^2+10x+25\)
b, \(\left(x-3y\right)^2=x^2-6xy+9y^2\)
c, \(\left(x^2-6z\right)\left(x^2+6z\right)=x^4-36z^2\)
d, \(\left(x+3y\right)^3=x^3+9x^2y+27xy^2+27y^3\)
e, \(27x^3-9y^2+y-\frac{1}{27}=\left(3x-\frac{1}{3}\right)^3\)
g, \(8x^6+12x^4y+6x^2y^2+y^3=\left(2x^2+y\right)\)
h, \(4x^2+12x^4y+6x^22y^2+y^3=\left(\sqrt[3]{4x^2}+y\right)\)