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4 tháng 10 2021
1) \(\dfrac{1}{27}+a^3=\left(\dfrac{1}{3}+a\right)\left(\dfrac{1}{9}-\dfrac{a}{3}+a^2\right)\)
2) \(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
3) \(=\left(\dfrac{1}{2}x+2y\right)\left(\dfrac{1}{4}x-xy+4y^2\right)\)
4) \(=\left(x^2+1\right)\left(x^4-x^2+1\right)\)
5) \(=\left(x^3+1\right)\left(x^6-x^3+1\right)\)
6) \(=\left(x-4\right)\left(x^2+4x+16\right)\)
7) \(=\left(x-5\right)\left(x^2+5x+25\right)\)
8) \(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)
9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)
10) \(=\left(\dfrac{1}{2}x-2\right)\left(\dfrac{1}{4}x^2+x+4\right)\)
11) \(=\left(x+2\right)^3\)
12) \(=\left(x+3\right)^3\)
12 tháng 1 2023
a: \(=\left(x+1+5\right)\left(x+1-5\right)=\left(x+6\right)\left(x-4\right)\)
b: =(1-2x)(1+2x)
c: \(=\left(2-3x\right)\left(4+6x+9x^2\right)\)
d: =(x+3)^3
e: \(=\left(2x-y\right)^3\)
f: =(x+2y)(x^2-2xy+4y^2)
15 tháng 11 2018
\(1-2y+y^2=\left(y-1\right)^2\)
\(\left(x+1\right)^2-25=\left(x-1\right)^2-5^2=\left(x-6\right)\left(x+4\right)\)
\(1-4x^2=1-\left(2x\right)^2=\left(1-2x\right)\left(1+2x\right)\)
\(8-27x^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
\(27+27x+9x^2+x^3=\left(x+3\right)^3\)
\(8x^3-12x^2y+6xy^2-y^3=\left(2x-y\right)^3\)
\(x^3+8y^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
Tham khảo nhé~
15 tháng 11 2018
Mấy cái này chỉ áp dụng HĐT thoyy nha!
\(a,1-2y+y^2=\left(1-y\right)^2\)
\(b,\left(x-1\right)^2-25=\left(x-1-5\right)\left(x-1+5\right)=\left(x-6\right)\left(x+4\right)\)
\(c,1-4x^2=\left(1-2x\right)\left(1+2x\right)\)
\(d,8-27x^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
\(e,27+27x+9x^2+x^3=\left(x+3\right)^3\)
\(f,8x^3-12x^2y+9xy^2-y^3=\left(2x-y\right)^2\)
\(g,x^3+8y^3=\left(x+2y\right)\left(x^2-2xy+y^2\right)=\left(x+2y\right)\left(x-y\right)^2\)
=.= hok tốt!!
BH
3
H9
HT.Phong (9A5)
CTVHS
31 tháng 7 2023
\(x^3+1\)
\(=x^3+1^3\)
\(=\left(x+1\right)\left(x^2-x+1\right)\)
______
\(8+x^3\)
\(=2^3+x^3\)
\(=\left(2+x\right)\left(4-2x+x^2\right)\)
______
\(27x^3-64y^3\)
\(=\left(3x\right)^3-\left(4y\right)^3\)
\(=\left(3x-4y\right)\left(9x^2+12xy+16y^2\right)\)
______
\(\dfrac{x^3}{64}-\dfrac{1}{125}\)
\(=\left(\dfrac{x}{4}\right)^3-\left(\dfrac{1}{5}\right)^3\)
\(=\left(\dfrac{x}{4}-\dfrac{1}{5}\right)\left(\dfrac{x^2}{16}+\dfrac{x}{20}+\dfrac{1}{25}\right)\)
30 tháng 7 2023
x^3+1=(x+1)(x^2-x+1)
x^3+8=(x+2)(x^2-2x+4)
27x^3-64y^3=(3x-4y)(9x^2+12xy+16y^2)
\(\dfrac{x^3}{64}-\dfrac{1}{25}=\left(\dfrac{1}{4}x-\sqrt[3]{\dfrac{1}{5}}\right)\left(\dfrac{1}{16}x^2+\dfrac{1}{4\sqrt[3]{5}}\cdot x+\dfrac{1}{\sqrt[3]{25}}\right)\)
TP
26 tháng 8 2018
Bài 1 : Phân tích các đa thức sau thành nhân tử :
a) 8x3 - 64
=(2x)3 + 43
=(2x+4)(4x2 - 8x + 16)
c) 125x3 + 1
=5x3 + 13
=(5x+1)(25x2 +5x+1)
d) 8x3 - 27
=(2x)3 - 33
=(2x - 3)(2x2 + 6x + 9)
e) 1 + 8x6y3
=1 + (2x2y)3
=(1 + 2x2y)(4x4y2 -2x2y + 1)
f) 125x3 + 27y3
=(5x)3 + (3y3)
=(5x + 3y)(25x2 - 15xy + 9y2)
DQ
26 tháng 8 2018
Bài 1
a) \(8x^3-64\)
\(=\left(2x\right)^3-4^3\)
\(=\left(2x-4\right)\left(4x^2+8x+16\right)\)
c) \(125x^3+1\)
\(=\left(5x\right)^3+1^3\)
\(=\left(5x+1\right)\left(25x^2-5x+1\right)\)
d) \(8x^3-27\)
\(=\left(2x\right)^3-3^3\)
\(=\left(2x-3\right)\left(4x^2+6x+9\right)\)
e) \(1+8x^6x^3\)
\(=1^3+\left(2x^2y\right)^3\)
\(=\left(1+2x^2y\right)\left(1-2x^2y+4x^4y^2\right)\)
f) \(125x^3+27y^3\)
\(=\left(5x\right)^3+\left(3y\right)^3\)
\(=\left(5x+3y\right)\left(25x^2-15xy+9x^2\right)\)
TB
1
LH
5 tháng 7 2018
1. x3 + 8 = (x + 2 )(x2 - x + 1)
2. 27 - 8y3 = ( 3 - 2y ) ( 9 + 6y + 4y2 )
3. y6 + 1 = (y2)3 + 1 = ( y2 + 1) ( y4 - y2 +1 )
4.64x3 - \(\dfrac{1}{8}\)y3 = ( 4x - \(\dfrac{1}{2}\)y ) ( 16x2 + 2xy + \(\dfrac{1}{4}\)y2)
5. 125x6 - 27y9 = (5x2)3 - (3y3)3
= ( 5x2 - 3y3)(25x4 +15x2y3 + 9y6)
x^3-1=(x-1)(x^2+x+1)
8x^3+1=(2x+1)(4x^2-2x+1)
x^3+1=(x+1)(x^2-x+1)
125-x^3=(5-x)(25+5x+x^2)
x^3+8y^3=x^3+(2y)^3
=(x+2y)(x^2-2xy+4y^2)
64y^3-125x^3
=(4y)^3-(5x)^3
=(4y-5x)(16y^2+20xy+25x^2)
\(27x^3-\dfrac{1}{8}=\left(3x\right)^3-\left(\dfrac{1}{2}\right)^3=\left(3x-\dfrac{1}{2}\right)\left(9x^2+\dfrac{3}{2}x+\dfrac{1}{4}\right)\)
\(a^6-b^3=\left(a^2\right)^3-b^3\)
\(=\left(a^2-b\right)\cdot\left(a^4+a^2b+b^2\right)\)