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1, \(x^3+3^3=\left(x+3\right)\left(x^2-3x+9\right)\)
2, đề sai
3, \(x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\)
4, \(x^3-64=\left(x-4\right)\left(x^2+4x+16\right)\)
5, \(1000-y^3=\left(10-y\right)=\left(100+10y+y^2\right)\)
tương tự ...
8, \(8x^3+27y^3=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
Câu 2 đề ko sai nha bạn.
2) x2 - (\(\sqrt{y^3}\))2 ( y>0)
= ( x -\(\sqrt{y^3}\)) ( x +\(\sqrt{y^3}\))
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)
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) \(125x^3+27y^3\)
\(=\left(5x\right)^3+\left(3y\right)^3\)
\(=\left(5x+3y\right)\left(25x^2-15xy+9x^2\right)\)
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)
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) \(=\left(x+3\right)^3\)
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)
\(27x^3-\dfrac{1}{8}=\left(3x\right)^3-\left(\dfrac{1}{2}\right)^3=\left(3x-\dfrac{1}{2}\right)\left(9x^2+\dfrac{3}{2}x+\dfrac{1}{4}\right)\)
\(a^6-b^3=\left(a^2\right)^3-b^3\)
\(=\left(a^2-b\right)\cdot\left(a^4+a^2b+b^2\right)\)
a: Đặt 2x+1=a; 3x-1=b
Phương trình trở thành \(\left(a+b\right)^3-a^3-b^3=0\)
\(\Leftrightarrow3ab\left(a+b\right)=0\)
=>5x(2x+1)(3x-1)=0
hay \(x\in\left\{0;-\dfrac{1}{2};\dfrac{1}{3}\right\}\)
c: Đặt x-3=a; x+1=b
Theo đề, ta có phương trình \(a^3+b^3=\left(a+b\right)^3\)
=>3ab(a+b)=0
=>(x-3)(x+1)(2x-2)=0
hay \(x\in\left\{3;1;-1\right\}\)
a) 8x3 - 64 = (2x)3 - 43
= (2x - 4)\([\)(2x)2 + 2x.4 + 42\(]\)
= (2x - 4)(4x2 + 8x + 16)
b) 1 + 8x6y3
= 13 + (2x2y)3
= (1 + 2x2y)[(2x2y)2 - 2x2y.1 + 12]
= (1 + 2x2y)(4x4y2 - 2x2y + 1)
c) 27x3 + \(\frac{y^3}{8}\)
= (3x)3 + \(\left(\frac{y}{2}\right)^3\)
= \(\left(3x+\frac{y}{2}\right)\left[\left(3x\right)^2-3x.\frac{y}{2}+\left(\frac{y}{2}\right)^2\right]\)
= \(\left(3x-\frac{y}{2}\right)\left(9x^2-\frac{3xy}{2}+\frac{y^2}{4}\right)\)
d) 125x3 + 27y3
= (5x)3 + (3y)3
= (5x + 3y)[(5x)2 - 5x.3y + (3y)2]
= (5x + 3y)(25x2 - 15xy + 9y2)
a: \(\Leftrightarrow\left(\dfrac{1}{3}x-1\right)^3=\left(\dfrac{1}{5}x-1\right)^3\)
=>1/3x-1=1/5x-1
=>2/15x=0
hay x=0
b: Đặt 2x+1=a; 3x-1=b
Theo đề, ta có \(\left(a+b\right)^3-a^3-b^3=0\)
=>3ab(a+b)=0
=>5x(2x+1)(3x-1)=0
hay \(x\in\left\{0;-\dfrac{1}{2};\dfrac{1}{3}\right\}\)
c: Đặt x-3=a; x+1=b
Theo đề, ta có: \(\left(a+b\right)^3=a^3+b^3\)
=>3ab(a+b)=0
=>(x-3)(x+1)(2x-2)=0
hay \(x\in\left\{3;-1;1\right\}\)
1. \(125x^3-1=\left(5x\right)^3-1^3=\left(5x-1\right)\left(25x^2+5x+1\right)\)
2. \(8x^3+125=\left(2x\right)^3+5^3=\left(2x+5\right)\left(4x^2-10x+25\right)\)
3. \(x^3+\frac{y^3}{8}=x^3+\left(\frac{y}{2}\right)^3=\left(x+\frac{y}{2}\right)\left(x^2-\frac{xy}{2}+\frac{y^2}{4}\right)\)