27+27x+9x62+x^3
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\(1-27x^3\)
\(=1-\left(3x\right)^3\)
\(=\left(1-3x\right)\left(1+3x+9x^2\right)\)
\(---\)
\(x-3^3+27\)
\(=x-27+27=x\)
\(---\)
\(27x^3+27x^2+9x+1\)
\(=\left(3x\right)^3+3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2+1^3\)
\(=\left(3x+1\right)^3\)
\(---\)
\(\dfrac{x^6}{27}-\dfrac{x^4y}{3}+x^2y^2-y^3\) (sửa đề)
\(=\left(\dfrac{x^2}{3}\right)^3-3\cdot\left(\dfrac{x^2}{3}\right)^2\cdot y+3\cdot\dfrac{x^2}{3}\cdot y^2-y^3\)
\(=\left(\dfrac{x^2}{3}-y\right)^3\)
#Ayumu
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)
b: \(x^3+\dfrac{1}{27}=\left(x+\dfrac{1}{3}\right)\left(x^2-\dfrac{1}{3}x+\dfrac{1}{9}\right)\)
c: \(x^3-3x^2+3x-1=\left(x-1\right)^3\)
e: \(a^2y^2-2axby+b^2x^2\)
\(=\left(ay\right)^2-2\cdot ay\cdot bx+\left(bx\right)^2\)
\(=\left(ay-bx\right)^2\)
f: \(100-\left(3x-y\right)^2\)
\(=\left(10-3x+y\right)\left(10+3x-y\right)\)
g: \(64x^2-\left(8a+b\right)^2\)
\(=\left(8x\right)^2-\left(8a+b\right)^2\)
\(=\left(8x-8a-b\right)\left(8x+8a+b\right)\)
a: \(27-27x+9x^2-x^3=\left(3-x\right)^3\)
b: \(125-x^3=\left(5-x\right)\left(25+5x+x^2\right)\)
\(x^3-9x^2+27x-27\)
=\(x^3-3.x^2.3+3.x.3^3-3^3\)
\(\left(x-3\right)^3\)
Thay x=1, ta có:
(1-3)3=-23=-8
Đề yêu cầu gì vậy bạn???
\(27+27x+9x^2+x^3\)
\(=x^3+3\cdot x^2\cdot3+3\cdot x\cdot3^2+3^3\)
\(=\left(x+3\right)^3\)