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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
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: \(64x^2-\left(8a+b\right)^2\)
\(=\left(8x\right)^2-\left(8a+b\right)^2\)
\(=\left(8x-8a-b\right)\left(8x+8a+b\right)\)
\(8-27x^3\)
\(=2^3-\left(3x\right)^3\)
\(=\left(2-3x\right)\left(4+6x+9x^2\right)\)
a) \(8-27x^3=\left(2-x\right)\left(4+6x+9x^2\right)\)
b) \(27+27x+9x^2+x^3=\left(3+x\right)^3\)
c) \(x^3+8y^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
\(x^5-27+x^3-27x^2=0\)
\(< =>\left(x^5+x^3\right)-\left(27x^2+27\right)=0\)
\(< =>x^3\left(x^2+1\right)-27\left(x^2+1\right)=0\)
\(< =>\left(x^2+1\right)\left(x^3-27\right)=0\)
\(< =>\left[{}\begin{matrix}x^2+1=0\\x^3-27=0\end{matrix}\right.< =>\left[{}\begin{matrix}x^2=-1\\x^3=3^3\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}x-v\text{ô}-nghi\text{ệ}m\\x=3\end{matrix}\right.\)
S=\(\left\{R,3\right\}\)
xin lỗi nhé mình kết luận nhầm như thế này mới đúng
S=\(\left\{\varnothing,3\right\}\)
\(x^3+9x^2+27x+27=0\)
\(\Leftrightarrow\left(x+3\right)^3=0\)
\(\Leftrightarrow x+3=0\)
\(\Leftrightarrow x=-3\)
KL:..............
x3 + 9x2 + 27x + 27 = 0
<=> x2 + 3.3.x2 + 3.32.x + 33 = 0
<=> (x + 3)3 = 0
<=> x + 3 = 0
<=> x = -3
x3 + 9x2 + 27x +27 = 0
=> x3 + 3.x2.3 + 3.x.32 + 33 = 0
=>( x + 3)3 = 0
=> x + 3 = 0
=> x = -3
Vậy x = -3
Đề yêu cầu gì vậy bạn???
\(27+27x+9x^2+x^3\)
\(=x^3+3\cdot x^2\cdot3+3\cdot x\cdot3^2+3^3\)
\(=\left(x+3\right)^3\)