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a, \(x^3-3x^2+3x-1=\left(x-1\right)^3\)
b, \(1-9x+27x^2-27x^3=-\left(3x-1\right)^3\)
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\(27x^3-27x^2+9x-1=\left(3x\right)^3-3.\left(3x^2\right).1+3.3x.1+1^3=\left(3x-1\right)^3\)
b) không hiểu đề
Tham khảo~
a/ đề sai chữa lại nha :
\(8+12x+6x^2+x^3=2^3+3.2^2.x+3.2.x^2+x^3=\left(2+x\right)^3\)
b/ đề bị lộn dấu ngay chỗ 3x và 3x^2
\(-x^3-3x^2+3x+1=1+3x-3x^2-x^3=1+3.\left(-1\right)^2.x+3x^2.\left(-1\right)+\left(-x\right)^3=\left(1-x\right)^3\)
c/ \(x^3+9x^2+27x+27=x^3+3.3.x^2+3.3^2.x+3^3=\left(x+3\right)^3\)
T I C K ủng hộ nha
CHÚC BẠN HỌC TỐT
Bài 1:
\(a,27x^3+27x^2+9x+1\)
\(=\left(3x\right)^3+3.\left(3x\right)^2.1+3.3x.1^2+1^3\)
\(=\left(3x+1\right)^3\)
\(b,x^3+3\sqrt{2}x^2y+6xy^2+2\sqrt{2}y^3\)
\(=x^3+3.x^2.\sqrt{2}y+3.x.\left(\sqrt{2}y\right)^2+\left(\sqrt{2}y\right)^3\)
\(=\left(x+\sqrt{2}y\right)^3\)
Bài 2:
\(a,x^3+9x^2+27x+27=0\)
\(\Leftrightarrow\left(x+3\right)^3=0\)
\(\Leftrightarrow x+3=0\Leftrightarrow x=-3\)
\(b,\left(x+1\right)^3-x\left(x-2\right)^2+x-1=0\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3-4x^2+4x+x-1=0\)
\(\Leftrightarrow-x^2+8x=0\)
\(\Leftrightarrow-x\left(x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)
1)
a) = (3x+1)3
b) (x+\(\sqrt{2}\) )3
2)
a)\(x^3+9x^2+27x+27=0\\ \left(x+3\right)^3=0\\ =>x=-3\)
b) Bài cuối bạn tự làm nhé! Mình mắc học bài
# Chúc bạn học tốt !
Lời giải:
a) $x^3+3x^2y+x+3xy^2+y+y^3$
$=(x^3+3x^2y+3xy^2+y^3)+(x+y)$
$=(x+y)^3+(x+y)=(x+y)[(x+y)^2+1]$
b) $x^3+y(1-3x^2)+x(3y^2-1)-y^3$
$=(x^3-3x^2y+3xy^2-y^3)-(x-y)$
$=(x-y)^3-(x-y)=(x-y)[(x-y)^2-1]=(x-y)(x-y-1)(x-y+1)$
c)
$27x^3+27x^2+9x+1=(3x+1)^3$
d)
$x(x+1)^2+x(x-5)-5(x+1)^2$
$=x(x+1)^2-5(x+1)^2+x(x-5)$
$=(x-5)(x+1)^2+x(x-5)=(x-5)[(x+1)^2+x]$
$=(x-5)(x^2+3x+2)=(x-5)(x+1)(x+2)$
a) Ta có: \(x^2+2x+1\)
\(=x^2+2\cdot x\cdot1+1^2\)
\(=\left(x+1\right)^2\)
b) Ta có: \(1-2y+y^2\)
\(=y^2-2\cdot y\cdot1+1^2\)
\(=\left(y-1\right)^2\)
c) Ta có: \(x^3-3x^2+3x-1\)
\(=x^3-x^2-2x^2+2x+x-1\)
\(=x^2\left(x-1\right)-2x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-2x+1\right)\)
\(=\left(x-1\right)^3\)
d) Ta có: \(27+27x+9x^2+x^3\)
\(=x^3+3x^2+6x^2+18x+9x+27\)
\(=x^2\left(x+3\right)+6x\left(x+3\right)+9\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2+6x+9\right)\)
\(=\left(x+3\right)^3\)
e) Ta có: \(8-125x^3\)
\(=2^3-\left(5x\right)^3\)
\(=\left(2-5x\right)\left(4+10x+25x^2\right)\)
f) Ta có: \(64x^3+\frac{1}{8}\)
\(=\left(4x\right)^3+\left(\frac{1}{2}\right)^3\)
\(=\left(4x+\frac{1}{2}\right)\left(16x^2-2x+\frac{1}{4}\right)\)
g) Ta có: \(1-x^2y^4\)
\(=1^2-\left(xy^2\right)^2\)
\(=\left(1-xy^2\right)\left(1+xy^2\right)\)
a) \(x^2+2x+1=x^2+2x.1+1^2=\left(x+1\right)^2\)
b) \(1-2y+y^2=1^2-2y.1+y^2=\left(1-y\right)^2\)
c) \(x^3-3x^2+3x-1=\left(x-1\right)^3\)
d) \(27+27x+9x^2+x^3=3^3+3.3^2x+3.3x^2+x^3=\left(3+x\right)^3\)
e) \(8-125x^3=2^3-\left(5x\right)^3=\left(2-5x\right)\left[2^2+2.5x+\left(5x\right)^2\right]=\left(2-5x\right)\left(4+10x+25x^2\right)\)
f) \(64x^3+\frac{1}{8}=\left(4x\right)^3+\left(\frac{1}{2}\right)^3=\left(4x+\frac{1}{2}\right)\left[\left(4x\right)^2-4x.\frac{1}{2}+\left(\frac{1}{2}\right)^2\right]=\left(4x+\frac{1}{2}\right)\left(16x^2-2x+\frac{1}{4}\right)\)
Ko chắc ạ!
a) \(x^3+6x^2+12x+8\)
\(=\left(x+2\right)^3\)
b) \(x^3-3x^2+3x-1\)
\(=\left(x-1\right)^3\)
c) \(1-9x+27x^2-27x^3\)
\(=-\left(27x^3-27x^2+9x-1\right)\)
\(=-\left(3x-1\right)^3\)
Bài 1:
\(B=\dfrac{1}{9}x^2-2x+9\)
\(=\left(\dfrac{1}{3}x\right)^2-2\cdot\dfrac{1}{3}x\cdot3+3^2=\left(\dfrac{1}{2}x-3\right)^2\)
\(C=x^3-9x^2+27x-27=\left(x-3\right)^3\)
\(D=27x^3+27x^2+9x+1=\left(3x+1\right)^3\)
\(E=\left(x-2y\right)^3\)
a) (3x-1)3 Xem lại đề câu a là -1
b) (x-1)3
c) (1/3 + x)(1/9 -x.1/3 +x2)
d) (0,1-10x)(0,01 + x +100x2)