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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, \(x^3-3x^2+3x-1=\left(x-1\right)^3\)
b, \(1-9x+27x^2-27x^3=-\left(3x-1\right)^3\)
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e:
Tham khảo: 
a: \(\Leftrightarrow x^2-2x+1+4x^2+4x+4-5x^2+5=0\)
\(\Leftrightarrow2x+10=0\)
hay x=-5
\(x^3+\frac{1}{x^3}=x^3+\left(\frac{1}{x}\right)^3=\left(x+\frac{1}{x}\right)\left(x^2-x+\frac{1}{x^2}\right)\)( x khác 0 )
\(-x^3+9x^2-27x+27=-\left(x^3-9x^2+27x-27\right)=-\left(x-3\right)^3\)
\(\left(xy+1\right)^2-\left(x-y\right)^2=\left(xy+1-x+y\right)\left(xy+1+x-y\right)\)
a.
\(\frac{x^2}{4}+x+3=\frac{x^2}{4}+x+1+2=\left(\frac{x}{2}+1\right)^2+2>0;\forall x\)
b.
\(A=-3x^2+2x-5=-3\left(x^2-2.\frac{1}{3}x+\frac{1}{9}\right)-\frac{14}{3}=-3\left(x-\frac{1}{3}\right)^2-\frac{14}{3}\le-\frac{14}{3}\)
\(A_{max}=-\frac{14}{3}\) khi \(x=\frac{1}{3}\)
c.
Đề thiếu (để ý 2 số hạng cuối)
\(A=x^4-2x^3+x^2+3x^2-6x+3-1\)
\(=\left(x^2-x\right)^2+3\left(x-1\right)^2-1\ge-1\)
\(A_{min}=-1\) khi \(x=1\)
d.
\(27x^2-\frac{9}{2}x+\frac{3}{16}=3\left(9x^2-\frac{3}{2}x+\frac{1}{16}\right)=3\left(3x-\frac{1}{4}\right)^2\)
e.
\(=\left[\left(b+c\right)+a\right]^2+\left[\left(b+c\right)-a\right]^2+\left[a-\left(b-c\right)\right]^2+\left[a+\left(b-c\right)\right]^2\)
\(=2\left(b+c\right)^2+2a^2+2a^2+2\left(b-c\right)^2\)
\(=4a^2+2b^2+4bc+2c^2+2b^2-4bc+2c^2\)
\(=4\left(a^2+b^2+c^2\right)\)
f.
\(\left(a^2+b^2\right)\left(c^2+d^2\right)=a^2c^2+b^2d^2+a^2d^2+b^2c^2\)
\(=\left(a^2c^2+b^2d^2+2ac.bd\right)+\left(a^2d^2+b^2c^2-2ad.bc\right)\)
\(=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)
7, \(27x^3+y^3=\left(3x+y\right)\left(9x^2-3xy+y^2\right)\)
8, \(8x^3-\frac{1}{125}y^3=\left(2x-\frac{1}{5}y\right)\left(4x^2+\frac{2}{5}xy+\frac{1}{25}y^2\right)\)
9, ĐK x >= 0
\(x-2\sqrt{x}-3=x-3\sqrt{x}+\sqrt{x}-3\)
\(=\sqrt{x}\left(\sqrt{x}+1\right)-3\left(\sqrt{x}+1\right)=\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)\)
10, \(-4x^2-4x+10=-\left(4x^2+4x+1\right)+11\)
\(=-\left[\left(2x+1\right)^2-11\right]=-\left(2x+1-\sqrt{11}\right)\left(2x+1+\sqrt{11}\right)\)
11;12 xem lại đề
13, \(-x^3+6xy^2-12xy^2+8y^3=-\left(x^3-6xy^2+12xy^2-8y^3\right)=-\left(x-2y\right)^3\)
Trả lời:
7, \(27x^3+y^3=\left(3x+y\right)\left(9x^2-3xy+y^2\right)\)
8, \(8x^3-\frac{1}{125}y^3=\left(2x-\frac{1}{5}y\right)\left(4x^2+\frac{2}{5}xy+\frac{1}{25}y^2\right)\)
9, \(x-2\sqrt{x}-3\left(ĐK:x\ge0\right)\)
\(=x-3\sqrt{x}+\sqrt{x}-3=\sqrt{x}\left(\sqrt{x}-3\right)+\left(\sqrt{x}-3\right)=\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)\)
10, \(10-4x-4x^2=-\left(4x^2+4x-10\right)=-\left(4x^2+4x+1-11\right)=-\left[\left(2x+1\right)^2-11\right]\)
\(=-\left(2x+1\right)^2+11=-\left[\left(2x+1\right)^2-11\right]=-\left(2x+1-\sqrt{11}\right)\left(2x+1+\sqrt{11}\right)\)
11,sửa đề: \(15x\left(x-3y\right)+20y\left(3y-x\right)=15x\left(x-3y\right)-20y\left(x-3y\right)=5\left(x-3y\right)\left(3x-4y\right)\)
12, \(25x^2-2=\left(5x-\sqrt{2}\right)\left(5x+\sqrt{2}\right)\)
13, sửa đề: \(-x^3+6x^2y-12xy^2+8y^3=-\left(x^3-6x^2y+12xy^2-8y^3\right)=-\left(x-2y\right)^3\)
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!
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
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