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Ta có
a, x2-x-y2-y
=x2-y2-(x+y)
=(x-y)(x+y) - (x+y)
=(x+y)(x-y-1)
b, x2-2xy+y2-z2
=(x-y)2-z2
=(x-y-z)(x-y+z)
![](https://rs.olm.vn/images/avt/0.png?1311)
Dài 166
b) 2x2+3x-27=2x2-6x+9x-27=2x(x-3)+9(x-3)=(x-3)(2x+9)
![](https://rs.olm.vn/images/avt/0.png?1311)
( 3x - 1 )2 - 16
= ( 3x - 1 )2 - 42
= ( 3x - 1 - 4 )( 3x - 1 + 4 )
= ( 3x - 5 )( 3x + 3 )
= 3( 3x - 5 )( x + 1 )
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\left(3x-1\right)^2-4^2=0\)
\(\Leftrightarrow\left(3x-1-4\right)\left(3x-1+4\right)=0\)
\(\Leftrightarrow\left(3x-5\right)\left(3x+3\right)=0\Leftrightarrow\orbr{\begin{cases}3x-5=0\\3x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{3}\\x=-1\end{cases}}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(m^3+27\)
\(\Leftrightarrow m^3+3^3\)
\(\Leftrightarrow\left(m+3\right)\left(m^2-m.3+3^2\right)\)
\(\Leftrightarrow\left(m+3\right)\left(m^2-3m+9\right)\)
b,\(\frac{1}{27}+a^3\)
\(\Leftrightarrow\frac{1}{27}\left(1+27a^3\right)\)
\(\Leftrightarrow\frac{1}{27}.\left(1+3a\right)\left(1-3a+9a^2\right)\)
c,\(\left(a+b\right)^3-c^3\)
\(\Leftrightarrow\left(a+b-c\right)\left[\left(a+b\right)^2+\left(a+b\right)c+c^2\right]\)
\(\Leftrightarrow\left(a+b-c\right)\left(a^2+2ab+b^2+ac+bc+c^2\right)\)
d,\(x^9+1\)
\(\Leftrightarrow\left(x^3+1\right)\left(x^6-x^3+1\right)\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)\left(x^6-x^3+1\right)\)
e,\(x^3+9x^2+27x+27\)
\(\Leftrightarrow x^3+3.x^2.3+3x.9+3^3\)
\(\Leftrightarrow x^3+3x^2.3+3x+3^2+3^3\)
\(\Leftrightarrow\left(x+3\right)^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(x^3-x^2-x-2=x^3-2x^2+x^2-2x+x-2\)
\(=x^2\left(x-2\right)+x\left(x-2\right)+x-2=\left(x-2\right)\left(x^2+x+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
(x2 - 3)2 + 16
= (x2 - 3)2 + 42
= (x2 - 3 + 4)(x2 - 3 - 4)
= (x2 + 1)(x2 - 7)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^3-9x^2+27x-27=-8\)
\(\Rightarrow\left(x-3\right)^3=-8\)
\(\Rightarrow x-3=-2\Rightarrow x=1\)
bn áp dụng HĐT sẽ ra
=(x+3)^3
=x3 + 3x2.3 + 3x.32 + 33