voi gia tri nao cua a,b thi da thuc x^3+ax^2+2x+b chia het cho x^2+x+1
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a) \(5x\left(x-1\right)=x-1\)
\(\Rightarrow5x\left(x-1\right)-\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right).\left(5x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\5x-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
Vậy \(x=1\) hoặc \(x=\frac{1}{5}\)
b) \(2\left(x+5\right)-x^2-5x\)
\(\Rightarrow2\left(x+5\right)-x\left(x+5\right)\)
\(\Rightarrow\left(x+5\right).\left(2-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+5=0\\2-x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-5\\x=2\end{cases}}\)
Vậy \(x=-5\)hoặc \(x=2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a+b+c=0=>a+b=-c=>\left(a+b\right)^3=-c^3\)
\(=>\left(a+b\right)^3-\left(-c^3\right)=0=>\left(a+b\right)^3+c^3=0\)
\(=>a^3+b^3+3a^2b+3ab^2+c^3=0=>a^3+b^3+c^3+3ab\left(a+b\right)=0\)
Mà a+b=-c
\(=>a^3+b^3+c^3+3ab.\left(-c\right)=0=>a^3+b^3+c^3-3abc=0\)
\(=>a^3+b^3+c^3=3abc\left(đpcm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tìm x
a) (12x-5)(3x-1)-(18x-1)(2x+3)=5
b) (x+2)(x-3)-(x-2)(x+5)=2(x+3)
c) (2x+3)(2x-1)-(2x+5)-(2x-3)=12
![](https://rs.olm.vn/images/avt/0.png?1311)