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\(a,a^3+3a^2+3a+1-27b^3\\ =\left(a+1\right)^3-\left(3b\right)^3\\ =\left(a+1-3b\right)\left[\left(a+1\right)^2+\left(a+1\right)\left(3b\right)+\left(3b\right)^2\right]\\ =\left(a+1-3b\right)\left(a^2+2a+1+3ab+3b+9b^2\right)\)
\(c,x^6-x^4+2x^3+2x^2\\ =x^4\left(x^2-1\right)+2x^2\left(x+1\right)\\ =x^4\left(x+1\right)\left(x-1\right)+2x^2\left(x+1\right)\\ =x^2\left(x+1\right)\left[x^2\left(x-1\right)+2\right]\\ =x^2\left(x+1\right)\left(x^3-x^2+2\right)\)
a)\(x^2+7x+6\)
\(=x^2+6x+x+6\)
\(=x\left(x+6\right)+\left(x+6\right)\)
\(=\left(x+1\right)\left(x+6\right)\)
b)\(x^4+2016x^2+2015x+2016\)
\(=x^4+2016x^2+\left(2016x-x\right)+2016\)
\(=\left(x^4-x\right)+\left(2016x^2+2016x+2016\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2016\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2016\right)\)
Bài 3:
Từ \(a^2+b^2+c^2+3=2\left(a+b+c\right)\)
\(\Rightarrow a^2+b^2+c^2+3-2a-2b-2c=0\)
\(\Rightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)
\(\Rightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\) (1)
Ta thấy:\(\begin{cases}\left(a-1\right)^2\ge0\\\left(b-1\right)^2\ge0\\\left(c-1\right)^2\ge0\end{cases}\)
\(\Rightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\) (2)
Từ (1) và (2) \(\Rightarrow\begin{cases}\left(a-1\right)^2=0\\\left(b-1\right)^2=0\\\left(c-1\right)^2=0\end{cases}\)
\(\Rightarrow\begin{cases}a-1=0\\b-1=0\\c-1=0\end{cases}\)\(\Rightarrow\begin{cases}a=1\\b=1\\c=1\end{cases}\)
\(\Rightarrow a=b=c=1\Rightarrow H=1\cdot1\cdot1+1^{2014}+1^{2015}+1^{2016}=1+1+1+1=4\)
1 ) \(a\left(m+n\right)+b\left(m+n\right)\)
\(=\left(a+b\right)\left(m+n\right)\)
2 ) \(a^2\left(x+y\right)-b^2\left(x+y\right)\)
\(=\left(a^2-b^2\right)\left(x+y\right)\)
\(=\left[\left(a-b\right).\left(a+3\right)\right]\left(x+y\right)\)
3 ) \(6a^2-3a+12ab\)
\(=3a.2a-3a+3a.4b\)
\(=3a.\left(2a-1+4b\right)\)
4 ) \(2x^2y^4-2x^4y^2+6x^3y^3\)
\(=2x^2y^2.y^2-2x^2y^2.x^2+2x^2y^2.3xy\)
\(=2x^2y^2\left(y^2-x^2+3xy\right)\)
5 ) \(\left(x+y\right)^3-x\left(x+y\right)^2\)
\(=\left(x+y\right)^2.\left(x+y-x\right)\)
\(=\left(x+y\right)^2.y\)
1)a(m+n)+b(m+n)
=(a+b)(m+n)
2)a2(x+y)-b2(x+y)
=(a2-b2)(x+y)
3)6a2-3a+12ab
=3a.2a-3a.(1-4b)
=3a.(2a-1+4b)
5)(x+y)3-x(x+y)2
=(x+y)(x+y)2-x(x+y)2
=(x+y)2(x+y-x)
\(x^3-x^2-14x+24\)
\(=x^3-2x^2+x^2-2x-12x+24\)
\(=x^2\left(x-2\right)+x\left(x-2\right)-12\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+x-12\right)\)
\(=\left(x-2\right).\left[x^2+4x-3x-12\right]\)
\(=\left(x-2\right).\left[x\left(x+4\right)-3\left(x+4\right)\right]\)
\(=\left(x-2\right)\left(x+4\right)\left(x-3\right)\)
\(x^4+x^3+2x-4\)
\(=x^4-x^3+2x^3-2x^2+2x^2-2x+4x-4\)
\(=x^3\left(x-1\right)+2x^2\left(x-1\right)+2x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(x-1\right)\left(x^3+2x^2+2x+4\right)\)
\(=\left(x-1\right).\left[x^2\left(x+2\right)+2\left(x+2\right)\right]\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+2\right)\)
\(8x^4-2x^3-3x^2-2x-1\)
\(=8x^4-8x^3+6x^3-6x^2+3x^2-3x+x-1\)
\(=8x^3\left(x-1\right)+6x^2\left(x-1\right)+3x\left(x-1\right)+x-1\)
\(=\left(x-1\right)\left(8x^3+6x^2+3x+1\right)\)
\(=\left(x-1\right)\left[\left(8x^3+1\right)+\left(6x^2+3x\right)\right]\)
\(=\left(x-1\right)\left[\left(2x+1\right)\left(4x^2-2x+1\right)+3x\left(2x+1\right)\right]\)
\(=\left(x-1\right)\left(2x+1\right)\left(4x^2+x+1\right)\)
\(3x^2-7x+2\)
\(=3x^2-6x-x+2\)
\(=3x\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-2\right)\left(3x-1\right)\)
Chúc bạn học tốt.
a, 2x-1 thuộc ước của 2,rồi giải ra
b,c tương tự
d\(\frac{x^2-64-123}{x+8}=\frac{\left(x+8\right)\left(x-8\right)-123}{x+8}=x-8-\frac{123}{X+8}\) .........rồi làm tương tự như câu a,,,,,,,,,,,,còn câu e cũng gần giống câu d
c, \(x^6-x^4+2x^3+2x^2\)
\(=x^2\left(x^4-x^2+2x+2\right)\)
\(=x^2[x^2\left(x-1\right)\left(x+1\right)+2\left(x+1\right)]\)
\(=x^2\left(x+1\right)\left(x^3-x^2+2\right)\)
\(=x^2\left(x+1\right)[x^2\left(x+1\right)-2x\left(x+1\right)+2\left(x+1\right)]\)
\(=x^2\left(x+1\right)^2\left(x^2-2x+2\right)\)
d,
\(2x^3-x^2-1\)
\(=2x^3-2x^2+x^2-x+x-1\)
\(=2x^2\left(x-1\right)+x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(2x^2+x+1\right)\)