1)
a) \(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\) (1)
Đặt \(x^2+x=t\), khi đó (1) trở thành:
\(t^2+4t=12\)
\(\Leftrightarrow t^2+4t-12=0\)
\(\Leftrightarrow t^2-2t+6t-12=0\)
\(\Leftrightarrow t\left(t-2\right)+6\left(t-2\right)=0\)
\(\Leftrightarrow\left(t-2\right)\left(t+6\right)=0\)
\(\Rightarrow\left(x^2+x-2\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow\left(x^2-x+2x-2\right)\left[\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\right]=0\)
\(\Leftrightarrow x\left(x-1\right)+2\left(x-1\right)=0\) (vì \(\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\ne0;\forall x\))
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)
b) \(\dfrac{x+1}{2008}+\dfrac{x+2}{2007}+\dfrac{x+3}{2006}=\dfrac{x+4}{2005}+\dfrac{x+5}{2004}+\dfrac{x+6}{2003}\)
\(\Leftrightarrow\dfrac{x+1}{2008}+\dfrac{x+2}{2007}+\dfrac{x+3}{2006}-\dfrac{x+4}{2005}-\dfrac{x+5}{2004}-\dfrac{x+6}{2003}=0\)
\(\Leftrightarrow\left(\dfrac{x+1}{2008}+1\right)+\left(\dfrac{x+2}{2007}+1\right)+\left(\dfrac{x+3}{2006}+1\right)-\left(\dfrac{x+4}{2005}-1\right)-\left(\dfrac{x+5}{2004}+1\right)-\left(\dfrac{x+6}{2003}+1\right)=0\)
\(\Leftrightarrow\dfrac{x+2009}{2008}+\dfrac{x+2009}{2007}+\dfrac{x+2009}{2006}-\dfrac{x+2009}{2005}-\dfrac{x+2009}{2004}-\dfrac{x+2009}{2003}=0\)
\(\Leftrightarrow\left(x+2009\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}-\dfrac{1}{2005}-\dfrac{1}{2004}-\dfrac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2009=0\) (vì \(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}-\dfrac{1}{2005}-\dfrac{1}{2004}-\dfrac{1}{2003}\ne0\))
\(\Leftrightarrow x=-2009\)
c) \(\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}=2012\)
\(\Leftrightarrow\left(\dfrac{x-1}{2012}-1\right)+\left(\dfrac{x-2}{2011}-1\right)+\left(\dfrac{x-3}{2010}-1\right)+...+\left(\dfrac{x-2012}{1}-1\right)=0\)
\(\Leftrightarrow\dfrac{x-2013}{2012}+\dfrac{x-2013}{2011}+\dfrac{x-2013}{2010}+...+\dfrac{x-2013}{1}=0\)
\(\Leftrightarrow\left(x-2013\right)\left(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}+...+1\right)=0\)
\(\Leftrightarrow x-2013=0\) (vì \(\dfrac{1}{2012}+\dfrac{1}{2011}+\dfrac{1}{2010}+...+1\ne0\))
\(\Leftrightarrow x=2013\)
d) \(\dfrac{x+7}{2003}+\dfrac{x+6}{2004}+\dfrac{x+5}{2005}+\dfrac{x+2025}{5}=0\)
\(\Leftrightarrow\left(\dfrac{x+7}{2003}+1\right)+\left(\dfrac{x+6}{2004}+1\right)+\left(\dfrac{x+5}{2005}+1\right)+\left(\dfrac{x+2025}{5}-3\right)=0\)
\(\Leftrightarrow\dfrac{x+2010}{2003}+\dfrac{x+2010}{2004}+\dfrac{x+2010}{2005}+\dfrac{x+2010}{5}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\dfrac{1}{2003}+\dfrac{1}{2004}+\dfrac{1}{2005}+\dfrac{1}{5}\right)=0\)
\(\Leftrightarrow x+2010=0\) (vì \(\dfrac{1}{2003}+\dfrac{1}{2004}+\dfrac{1}{2005}+\dfrac{1}{5}\ne0\))
\(\Leftrightarrow x=-2010\)
\(\mathtt{Toru}\)
Bài 2:
a: \(x^3-5x^2+8x-4\)
\(=x^3-x^2-4x^2+4x+4x-4\)
\(=x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-4x+4\right)=\left(x-1\right)\left(x-2\right)^2\)
b: \(x^4-30x^2+31x-30\)
\(=x^4-5x^3+5x^3-25x^2-5x^2+25x+6x-30\)
\(=x^3\left(x-5\right)+5x^2\left(x-5\right)-5x\left(x-5\right)+6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^3+5x^2-5x+6\right)\)
\(=\left(x-5\right)\left(x^3+6x^2-x^2-6x+x+6\right)\)
\(=\left(x-5\right)\left(x+6\right)\left(x^2-x+1\right)\)
c: \(x^8+x^6+x^4+x^2+1\)
\(=x^8+x^7+x^6+x^5+x^4-x^7-x^6-x^5-x^4-x^3+x^6+x^5+x^4+x^3+x^2-x^5-x^4-x^3-x^2-x+x^4+x^3+x^2+x+1\)
\(=\left(x^4+x^3+x^2+x+1\right)\left(x^4-x^3+x^2-x+1\right)\)
d: \(x^2\left(x^4-1\right)\left(x^2+2\right)+1\)
\(=x^2\left(x^2-1\right)\left(x^2+1\right)\left(x^2+2\right)+1\)
\(=\left(x^4+x^2\right)\left(x^4+x^2-2\right)+1\)
\(=\left(x^4+x^2\right)^2-2\left(x^4+x^2\right)+1=\left(x^4+x^2-1\right)^2\)
e: \(\left(a+1\right)\left(a+3\right)\left(a+5\right)\left(a+7\right)+15\)
\(=\left(a^2+8a+7\right)\left(a^2+8a+15\right)+15\)
\(=\left(a^2+8a\right)^2+22\left(a^2+8a\right)+120\)
\(=\left(a^2+8a+10\right)\left(a^2+8a+12\right)=\left(a^2+8a+10\right)\left(a+2\right)\left(a+6\right)\)
Bài 1:
a: \(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)
=>\(\left(x^2+x\right)^2+4\left(x^2+x\right)-12=0\)
=>\(\left(x^2+x+6\right)\left(x^2+x-2\right)=0\)
mà \(x^2+x+6=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}>0\forall x\)
nên \(x^2+x-2=0\)
=>(x+2)*(x-1)=0
=>\(\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
b: \(\dfrac{x+1}{2008}+\dfrac{x+2}{2007}+\dfrac{x+3}{2006}=\dfrac{x+4}{2005}+\dfrac{x+5}{2004}+\dfrac{x+6}{2003}\)
=>\(\dfrac{x+1}{2008}+1+\dfrac{x+2}{2007}+1+\dfrac{x+3}{2006}+1=\dfrac{x+4}{2005}+1+\dfrac{x+5}{2004}+1+\dfrac{x+6}{2003}+1\)
=>\(\dfrac{x+2009}{2008}+\dfrac{x+2009}{2007}+\dfrac{x+2009}{2006}=\dfrac{x+2009}{2005}+\dfrac{x+2009}{2004}+\dfrac{x+2009}{2003}\)
=>x+2009=0
=>x=-2009
c: \(\dfrac{x-1}{2012}+\dfrac{x-2}{2011}+\dfrac{x-3}{2010}+...+\dfrac{x-2012}{1}=2012\)
=>\(\left(\dfrac{x-1}{2012}-1\right)+\left(\dfrac{x-2}{2011}-1\right)+...+\left(\dfrac{x-2012}{1}-1\right)=0\)
=>\(\dfrac{x-2013}{2012}+\dfrac{x-2013}{2011}+\dfrac{x-2013}{2010}+...+\dfrac{x-2013}{1}=0\)
=>\(\left(x-2013\right)\left(\dfrac{1}{2012}+\dfrac{1}{2011}+...+1\right)=0\)
=>x-2013=0
=>x=2013
d: \(\dfrac{x+7}{2003}+\dfrac{x+6}{2004}+\dfrac{x+5}{2005}+\dfrac{x+2025}{5}=0\)
=>\(\left(\dfrac{x+7}{2003}+1\right)+\left(\dfrac{x+6}{2004}+1\right)+\left(\dfrac{x+5}{2005}+1\right)+\left(\dfrac{x+2025}{5}-3\right)=0\)
=>\(\dfrac{x+2010}{2003}+\dfrac{x+2010}{2004}+\dfrac{x+2010}{2005}+\dfrac{x+2010}{5}=0\)
=>x+2010=0
=>x=-2010
3) \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+2008\)
\(=\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]+2008\)
\(=\left(x^2+10x+16\right)\left(x^2+10x+24\right)+2008\) (1)
Đặt \(x^2+10x+21=y\), khi đó (1) trở thành:
\(\left(y-5\right)\left(y+3\right)+2008\)
\(=y^2-2y-15+2008\)
\(=y^2-2y+1993\)
\(=\left(x^2+10x+21\right)^2-2\left(x^2+10x+21\right)+1993\)
Vì \(\left(x^2+10x+21\right)^2-2\left(x^2+10x+21\right)⋮\left(x^2+10x+21\right)\)
nên \(\left(x^2+10x+21\right)^2-2\left(x^2+10x+21\right)+1993\) chia \(x^2+10x+21\) dư 1993
hay số dư phép chia \(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+2008\) cho \(x^2+10x+21\) là 1993.
\(\text{#}\mathtt{Toru}\)
