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a) \(x^5-2x^4+3x^3-4x^2+2\)
\(=x^5-x^4-x^4+x^3+2x^3-2x^2-2x^2+2\)
\(=x^4\left(x-1\right)-x^3\left(x-1\right)+2x^2\left(x-1\right)-2\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x^4-x^3+2x^2-2x-2\right)\)
b) \(x^4+1997x^2+1996x+1997\)
\(=\left(x^4+x^2+1\right)+1996\left(x^2+x+1\right)\)
\(=\left(x^2-x+1\right)\left(x^2+x+1\right)+1996\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1997\right)\)
c) \(x^8+x^4+1\)
\(=x^8+2x^4+1-x^4\)
\(=\left(x^4+1\right)-x^4\)
\(=\left(x^4-x^2+1\right)\left(x^4+x^2+1\right)\)
\(=\left(x^4-x^2+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)\)
c) \(x^5+x+1\)
\(=x^5-x^2+x^2+x+1\)
\(=x^2\left(x^3-1\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
a) x4 + 1997x2 + 1996x +1997
= x4 + 1997x2 + 1997x - x +1997
=(x4-x) + (1997x2 +1997x+1997)
=x(x3-1) + 1997(x2+x+1)
=x(x-1)(x2+x+1) + 1997(x2+x+1)
=(x2+x+1)(x2-x) + 1997(x2+x+1)
=(x2+x+1)(x2-x+1997)
b) x2 -x -2001.2002
=x2 - x -20022 +2002
=(x2-20022)-(x-2002)
=(x-2002)(x+2002) - (x-2002)
=(x-2002)(x+2002+1)
=(x-2002)(x+2003)
c)x8 + 98x4 +1
= (x8+2x4+1) + 96x4
= (x4+1)2 + 96x4
=[(x4+1)2 + 2.(x4+1).8 + 64x4 ]+[32x4 - 16x2(x4+1)]
=(x4+1+8x2)-16x2(-2x2+x4+1)
=(x4+8x2+1)2- 16x2(x2-1)2
=(x4 + 8x2 +1)2- [4x(x2-1)]2
=(x4+8x2+1)2 - (4x3-4x)2
=(x4-4x3+8x2+4x+1)(x4+4x3+8x2-4x+1)
Bạn tự làm cho trung thực đừng dựa vào người khác
Nếu ai thấy những gì mình nói là đúng thì nhớ k nha
Thanks
a) \(x^4+1997x^2+1996x+1997\)
\(=\left(x^4-x\right)+\left(1997x^2+1997x+1997\right)\)
\(=x\left(x^3-1\right)+1997\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+1997\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left[x\left(x-1\right)+1997\right]\)
\(=\left(x^2+x+1\right)\left(x^2-x+1997\right)\)
b) \(x^2-x-2015.2016\)
\(=x^2-2016x+2015x-2015.2016\)
\(=\left(x^2-2016x\right)+\left(2015x-2015.2016\right)\)
\(=x\left(x-2016\right)+2015\left(x-2016\right)\)
\(=\left(x-2016\right)\left(x+2015\right)\)
Bài 3 :
Ta có : \(A=x^2+x+2012\)
=> \(A=x^2+x+\left(\frac{1}{2}\right)^2+\frac{8047}{4}\)
=> \(A=\left(x+\frac{1}{2}\right)^2+\frac{8047}{4}\)
- Ta thấy : \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
=> \(\left(x+\frac{1}{2}\right)^2+\frac{8047}{4}\ge\frac{8047}{4}\forall x\)
- Dấu "=" xảy ra <=> \(x+\frac{1}{2}=0\)
<=> \(x=-\frac{1}{2}\)
Vậy MinA = \(\frac{8047}{4}\) <=> x = \(-\frac{1}{2}\) .
Bài 1 :
a, Ta có : \(\left(3x-2\right)\left(4+5x\right)=0\)
=> \(\left[{}\begin{matrix}3x-2=0\\4+5x=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}3x=2\\5x=-4\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{2}{3}\\x=-\frac{4}{5}\end{matrix}\right.\)
Vậy phương trình có nghiệm là x = \(\frac{2}{3}\), x = \(-\frac{4}{5}\) .
b,- ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ne0\\x+1\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)
=> \(x\ne\pm1\)
Ta có : \(\frac{x+1}{x-1}-\frac{4}{x+1}=\frac{3-x^2}{1-x^2}\)
=> \(\frac{\left(x+1\right)^2}{x^2-1}-\frac{4\left(x-1\right)}{x^2-1}=\frac{x^2-3}{x^2-1}\)
=> \(\left(x+1\right)^2-4\left(x-1\right)=x^2-3\)
=> \(x^2+2x+1-4x+4=x^2-3\)
=> \(-2x=-3-5\)
=> \(x=4\left(TM\right)\)
Vậy phương trình có nghiệm là x = 4 .
c, Ta có : \(\frac{10x+3}{2009}+\frac{10x-1}{2013}=\frac{10x+1}{2011}-\frac{2-10x}{2014}\)
=> \(\frac{10x+3}{2009}+\frac{10x-1}{2013}=\frac{10x+1}{2011}+\frac{10x-2}{2014}\)
=> \(\frac{10x+3}{2009}+1+\frac{10x-1}{2013}+1=\frac{10x+1}{2011}+1+\frac{10x-2}{2014}+1\)
=> \(\frac{10x+3}{2009}+\frac{2009}{2009}+\frac{10x-1}{2013}+\frac{2013}{2013}=\frac{10x+1}{2011}+\frac{2011}{2011}+\frac{10x-2}{2014}+\frac{2014}{2014}\)
=> \(\frac{10x+2012}{2009}+\frac{10x+2012}{2013}=\frac{10x+2012}{2011}+\frac{10x+2012}{2014}\)
=> \(\frac{10x+2012}{2009}+\frac{10x+2012}{2013}-\frac{10x+2012}{2011}-\frac{10x+2012}{2014}=0\)
=> \(\left(10x+2012\right)\left(\frac{1}{2009}+\frac{1}{2013}-\frac{1}{2011}-\frac{1}{2014}\right)=0\)
=> \(10x+2012=0\)
=> \(x=-\frac{2012}{10}\)
Vậy phương trình có nghiệm là x = \(-\frac{2012}{10}\) .
Bài 3:
Giải:
Ta có : A = x2 + x + 2012
= x2 + 2.\(\frac{1}{2}\).x + \(\frac{1}{4}\) + \(\frac{8047}{4}\)
= (x + \(\frac{1}{2}\))2 + \(\frac{8047}{4}\) ≥ \(\frac{8047}{4}\)
⇒ Amin = \(\frac{8047}{4}\) ⇔ (x + \(\frac{1}{2}\))2 = 0 ⇔ x = \(-\frac{1}{2}\)
Vậy Amin = \(\frac{8047}{4}\) tại x = \(-\frac{1}{2}\)
Chúc bạn học tốt@@
a)\(2+\frac{3}{x-5}=1\)
\(\Rightarrow\frac{3}{x-5}=-1\)
\(\Rightarrow3=-x+5\)
\(\Leftrightarrow x+3=5\)
\(\Rightarrow x=2\)
\(a,\frac{15-x}{2000}+\frac{14-x}{2001}=\frac{13-x}{2002}+\frac{12-x}{2003}\)
\(\Leftrightarrow\frac{15-x}{2000}+1+\frac{14-x}{2001}+1=\frac{13-x}{2002}+1+\frac{12-x}{2003}+1\)
\(\Leftrightarrow\frac{15-x+2000}{2000}+\frac{14-x+2001}{2001}=\frac{13-x+2002}{2002}+\frac{12-x+2003}{2003}\)
\(\Leftrightarrow\frac{2015-x}{2000}+\frac{2015-x}{2001}=\frac{2015}{2002}+\frac{2015-x}{2003}\)
\(\Leftrightarrow\left(2015-x\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
mà \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}>0\)
\(\Leftrightarrow2015-x=0\)
\(\Leftrightarrow x=2015\)
KL : PT có nghiệm \(S=\left\{2015\right\}\)
\(x^4+1997x^2+1996x+1997=0\)
\(\Leftrightarrow\left(x^4-x\right)+1997\left(x^2+x+1\right)=0\)
\(\Leftrightarrow x\left(x^3-1\right)+1997\left(x^2+x+1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x^2+x+1\right)+1997\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x^2-x+1997\right)\left(x^2+x+1\right)=0\)
\(\hept{\begin{cases}x^2-x+1997>0\\x^2+x+1>0\end{cases}}\Rightarrow ptvn\)
\(x^2-x+2011.2012=0\)
\(\Leftrightarrow x^2+2011x-2012x+2011.2012=0\)
\(\Leftrightarrow x\left(x+2011\right)-2012\left(x+2011\right)=0\Leftrightarrow\left(x-2012\right)\left(x+2011\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2012=0\\x+2011=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2012\\x=-2011\end{cases}}\)
câu b) đề sai nhé,ở trên mk nhầm
c)
\(x^5=x^4+x^3+x^2+x+2\)
\(\Leftrightarrow x^5-x^4-x^3-x^2-x-2=0\)
\(\Leftrightarrow x^5-2x^4+x^4-2x^3+x^3-2x^2+x^2-2x+x-2=0\)
\(\Leftrightarrow x^4\left(x-2\right)+x^3\left(x-2\right)+x^2\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^4+x^3+x^2+1\right)\left(x-2\right)=0\Leftrightarrow x=2\)