giải phương trình
\(a,x^4+1997x^2+1996x+1997=0\)
\(b,x^2-x+2011\cdot2012=0\)
\(c,x^5=x^4+x^3+x^2+x+2\)
K
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KT
31 tháng 1 2018
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) \(x^5+x+1\)
\(=x^5-x^2+x^2+x+1\)
\(=x^2\left(x^3-1\right)+\left(x^2+x+1\right)\)
\(=x^2\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
3 tháng 11 2018
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)
29 tháng 10 2017
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
ST
1
12 tháng 7 2017
bn chép lại đề nha
\(=x^4-x+1997\left(x^2+x+1\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+1997\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1997\right)\)
xong nha. chúc bn hc tốt
YO
1
NN
12 tháng 12 2017
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) \(x^2-x-2015.2016\)
\(=x^2-2016x+2015x-2015.2016\)
\(=\left(x^2-2016x\right)+\left(2015x-2015.2016\right)\)
\(=x\left(x-2016\right)+2015\left(x-2016\right)\)
\(=\left(x-2016\right)\left(x+2015\right)\)
NV
2
8 tháng 9 2017
=(x4+x2+1)+(1996x2+1996x+1996)
=(x2+x+1)(x2-x+1)+1996(x2+x+1)
=(x2+x+1)(x2-x+1+1996)
3 tháng 5 2021
=) vào ngay quả bảng phá dấu GTTĐ, cay thế :<
a, \(3x+\frac{2x}{3}-3=\frac{5}{2}x-2\Leftrightarrow\frac{18x+4x-18}{6}=\frac{15x-12}{6}\)
\(\Rightarrow22x-18=15x-12\Leftrightarrow7x=6\Leftrightarrow x=\frac{6}{7}\)
Vậy pt có nghiệm x = 6/7
b, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}+\frac{x+1}{3}=\frac{x+7}{12}\)
\(\Leftrightarrow\frac{9\left(2x+1\right)-2\left(5x+3\right)+4\left(x+1\right)}{12}=\frac{x+7}{12}\)
\(\Rightarrow18x+9-10x-6+4x+4=x+7\)
\(\Leftrightarrow12x+7=x+7\Leftrightarrow11x=0\Leftrightarrow x=0\)
Vậy pt có nghiệm là x = 0
c, \(\frac{3x}{x-3}-\frac{x-3}{x+3}=2\)ĐK : \(x\ne\pm3\)
\(\Leftrightarrow\frac{3x\left(x+3\right)-\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\frac{2\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow3x^2+9x-x^2+6x-9=2\left(x^2-9\right)\)
\(\Leftrightarrow2x^2+15x-9=2x^2-18\Leftrightarrow15x+9=0\Leftrightarrow x=-\frac{9}{15}=-\frac{3}{5}\)
Vậy pt có nghiệm là x = -3/5
d, Sửa đề : \(\frac{x+10}{2003}+\frac{x+6}{2007}+\frac{x+2}{2011}+3=0\)
\(\Leftrightarrow\frac{x+10}{2003}+1+\frac{x+6}{2007}+1+\frac{x+2}{2011}+1=0\)
\(\Leftrightarrow\frac{x+2013}{2003}+\frac{x+2013}{2007}+\frac{x+2013}{2011}=0\)
\(\Leftrightarrow\left(x+2013\right)\left(\frac{1}{2003}+\frac{1}{2007}+\frac{1}{2011}\ne0\right)=0\Leftrightarrow x=-2013\)
Vậy pt có nghiệm là x = -2013
3 tháng 5 2021
e, \(4\left(x+5\right)-3\left|2x-1\right|=10\)
\(\Leftrightarrow4x+20-3\left|2x-1\right|=10\Leftrightarrow-3\left|2x-1\right|=-10-4x\)
\(\Leftrightarrow\left|2x-1\right|=\frac{10+4x}{3}\)
ĐK : \(\frac{10+4x}{3}\ge0\Leftrightarrow10+4x\ge0\Leftrightarrow x\ge-\frac{10}{4}=-\frac{5}{2}\)
TH1 : \(2x-1=\frac{10+4x}{3}\Rightarrow6x-3=10+4x\Leftrightarrow2x=13\Leftrightarrow x=\frac{13}{2}\)( tm )
TH2 : \(2x-1=\frac{-10-4x}{3}\Rightarrow6x-3=-10-4x\Leftrightarrow10x=-7\Leftrightarrow x=-\frac{7}{10}\)( tm )
f, để mình xem lại đã, quên cách phá GTTĐ rồi :v :>
\(x^4+1997x^2+1996x+1997=0\)
\(\Leftrightarrow\left(x^4-x\right)+1997\left(x^2+x+1\right)=0\)
\(\Leftrightarrow x\left(x^3-1\right)+1997\left(x^2+x+1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x^2+x+1\right)+1997\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x^2-x+1997\right)\left(x^2+x+1\right)=0\)
\(\hept{\begin{cases}x^2-x+1997>0\\x^2+x+1>0\end{cases}}\Rightarrow ptvn\)
\(x^2-x+2011.2012=0\)
\(\Leftrightarrow x^2+2011x-2012x+2011.2012=0\)
\(\Leftrightarrow x\left(x+2011\right)-2012\left(x+2011\right)=0\Leftrightarrow\left(x-2012\right)\left(x+2011\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2012=0\\x+2011=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2012\\x=-2011\end{cases}}\)
câu b) đề sai nhé,ở trên mk nhầm
c)
\(x^5=x^4+x^3+x^2+x+2\)
\(\Leftrightarrow x^5-x^4-x^3-x^2-x-2=0\)
\(\Leftrightarrow x^5-2x^4+x^4-2x^3+x^3-2x^2+x^2-2x+x-2=0\)
\(\Leftrightarrow x^4\left(x-2\right)+x^3\left(x-2\right)+x^2\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^4+x^3+x^2+1\right)\left(x-2\right)=0\Leftrightarrow x=2\)