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SD
1
N
10 tháng 10 2017
x^4-30x^2+31x^2-30=0
=>x^4+x-30x^2+30x-30=0
=>x(x^3+1)-30(x^2-x+1)=0
=>x(x+1)(x^2-x+1)-30(x^2-x+1)=0
=>(x^2-x+1)(x^2+x-30)=0
=>(x^2-x+1)(x^2-5x+6x-30)=0
=>(x^2-x+1)[x(x-5)+6(x-5)]=0
=>(x^2-x+1)(x-5)(x+6)=0
vì x^2-x+1=x^2-2x.(1/2)+1/4+3/4
=(x-1/2)^2+3/4>0 với mọi x
Do đó x-5=0 hoac x+6=0
=>x=5 hoặc x=-6
NN
22 tháng 11 2017
\(x^4-30x^2+31x-30=0\)
\(\Leftrightarrow x^4+x-30x^2+30x-30=0\)
\(\Leftrightarrow\left(x^4+x\right)-\left(30x^2-30x+30\right)=0\)
\(\Leftrightarrow x\left(x^3+1\right)-30\left(x^2-x+1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left[x\left(x+1\right)-30\right]=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2+x-30\right)=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2-5x+6x-30\right)=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left[\left(x^2-5x\right)+\left(6x-30\right)\right]=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left[x\left(x-5\right)+6\left(x-5\right)\right]=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x-5\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-x+1=0\\x-5=0\\x+6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\left(loai\right)\\x=5\\x=-6\end{matrix}\right.\)
Vậy x=5 hoặc x=-6
BD
3
ND
3 tháng 4 2018
\(x^4-30x^2+31x-30=0\)
\(\Leftrightarrow x^4-5x^3+5x^3-25x^2-5x+25x+6x-30=0\)
\(\Leftrightarrow\left(x^4-5x^3\right)+\left(5x^3-25x^2\right)-\left(5x^2-25x\right)+\left(6x-30\right)=0\)
\(\Leftrightarrow x^3\left(x-5\right)+5x^2\left(x-5\right)-5x\left(x-5\right)+6\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^3+5x^2-5x+6\right)=0\)
12 tháng 7 2017
c) Ta có : \(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Rightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\)\(\left(\frac{x+6}{2003}+1\right)\)
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
\(\Leftrightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)
Mà : \(\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)\ne0\)
Nên x + 2009 = 0 => x = -2009
WI
1
HS
31 tháng 7 2018
\(x^4-30x^2+31x-30=0\)
<=>\(x^4-30x^2+30x+x-30=0\)
<=>\(\left(x^4+x\right)-\left(30x^2-30x+30\right)=0\)
<=>\(x\left(x^3+1\right)-30\left(x^2-x+1\right)=0\)
<=>\(x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)
<=>\(\left(x^2-x+1\right)\left(x^2+x-30\right)=0\)
<=>\(\left(x^2-x+1\right)\left[\left(x^2+6x\right)-5\left(x+30\right)\right]=0\)
<=>\(x^2\left(-x+1\right)\left[x\left(x+6\right)-5\left(x+6\right)\right]=0\)
<=>\(\left(x^2-x+1\right)\left(x+6\right)\left(x-5\right)=0\)
=>\(x+6=0hoặcx-5=0\) vì\(\left[x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\right]\)
<=> x=-6 hoặc x=5
Vậy......
TK
2
11 tháng 7 2017
\(x^4-30x^2+31x-30=0\)
\(\Leftrightarrow x^4-5x^3+5x^3-25x^2-5x^2+25x+6x-30=0\)
\(\Leftrightarrow x^3\left(x-5\right)+5x^2\left(x-5\right)-5x\left(x-5\right)+6\left(x-5\right)=0\)
\(\Leftrightarrow\left(x^3+5x^2-5x+6\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left(x^3+6x^2-x^2-6x+x+6\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[x^2\left(x+6\right)-x\left(x+6\right)+\left(x+6\right)\right]\left(x-5\right)=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x+6\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]\left(x+6\right)\left(x-5\right)=0\)
Vì \(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
\(\Rightarrow\left[{}\begin{matrix}x+6=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-6\\x=5\end{matrix}\right.\)
Vậy x = -6 hoặc x = 5
AT
11 tháng 7 2017
\(x^4-30x^2+31x-30=0\)
\(\Leftrightarrow x^4-30x^2+30x-30+x=0\)
\(\Leftrightarrow x^4+x-30x^2+30x-30=0\)
\(\Leftrightarrow x\left(x^3+1\right)-30\left(x^2-x+1\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left[x\left(x+1\right)-30\right]=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(x^2+x-30\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-x+1=0\\x^2+x-30=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vl\right)\\\left(x-5\right)\left(x+6\right)=0\end{matrix}\right.\)
=> x = 5 hoặc x = -6.
p/s: ***** = vô lý :V
NT
2
VM
4 tháng 3 2015
= x^4+x^2+1-31x^2+31x-31
= (x^2+x+1)(x^2-x+1)-31(x^2-x+1)
= (x^2-x+1)(x^2+x+1-31)
= (x^2-x-1)(x^2+x-30)
= (x^2-x+1)(x^2+6x-5x-30)
= (x^2-x+1)(x-5)(x+6)
NH
Giải các phương trình:
\(a,\left(x^2-5x\right)^2+10\left(x^2-5x\right)+24=0\)
\(b,x^4-30x^2+31x-30=0\)
2
3 tháng 5 2019
a, Đặt \(x^2-5x=a\)
\(\Rightarrow\)\(a^2+10a+24=0\)
\(\Rightarrow a^2+4a+6a+24=0\)
\(\Rightarrow\left(a+4\right)\left(a+6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}a+4=0\\a+6=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2-5x+4=0\left(1\right)\\x^2-5x+6=0\left(2\right)\end{cases}}}\)
Giải pt (1) ta có : \(x^2-5x+4=0\)
\(\Rightarrow x^2-4x-x+4=0\)
\(\Rightarrow\left(x-4\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=4\end{cases}}\)
Giải pt (2) ta có : \(x^2-5x+6=0\)
\(\Rightarrow x^2-2x-3x+6=0\)
\(\Rightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
Vậy \(S=\left\{1;2;3;4\right\}\)
3 tháng 5 2019
\(x^4-30x^2+31x-30=0\)
\(\Rightarrow x^4-30x^2+x+30x-30=0\)
\(\Rightarrow\left(x^4+x\right)-\left(30x^2-30x+30\right)=0\)
\(\Rightarrow x\left(x^3+1\right)-30\left(x^2-x+1\right)\)
\(\Rightarrow x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)\)
\(\Rightarrow\left(x^2-x+1\right)\left(x^2+x-30\right)=0\)
Mà \(x^2-x+1>0\)với \(\forall\)\(x\)
\(\Rightarrow x^2+x-30=0\)
\(\Rightarrow x^2-5x+6x-30=0\)
\(\Rightarrow x\left(x-5\right)+6\left(x-5\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x+6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=-6\end{cases}}\)
Vậy \(S=\left\{5;-6\right\}\)
\(x^4-30x^2+31x-30=0\)
\(\left(x^4+x\right)-30\left(x^2-x+1\right)=0\)
\(x\left(x^3+1\right)-30\left(x^2-x+1\right)=0\)
\(x\left(x+1\right)\left(x^2-x+1\right)-30\left(x^2-x+1\right)=0\)
\(\left(x^2-x+1\right)\left[x\left(x+1\right)-30\right]=0\)
Ta có: \(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}>0\)
\(\Rightarrow x^2+x-30=0\left(x^2-x+1\ne0\right)\)
\(\left(x^2-5x\right)+\left(6x-30\right)=0\)
\(x\left(x-5\right)+6\left(x-5\right)=0\)
\(\left(x-5\right)\left(x+6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-6\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=5\\x=-6\end{cases}}\)
Thanks bạn nha