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\(x^4-3x^3+4x^2-3x-1=0\)
\(\Leftrightarrow x^4+x^3+2x^3+2x^2+2x^2+2x+x+1=0\)
\(\Leftrightarrow x^3\left(x+1\right)+2x^2\left(x+1\right)+2x\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x^3+2x^2+2x+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x^3+2x^2+2x+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow(x^3+x^2+x^2+x+x+1)\left(x+1\right)=0\)
\(\Leftrightarrow[x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)]\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}(x+1)^2=0\\x^2+x+1=0\end{cases}}\Rightarrow\hept{\begin{cases}x+1=0\\\varnothing\end{cases}}\Rightarrow x=-1\)
1) \(x^4-6x^3-x^2+54x-72=0\)
\(\Leftrightarrow x^3\left(x-2\right)-4x^2\left(x-2\right)-9x\left(x-2\right)+36\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-4x^2-9x+36\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x-4\right)-9\left(x-4\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x-3\right)\left(x+3\right)=0\)
Tự làm nốt...
2) \(x^4-5x^2+4=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)-4\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)=0\)
Tự làm nốt...
\(x^4-2x^3-6x^2+8x+8=0\)
\(\Leftrightarrow x^3\left(x-2\right)-6x\left(x-2\right)-4\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-6x-4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x+2\right)-2x\left(x+2\right)-2\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x^2-2x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left[\left(x-1\right)^2-\left(\sqrt{3}\right)^2\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-1-\sqrt{3}\right)\left(x-1+\sqrt{3}\right)=0\)
...
\(2x^4-13x^3+20x^2-3x-2=0\)
\(\Leftrightarrow2x^3\left(x-2\right)-9x^2\left(x-2\right)+2x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^3-9x^2+2x+1\right)=0\)
Bí
1) \(x^4-8x^3+11x^2+8x-12=0\)
\(\Leftrightarrow x^4-x^3-7x^3+7x^2+4x^2-4x+12x-12=0\)
\(\Leftrightarrow x^3\left(x-1\right)-7x^2\left(x-1\right)+4x\left(x-1\right)+12\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3-7x^2+4x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2-8x^2-8x+12x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x+1\right)-8x\left(x+1\right)+12\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^2-8x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^2-2x-6x+12\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left[x\left(x-2\right)-6\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x-2=0\\x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\\x=6\end{matrix}\right.\)
Vậy ...
\(a,x^4-16x^2+32x-16=0\)
\(\Leftrightarrow\left(x^4-16\right)-16x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^4+4\right)\left(x-2\right)\left(x+2\right)-16x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+2x^2-12x+8\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-2x^2+4x^2-8x-4x+8\right)=0\)\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x-2\right)+4x\left(x-2\right)-4\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-2\right)\left(x^2+4x-4\right)=0\)
\(\Leftrightarrow\left(x-2\right)^2\left[\left(x+2\right)^2-8\right]=0\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\\left(x+2\right)^2-8=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-2=0\\\left(x+2\right)^2=8\Rightarrow\left[{}\begin{matrix}x+2=\sqrt{8}\\x+2=-\sqrt{8}\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{8}-2\\x=-\sqrt{8}-2\end{matrix}\right.\)
a)9x2 - 3 = ( 3x + 1 )( 2x - 3 )
<=> 9x2 - 3 = 6x2 - 7x - 3
<=> 3x2 + 7x = 0
<=> x( 3x + 7 ) = 0
<=> x = 0 hoặc x = -7/3
b) 6x2 - 13x + 6 = 0
<=> 6x2 - 9x - 4x + 6 = 0
<=> 3x( 2x - 3 ) - 2( 2x - 3 ) = 0
<=> ( 2x - 3 )( 3x - 2 ) = 0
<=> x = 3/2 hoặc x = 2/3
c) \(\frac{3}{x-1}=\frac{3x+2}{1-x^2}-\frac{4}{x+1}\)( ĐKXĐ : x ≠ ±1 )
<=> \(\frac{3\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{-3x-2}{\left(x-1\right)\left(x+1\right)}-\frac{4\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
=> 3x + 3 = -3x - 2 - 4x + 4
<=> 10x = -1 <=> x = -1/10 (tm)
a, \(9x^2-3=\left(3x+1\right)\left(2x-3\right)\Leftrightarrow9x^2-3=6x^2-9x+2x-3\)
\(\Leftrightarrow9x^2-3=6x^2-7x-3\Leftrightarrow3x^2+7x=0\Leftrightarrow x\left(3x+7\right)=0\Leftrightarrow x=0;x=-\frac{7}{3}\)
Vậy tập nghiệm của phương trình là S = { -7/3 ; 0 }
b, \(6x^2-13x+6=0\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=0\Leftrightarrow x=\frac{2}{3};x=\frac{3}{2}\)
Vậy tập nghiệm của phương trình là S = { 2/3 ; 3/2 }
c, \(\frac{3}{x-1}=\frac{3x+2}{1-x^2}-\frac{4}{x+1}ĐK:x\ne\pm1\)
\(\Leftrightarrow\frac{3\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{-3x-2}{\left(x-1\right)\left(x+1\right)}-\frac{4\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(\Leftrightarrow3x+3=-3x-2-4x+4\Leftrightarrow3x+3=-7x+2\)
\(\Leftrightarrow10x=-1\Leftrightarrow x=-\frac{1}{10}\)Vậy tập nghiệm của phương trình là S = { -1/10 }
\(j,3x^2+7x+2=0\)
\(\Leftrightarrow3x^2+6x+x+2=0\)
\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{3}\\x=-2\end{matrix}\right.\)
Vậy...............................
\(m,3x^2+4x-4=0\)
\(\Leftrightarrow3x^2+6x-2x-4=0\)
\(\Leftrightarrow3x\left(x+2\right)-2\left(x+2\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=-2\end{matrix}\right.\)
Câu 2 sai đề nhé
Phải là:(x-999)/99+(x-896)/101+(x-789/103)=6
Câu 1:
Đặt \(x+1=a\). Khi đó \(x+3=a+2; x-1=a-2\).
PT đã cho tương đương với:
\((a+2)^4+(a-2)^4=626\)
\(\Leftrightarrow 2a^4+48a^2+32=626\)
\(\Leftrightarrow a^4+24a^2-297=0\)
\(\Leftrightarrow (a^2+12)^2=441\)
\(\Rightarrow a^2+12=\sqrt{441}=21\) (do \(a^2+12>0)\)
\(\Rightarrow a^2=9\Rightarrow a=\pm 3\)
Nếu $a=3$ thì \(x=a-1=2\)
Nếu $a=-3$ thì $x=a-1=-4$
Câu 2:
Đặt \(2x-1=a; x-1=b\). PT đã cho tương đương với:
\(a^3+b^3+(-a-b)^3=0\)
\(\Leftrightarrow a^3+b^3-(a+b)^3=0\)
\(\Leftrightarrow a^3+b^3-[a^3+b^3+3ab(a+b)]=0\)
\(\Leftrightarrow ab(a+b)=0\Rightarrow \left[\begin{matrix} a=0\\ b=0\\ a+b=0\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} 2x-1=0\\ x-1=0\\ 3x-2=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{1}{2}\\ x=1\\ x=\frac{2}{3}\end{matrix}\right.\)