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\(b.6x^4+25x^3+12x^2-25x+6=0\\\Leftrightarrow 6x^4+12x^3+13x^3+26x^2-14x^2-28x+3x+6=0\\\Leftrightarrow 6x^3\left(x+2\right)+13x^2\left(x+2\right)-14x\left(x+2\right)+3\left(x+2\right)=0\\\Leftrightarrow \left(6x^3+13x^2-14x+3\right)\left(x+2\right)=0\\ \Leftrightarrow\left(6x^3+18x^2-5x^2-15x+x+3\right)\left(x+2\right)=0\\\Leftrightarrow \left[6x^2\left(x+3\right)-5x\left(x+3\right)+\left(x+3\right)\right]\left(x+2\right)=0\\ \Leftrightarrow\left(6x^2-5x+1\right)\left(x+3\right)\left(x+2\right)=0\\ \Leftrightarrow\left(6x^2-3x-2x+1\right)\left(x+3\right)\left(x+2\right)=0\\\Leftrightarrow \left[3x\left(2x-1\right)-\left(2x-1\right)\right]\left(x+3\right)\left(x+2\right)=0\\\Leftrightarrow \left(3x-1\right)\left(2x-1\right)\left(x+3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\2x-1=0\\x+3=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=\frac{1}{2}\\x=-3\\x=-2\end{matrix}\right.\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{\frac{1}{3};\frac{1}{2};-3;-2\right\}\)
\(2x^4-9x^3+14x^2-9x+2=0\\\Leftrightarrow 2x^4-2x^3-7x^3+7x^2+7x^2-7x-2x+2=0\\\Leftrightarrow 2x^3\left(x-1\right)-7x^2\left(x-1\right)+7x\left(x-1\right)-2\left(x-1\right)=0\\\Leftrightarrow \left(2x^3-7x^2+7x-2\right)\left(x-1\right)=0\\\Leftrightarrow \left[2\left(x^3-1\right)-7x\left(x-1\right)\right]\left(x-1\right)=0\\\Leftrightarrow \left(x-1\right)^2\left[2\left(x^2+x+1\right)-7x\right]=0\\\Leftrightarrow \left(2x^2+2x+2-7x\right)\left(x-1\right)^2=0\\\Leftrightarrow \left(2x^2-5x+2\right)\left(x-1\right)^2=0\\\Leftrightarrow \left(2x^2-x-4x+2\right)\left(x-1\right)^2=0\\\Leftrightarrow \left[x\left(2x-1\right)-2\left(2x-1\right)\right]\left(x-1\right)^2=0\\\Leftrightarrow \left(x-2\right)\left(2x-1\right)\left(x-1\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-1=0\\\left(x-1\right)^2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\2x=1\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{2}\\x=1\end{matrix}\right.\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{2;\frac{1}{2};1\right\}\)
\(2x^4-9x^3+14x^2-9x+2=0\)
\(\Leftrightarrow2x^4-4x^3-5x^3+10x^2+4x^2-8x-x+2=0\)
\(\Leftrightarrow2x^3\left(x-2\right)-5x^2\left(x-2\right)+4x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^3-5x^2+4x-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^3-2x^2-3x^2+3x+x-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[2x^2\left(x-1\right)-3x\left(x-1\right)+\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x-1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)
1) 2x4 - 9x3 + 14x2 - 9x + 2 = 0
<=> (2x4 - 4x3) - (5x3 - 10x2) + (4x2 - 8x) - (x - 2) = 0
<=> 2x3(x - 2) - 5x2(x - 2) + 4x(x - 2) - (x - 2) = 0
<=> (2x3 - 5x2 + 4x - 1)(x - 2) = 0
<=> [(2x3 - 2x2) - (3x2 - 3x) + (x - 1)](x - 2) = 0
<=> [2x2(x - 1) - 3x(x - 1) + (x - 1)](x - 2) = 0
<=> (2x2 - 2x - x + 1)(x - 1)(x - 2) = 0
<=> (2x - 1)(x - 1)2(x - 2) = 0
<=> 2x - 1=0
hoặc x - 1 = 0
hoặc x - 2 = 0
<=> x = 1/2
hoặc x = 1
hoặc x = 2
Vậy S = {1/2; 1; 2}
\(a,x^4-7x^2+6\)
\(=x^4-x^2-6x^2+6\)
\(=x^2\left(x^2-1\right)-6\left(x^2-1\right)\)
\(=\left(x^2-6\right)\left(x^2-1\right)\)
\(=\left(x+\sqrt{6}\right)\left(x-\sqrt{6}\right)\left(x+1\right)\left(x-1\right)\)
\(b,x^4+2x^2-3=x^4+3x^2-x^2-3\)
\(=x^2\left(x^2+3\right)-\left(x^2+3\right)\)
\(=\left(x^2-1\right)\left(x^2+3\right)\)
\(=\left(x+1\right)\left(x-1\right)\left(x^2+3\right)\)
a.) 2x2 - 7xy + 6y2 + 9x - 13y + 5
= (2x -3y)(x-2y) + 5(2x - 3y) -x +2y -5
= (2x - 3y)(x-2y + 5) - (x - 2y + 5)
=(x-2y+5)(2x-3y-1)
a)\(3x\left(x^2-2x\right)\)
\(=3x^3-6x^2\)
b) \(\left(27x^3-1\right):\left(9x^2+3x+1\right)\)
\(=\left(3x-1\right)\left(9x^2+3x+1\right):\left(9x^2+3x+1\right)\)
\(=3x-1\)
c) \(\dfrac{4y^3}{7x^2}.\dfrac{14x^3}{y}\)
\(=8xy^2\)
\(\)d)\(\dfrac{x^2-9}{2x+6}:\dfrac{3-x}{2}\)
\(=-\dfrac{\left(x-3\right)\left(x+3\right)}{2\left(x+3\right)}:\dfrac{x-3}{2}\)
\(=-\dfrac{\left(x+3\right)\left(x-3\right)}{2\left(x+3\right)}.\dfrac{2}{\left(x-3\right)}\)
\(=-1\)
a.3x(x2-2x)=3x3-6x2
b.(27x3-1) : (9x2+3x+1)=\([\left(3x\right)^3-1]:\left(9x^2+3x+1\right)=\left(3x-1\right)\left(9x^2+3x+1\right):\left(9x^2+3x+1\right)=3x-1\)
\(2x^4-9x^3+14x^2-9x+2=0\)
\(\Leftrightarrow2x^4-4x^3+2x^2-5x^3+10x^2-5x+2x^2-4x+2=0\)
\(\Leftrightarrow2x^2\left(x^2-2x+1\right)-5x\left(x^2-2x+1\right)+2\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(2x^2-5x+2\right)\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(2x^2-x-4x+2\right)\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left[x\left(2x-1\right)-2\left(2x-1\right)\right]\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)^2\left(2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\\2x-1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=2\\x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(2x^4-9x^3+14x^2-9x+2=0\)
\(\Leftrightarrow2x^4-2x^3-7x^3+7x^2+7x^2-7x-2x+2=0\)
\(\Leftrightarrow2x^3\cdot\left(x-1\right)-7x^2\cdot\left(x-1\right)+7x\cdot\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x^3-7x^2+7x-2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\cdot\left[2\left(x^3-1\right)-7x\cdot\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\cdot\left[2\left(x-1\right)\cdot\left(x^2+x+1\right)-7x\cdot\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left[2\left(x^2+x+1\right)-7x\right]=0\)
\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left(2x^2+2x+2-7x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left(2x^2-5x+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left(2x^2-x-4x+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left[x\cdot\left(2x-1\right)-2\left(2x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\cdot\left(x-1\right)\cdot\left(x-2\right)\cdot\left(2x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left(x-2\right)\cdot\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\x-2=0\\2x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(x_1=\dfrac{1}{2};x_2=1;x_3=2\)