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a/ \(x\left(x^2-2x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\pm\sqrt{3}\\\end{matrix}\right.\)
b/ \(\Leftrightarrow2x^3-4x^2+6x-x^2+2x-3=0\)
\(\Leftrightarrow2x\left(x^2-2x+3\right)-\left(x^2-2x+3\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2-2x+3\right)=0\)
c/ \(\Leftrightarrow3x^3-15x^2+9x+x^2-5x+3=0\)
\(\Leftrightarrow3x\left(x^2-5x+3\right)+\left(x^2-5x+3\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x^2-5x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=-\frac{1}{3}\\x=\frac{5\pm\sqrt{13}}{2}\end{matrix}\right.\)
d/ \(x\left(x^2+6x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\pm\sqrt{14}\end{matrix}\right.\)
a, \(x^4-5x^3+2x^2+10x+2=0\)
\(\Rightarrow x^4+x^3-6x^3-6x^2+8x^2+8x+2x+2=0\)
\(\Rightarrow x^3\left(x+1\right)-6x^2\left(x+1\right)+8x\left(x+1\right)+2\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x^3-6x^2+8x+2\right)=0\)
Vì \(x^3-6x^2+8x+2>0\) nên \(x+1=0\Rightarrow x=-1\)
Các câu còn lại tương tự!
Chúc bạn học tốt!!!
\(2x^4+3x^3-9x^2-3x+2\)
\(=2x^4+5x^3-2x^2-2x^3-5x^2+2x-2x^2-5x+2\)
\(=x^2\left(2x^2+5x-2\right)-x\left(2x^2+5x-2\right)-\left(2x^2+5x-2\right)\)
\(=\left(x^2-x-1\right)\left(2x^2+5x-2\right)\)
b/
\(x^4-3x^3-6x^2+3x+1\)
\(=x^4-4x^3-x^2+x^3-4x^2-x-x^2+4x+1\)
\(=x^2\left(x^2-4x-1\right)+x\left(x^2-4x-1\right)-\left(x^2-4x-1\right)\)
\(=\left(x^2+x-1\right)\left(x^2-4x-1\right)\)
c/
\(x^4-6x^3+12x^2-14x+3\)
\(=x^4-4x^3+x^2-2x^3+8x^2-2x+3x^2-12x+3\)
\(=x^2\left(x^2-4x+1\right)-2x\left(x^2-4x+1\right)+3\left(x^2-4x+1\right)\)
\(=\left(x^2-2x+3\right)\left(x^2-4x+1\right)\)
e/
Đề sai, sao có 2 hạng tử chứa \(x^4\) thế kia?
1
\(-3x\left(x-5\right)+5\left(x-1\right)+3x^2=4-x\)
=> \(-3x^2+15x+5x-5+3x^2=4-x\)
=> \(20x-5=4-x\)
=> \(21x=9\)
=> \(x=\dfrac{3}{7}\)
Vậy x = \(\dfrac{3}{7}\)
2,
\(7x\left(x-2\right)-5\left(x-1\right)=21x^2-14x^2+3\)
=> \(7x^2-14x-5x+5=7x^2+3\)
=> \(-14x-5x+5=3\)
=> \(-19x=-2\)
=> \(x=\dfrac{2}{19}\)
Vậy \(x=\dfrac{2}{19}\)
3,
\(3\left(5x-1\right)-x\left(x-2\right)+x^2-13x=7\)
=> \(15x-3-x^2+2x+x^2-13x=7\)
=> \(4x-3=7\)
=> 4x = 10
=> x = \(\dfrac{5}{2}\)
Vậy x = \(\dfrac{5}{2}\)
4,
\(\dfrac{1}{5}x\left(10x-15\right)-2x\left(x-5\right)=12\)
=> \(2x^2-3x-2x^2+10x=12\)
=> 7x = 12
=> x = \(\dfrac{12}{7}\)
Vậy x = \(\dfrac{12}{7}\)
Kkk
\(\left(x^2-2x+4\right)\left(x^2+3x+4\right)=14x^2\)
\(\Leftrightarrow x^4+3x^3+4x^2-2x^3-6x^2-8x+4x^2+12x+16=14x^2\)
\(\Leftrightarrow x^4+x^3+2x^2-14x^2+4x=0\)
\(\Leftrightarrow x^4+x^3-12x^2+4x=0\)
\(\Leftrightarrow x\left(x^3+x^2-12x+4\right)=0\)
\(\Leftrightarrow x=0\) ( vì \(x^3+x^2-12x+4\ne0\) )