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\(\Leftrightarrow\) \(6x^5-12x^4-17x^4+34x^3-7x^3+14x^2+13x^2-26x-3x+\)6 =0
\(6x^5-29x^4+27x^3+27x^2-29x+6=0\)
\(\Leftrightarrow\left(6x^5-18x^4\right)+\left(-11x^4+33x^3\right)+\left(-6x^3+18x^2\right)+\left(9x^2-27x\right)+\left(-2x+6\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(6x^4-11x^3-6x^2+9x-2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\left(6x^4-12x^3\right)+\left(x^3-2x^2\right)+\left(-4x^2+8x\right)+\left(x-2\right)\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(6x^3+x^2-4x+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(\left(6x^3+6x^2\right)+\left(-5x^2-5x\right)+\left(x+1\right)\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x+1\right)\left(6x^2-5x+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x+1\right)\left(\left(6x^2-3x\right)+\left(-2x+1\right)\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x+1\right)\left(2x-1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow x=\left(3;2;-1;\frac{1}{2};\frac{1}{3}\right)\)
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\(=\dfrac{3x^3-6x^2-4x^2+8x+5x-10}{x-2}=3x^2-4x+5\)
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Bài 10:
a) (x+2)2 -x(x+3) + 5x = -20
=> x2 + 4x + 4 - x2 - 3x + 5x = -20
=> 6x = -20 + (-4)
=> 6x = -24
=> x = -4
b) 5x3-10x2+5x=0
=>5x(x2-2x+1)=0
=>5x(x-1)2 =0
=> 5x=0 hoặc (x-1)2=0
=>x=0 hoặc x=1
c) (x2 - 1)3 - (x4 + x2 + 1)(x2 - 1) = 0
=> (x2 - 1)[(x2 - 1)2 - (x4 + x2 + 1)] = 0
<=> (x2 - 1)(x4 - 2x2 + 1 - x4 - x2 - 1) = 0
<=> (x2 - 1)(-3x2) = 0
<=> (x2 - 1)=0 hoặc (-3x2) =0
<=> x2=1 hoặc x2=0
<=> x=−1;1 hoặc x=0
d)
(x+1)3−(x−1)3−6(x−1)2=-19
⇔x3+3x2+3x+1−(x3−3x2+3x−1)−6(x2−2x+1)+19=0
⇔x3+3x2+3x+1−x3+3x2−3x+1−6x2+12x−6+19=0
⇔12x+13=0⇔12x+13=0
⇔12x=-13
⇔x=-23/12
Học tốt nhé:333
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a: =>x^3(x-2)+10x(x-2)=0
=>(x-2)(x^3+10x)=0
=>x(x-2)(x^2+10)=0
=>x(x-2)=0
=>x=0 hoặc x=2
b: =>x^2*(x-3)-16(x-3)=0
=>(x-3)(x^2-16)=0
=>(x-3)(x+4)(x-4)=0
=>\(x\in\left\{3;4;-4\right\}\)
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So sánh điều kiện, không có giá trị nào của x thỏa mãn.
Vậy không có giá trị x nào thỏa mãn yêu cầu đề bài.
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\(x=9\Leftrightarrow x+1=10\\ \Leftrightarrow M=x^{2012}-\left(x+1\right)x^{2012}+...-\left(x+1\right)x^2+\left(x+1\right)x-\left(x+1\right)\\ M=x^{2012}-x^{2013}-x^{2012}+...-x^3-x^2+x^2+x-x-1\\ \Leftrightarrow M=-x^{2013}-1=-9^{2013}-1\)
Bài 4 : Tìm y , biết :
a) y ( 2y - 7 ) - 4y + 14 = 0
b) ( y + 3 )( y2 - 3y + 9 ) - y( y2 - 3 ) = 18
GIẢI HỘ LUN Ạ
\(< =>10x^2-25x-4x+10=0\)
\(< =>\left(10x^2-25x\right)+\left(-4+10\right)=0\)
\(< =>5x\left(2x-5\right)-2\left(2x-5\right)=0\)
\(< =>\left(5x-2\right)\left(2x-5\right)=0\)
\(TH1:5x-2=0< =>5x=2< =>x=\frac{2}{5}\) \(TH2:2x-5=0< =>2x=5< =>x=\frac{5}{2}\)
\(S=\left\{\frac{2}{5};\frac{5}{2}\right\}\)