phân tích đa thức
x^4+6x^3+11x^2+6x+1
x^4+x^3+x^2+x+1
6x^4+5x^3-38x^2+5x+6
x^4+5x^3-12x^2+5x+1
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EC
3
TL
1
12 tháng 5 2023
c: =>(x+2)(x+3)(x-5)(x-6)=180
=>(x^2-3x-10)(x^2-3x-18)=180
=>(x^2-3x)^2-28(x^2-3x)=0
=>x(x-3)(x-7)(x+4)=0
=>\(x\in\left\{0;3;7;-4\right\}\)
c: =>(x-3)(x+2)(2x+1)(3x-1)=0
=>\(x\in\left\{3;-2;-\dfrac{1}{2};\dfrac{1}{3}\right\}\)
4 tháng 11 2017
\(1.x^4+6x^3+11x^2+6x+1\)
\(=x^4+6x^3+9x^2+2x^2+6x+1\)
\(=x^4+9x^2+1+6x^3+2x^2+6x\)
\(=\left(x^2\right)^2+\left(3x\right)^2+1^2+2.x^2.3x+2.x^2.1+2.3x.1\)
\(=\left(x^2+3x+1\right)^2\)
\(2,6x^4+5x^3-38x^2+5x+6\)
\(=6x^4+6x^3+2x^3-3x^3-36x^2+2x^2-3x^2-x^2-12x+18x-x+6\)
\(=\left(6x^4+2x^3\right)+\left(6x^3+2x^2\right)-\left(3x^3+x^2\right)-\left(36x^2+12x\right)+\left(18x+6\right)-\left(3x^2+x\right)\)
\(=2x^3\left(3x+1\right)+2x^2\left(3x+1\right)-x^2\left(3x+1\right)-12x\left(3x+1\right)+6\left(3x+1\right)-x\left(3x+1\right)\)
\(=\left(3x+1\right)\left(2x^3+2x^2-x^2-12x+6-x\right)\)
\(=\left(3x+1\right)\left[\left(2x^3-x^2\right)+\left(2x^2-x\right)-\left(12x-6\right)\right]\)
\(=\left(3x+1\right)\left[x^2\left(2x-1\right)+x\left(2x-1\right)-6\left(2x-1\right)\right]\)
\(=\left(3x+1\right)\left(2x-1\right)\left(x^2+x-6\right)\)
\(=\left(3x+1\right)\left(2x-1\right)\left(x^2+3x-2x-6\right)\)
\(=\left(3x+1\right)\left(2x-1\right)\left[\left(x^2+3x\right)-\left(2x+6\right)\right]\)
\(=\left(3x+1\right)\left(2x-1\right)\left[x\left(x+3\right)-2\left(x+3\right)\right]\)
\(=\left(3x+1\right)\left(2x-1\right)\left(x+3\right)\left(x-2\right)\)
4 tháng 11 2017
1. \(x^4+6x^3+11x^2+6x+1\)
\(=\left(x^2\right)^2+2.x^2.3x+\left(3x\right)^2+2x^2+6x+1\)
\(=\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1\)
\(=\left(x^2+3x+1\right)^2\)
3. \(x^4-7x^3+14x^2-7x+1\)
\(=x^2\left(x^2-7x+14-\dfrac{7}{x}+\dfrac{1}{x^2}\right)\)
\(=x^2\left[\left(x^2+\dfrac{1}{x^2}\right)-\left(7x+\dfrac{7}{x}\right)+14\right]\)
\(=x^2\left[\left(x+\dfrac{1}{x}\right)^2-7\left(x+\dfrac{1}{x}\right)+12\right]\)
\(=x^2\left[\left(x+\dfrac{1}{x}\right)^2-2\left(x+\dfrac{1}{x}\right).\dfrac{7}{2}+\dfrac{49}{4}-\dfrac{1}{4}\right]\)
\(=x^2\left[\left(x+\dfrac{1}{x}-\dfrac{7}{2}\right)^2-\dfrac{1}{4}\right]\)
\(=\left(x^2+1-\dfrac{7}{2}x\right)^2-\left(\dfrac{1}{2}x\right)^2\)
\(=\left(x^2-3x+1\right)\left(x^2-4x+1\right)\)
Có thể phân tích thành HĐT tiếp hoặc không.
26 tháng 6 2023
2: =(2x+1)^2-y^2
=(2x+1+y)(2x+1-y)
3: =x^2(x^2+2x+1)
=x^2(x+1)^2
4: =x^2+6x-x-6
=(x+6)(x-1)
5: =-6x^2+3x+4x-2
=-3x(2x-1)+2(2x-1)
=(2x-1)(-3x+2)
6: =5x(x+y)-(x+y)
=(x+y)(5x-1)
7: =2x^2+5x-2x-5
=(2x+5)(x-1)
8: =(x^2-1)*(x^2-4)
=(x-1)(x+1)(x-2)(x+2)
9: =x^2(x-5)-9(x-5)
=(x-5)(x-3)(x+3)
28 tháng 12 2017
a, \(x^4-6x^3+11x^2-6x+1=0\)
\(\Rightarrow\left(x^2-3x+1\right)^2=0\)
\(\Rightarrow x^2-3x+1=0\)
\(\Rightarrow x=\frac{\pm\sqrt{5}+3}{2}\)
Chúc bạn học tốt
28 tháng 12 2017
\(x^4-\left(6x^2-2x^2\right)+\left(9x^2-6x+1\right)=0\)
\(x^4-2x^2\left(3x-1\right)+\left(3x-1\right)^2=0\)
\(\left(x^2-3x+1\right)^2=0\)
tự làm
B) \(\left(6x^4-18x^3\right)+\left(13x^{^3}-39x^2\right)+\left(x-3x\right)-\left(2x-6\right)=0\)
\(6x^3\left(x-3\right)+13x^2\left(x-3\right)+x\left(x-3\right)-2\left(x-3\right)=0\)
\(\left(x-3\right)\left(6x^3+13x^2-2\right)=0\)
\(\left(x-3\right)\left(6x^3+12x^2+x^2+2x-x-2\right)\)
\(\left(x-3\right)\left\{6x^2\left(x+2\right)+x\left(x+2\right)-\left(x+2\right)\right\}\)
\(\left(x-3\right)\left(x+2\right)\left(6x^2-x-1\right)\)
\(\left(x-3\right)\left(x+2\right)\left(6x^2-3x+2x-1\right)\)
\(\left(x-3\right)\left(x+2\right)\left(3x\left(2x-1\right)+\left(2x-1\right)\right)\)
\(\left(x-3\right)\left(x+2\right)\left(2x-1\right)\left(3x+1\right)=0\)
câu C nghĩ đã
AN
27 tháng 12 2017
a, \(x^4-6x^3+11x^2-6x+1=0\)
=> \(x^4-6x^3+9x^2+2x^2-6x+1=0\)
=> \(x^2+3x+1=0\)
=> \(\Delta\) =\(b^2-4c\)
=\(3^2.4=5\)
Nên \(\sqrt{\Delta}=5\)
x= \(\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{-3+\sqrt{5}}{2}\)
hoặc x= \(\dfrac{b+\sqrt{\Delta}}{2a}=\dfrac{3+\sqrt{5}}{2}\)
L
1
7 tháng 10 2018
\(4x^4+4x^3+5x^2+2x+1\)
\(=\left(2x^2\right)^2+2.2x^2.x+x^2+4x^2+2x+1\)
\(=\left(2x^2+x\right)^2+2\left(2x^2+x\right)+1\)
\(=\left(2x^2+x+1\right)^2\)
\(x^4+6x^3+11x^2+6x+1\)
\(=\left(x^2\right)^2+2.x^2.3x+\left(3x\right)^2+2x^2+6x+1\)
\(=\left(x^2+3x\right)^2+2\left(x^2+3x\right)+1\)
\(=\left(x^2+3x+1\right)^2\)
Chúc bạn học tốt.
help me
a)\(x^4+6x^3+11x^2+6x+1\)
\(=x^4+9x^2+1+6x^3+6x+2x^2\)
\(=\left(x^2+3x+1\right)^2\)