a) \(6x^4+y^4\)
b) \(x^4+x^2+1\)
c) \(x^7+x^5+1\)
d) \(x^7+x^5-1\)
e) \(x^4+2016x^2+2015x+2016\)
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TN
17 tháng 1 2016
1)can(2)*(can(2)+1-can(3))
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3)120
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6)60
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8)72
9)47
NT
0
14 tháng 6 2022
a: \(\Leftrightarrow x^2+11x^2-7x+22x-14-4=0\)
\(\Leftrightarrow12x^2+15x^2-18=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-1\right)=0\)
=>x=-6 hoặc x=1
b: \(x^4+3x^2-4=0\)
\(\Leftrightarrow\left(x^2+4\right)\left(x^2-1\right)=0\)
=>x=1 hoặc x=-1
4 tháng 7 2017
a, \(4x\left(x-5\right)-7x\left(x-4\right)+3x^2=12\)
\(\Leftrightarrow4x^2-20x-7x^2+28x+3x^2=12\)
\(\Leftrightarrow8x=12\)
\(\Leftrightarrow x=\dfrac{3}{2}\)
Vậy...
b, \(-3x\left(x-5\right)+5\left(x-1\right)+3x^2=4-x\)
\(\Leftrightarrow-3x^2+15x+5x-5+3x^2=4-x\)
\(\Leftrightarrow21x=9\)
\(\Leftrightarrow x=\dfrac{3}{7}\)
Vậy...
c, \(\left(x-5\right)\left(x-4\right)-\left(x+1\right)\left(x-2\right)=7\)
\(\Leftrightarrow x^2-9x+20-x^2+x+2=7\)
\(\Leftrightarrow-8x=-15\Leftrightarrow x=\dfrac{15}{8}\)
Vậy...
d, \(-\left(x+3\right)\left(x-4\right)+\left(x-1\right)\left(x+1\right)=10\)
\(\Leftrightarrow-x^2+x+12+x^2-1=10\)
\(\Leftrightarrow x=-1\)
Vậy...
e, \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(5-x^2\right)=6x\)
\(\Leftrightarrow x^3-27+5x-x^3=6x\)
\(\Leftrightarrow x=-27\)
Vậy...
HL
4 tháng 7 2017
a) \(4x\left(x-5\right)-7x\left(x-4\right)+3x^2=12\)
\(4x^2-20x-7x^2+28x+3x^2-12=0\)
\(8x-12=0\)
\(4\left(2x-3\right)=0\)
\(2x-3=0\Rightarrow x=\dfrac{3}{2}\)
b) \(-3x\left(x-5\right)+5\left(x-1\right)+3x^2=4-x\)
\(-3x^2+15x+5x-5+3x^2-4+x=0\)
\(21x-9=0\)
\(3\left(7x-3\right)=0\)
\(\Rightarrow7x-3=0\Rightarrow x=\dfrac{3}{7}\)
c) \(\left(x-5\right)\left(x-4\right)-\left(x-1\right)\left(x-2\right)=7\)
\(x^2-4x-5x+20-x^2+2x+x-2-7=0\)
\(-6x+11=0\Rightarrow x=\dfrac{11}{6}\)
d) \(-\left(x-3\right)\left(x-4\right)+\left(x-1\right)\left(x+1\right)=10\)
\(-x^2+4x+3x-12+x^2-1-10=0\)
\(7x-23=0\)
\(x=\dfrac{23}{7}\)
e) \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(5-x^2\right)=6x\)
\(x^3-27+5x-x^3-6x=0\)
\(-x-27=0\Rightarrow x=-27\)
b: \(x^4+x^2+1\)
\(=x^4+2x^2+1-x^2\)
\(=\left(x^2+1\right)^2-x^2\)
\(=\left(x^2-x+1\right)\left(x^2+x+1\right)\)
c: \(x^7+x^5+1\)
\(=x^7+x^6+x^5-x^6-x^5-x^4+x^5+x^4+x^3-x^3+1\)
\(=x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^5-x^4+x^3-x+1\right)\)