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x2+3x2+3x+1-3x2-3x = 0
=> x3+1 = 0
=> x3 = 0-1
=> x3 = -1
=> x = -1
\(x^3+3x^2+3x+1-3x^2-3x=0\)0
\(\Leftrightarrow x^3+\left(3x^2-3x^2\right)+\left(3x-3x\right)+1=0\)
\(\Leftrightarrow x^3+1=0\)
\(\Leftrightarrow x^3=1\)
\(\Leftrightarrow x^3=1^3\)
\(\Rightarrow x=1\)
a) \(\left(x+3\right)^2-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)-2x^2=54\)
=> x2 + 6x + 9 - x(9x2 + 6x + 1) + (2x)3 + 13 - 2x2 = 54
=> x2 + 6x + 9 - 9x3 - 6x2 - x + 8x3 + 1 - 2x2 = 54
=> (-9x3 + 8x3) + (x2 - 6x2 - 2x2) + (6x - x) + (9 + 1) = 54
=> -x3 - 7x2 + 5x + 10 = 54
=> -(x3 + 7x2 - 5x - 10) = 54
=> phương trình vô nghiệm
b) (x + 3)3 - (x - 3)(x2 + 3x + 9) + 6(x + 1)2 + 3x = -33
=> x3 + 9x2 + 27x + 27 - (x3 - 33) + 6(x2 + 2x + 1) + 3x = -33
=> x3 + 9x2 + 27x + 27 - x3 + 27 + 6x2 + 12x + 6 + 3x = -33
=> (x3 - x3) + (9x2 + 6x2) + (27x + 12x + 3x) + (27 + 27 + 6) = -33
=> 15x2 + 42x + 60 = -33
=> 15x2 + 42x + 60 + 33 = 0
=> 15x2 + 42x + 93 = 0
=> 3(5x2 + 14x + 31) = 0
=> 5x2 + 14x + 31 = 0
=> không tìm được x
a) = x3 + 9x2 + 27x + 27 - 9x3 -6x2 - x + 8x3 +1 -3x2 =54
26x +28 = 54
26x = 54-28 = 26
x = 1
b) = x3 - 9x2 + 27x -27 - x3 +27 +6x2 + 12x + 6 +3x2 = -33
39x +6 = -33
39x = -33-6 = -39
x = -1
\(a.x\left(x-5\right)\left(x+5\right)-\left(x+2\right)\left(x^2-2x+4\right)=3\)
\(\Leftrightarrow x\left(x^2-5^2\right)-\left(x^3+2^3\right)=3\)
\(\Leftrightarrow x^3-25x-x^3-8=3\)
\(\Leftrightarrow x^3-x^3-25x=8+3\)
\(\Leftrightarrow x=\frac{11}{-25}\)
Vậy x có nghiệm là \(\frac{-11}{25}.\)
\(\)
\(5x-3x^2+3x^3-x^4=\left(x+1\right)^2\)
\(\Leftrightarrow5x-3x^2+3x^3-x^4=x^2+2x+1\)
\(\Leftrightarrow5x-3x^2+3x^3-x^4-1=x^2+2x+1-1\)
\(\Leftrightarrow5x-3x^2+3x^3-x^4-1=x^2+2x\)
\(\Leftrightarrow5x-3x^2+3x^3-x^4-1-2x=x^2+2x-2x\)
\(\Leftrightarrow-x^4+3x^3-3x^2+3x-1=x^2\)
\(\Leftrightarrow-x^4+3x^3-3x^2-1-x^2=x^2-x^2\)
\(\Leftrightarrow-x^4+3x^3-4x^2+3x-1=0\)
\(\Leftrightarrow-\left(x-1\right)^2\left(x^2-x+1\right)=0\)
\(\Rightarrow\hept{\begin{cases}x=1\\x=\frac{1}{2}+i\frac{\sqrt{3}}{2}\\x=\frac{1}{2}-i\frac{\sqrt{3}}{2}\end{cases}}\)
Mình ko chắc :(
\(x^3+3x^2=-3x-1\)
\(\Leftrightarrow x^3+3x^2+3x+1=0\)
\(\Leftrightarrow\left(x+1\right)^3=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)