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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
\(\Leftrightarrow x^3+\left(3x^2-3x^2\right)+\left(3x-3x\right)+1=0\)
\(\Leftrightarrow x^3+1=0\)
\(\Leftrightarrow x^3=1\)
\(\Leftrightarrow x^3=1^3\)
\(\Rightarrow x=1\)
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1)2x3+3x2+2x+3=0
=> (2x3+3x2)+(2x+3)=0
=> x2(2x+3)+(2x+3)=0
=> (2x+3)(x2+1)=0
=>\(\hept{\begin{cases}2x+3=0\\x^2+1=0\end{cases}}\)=>\(\hept{\begin{cases}2x=-3\\x^2=-1\end{cases}}\)=>\(\hept{\begin{cases}x=\frac{-3}{2}\\vo.nghiem\end{cases}}\)
Vậy x=-3/2
2)x2-3x-18=0
=> (x2+3x)-(6x+18)=0
=> x(x+3)-6(x+3)=0
=> (x+3)(x-6)=0
=> \(\hept{\begin{cases}x+3=0\\x-6=0\end{cases}}\)=>\(\hept{\begin{cases}x=-3\\x=6\end{cases}}\)
Vậy x=-3 hoặc x=6
3)Sai đề rồi bạn, 30 thành 30x mới đúng
x3-11x2+30x=0
=> x(x2-11x+30)=0
=> x[(x2-5x)-(6x-30)]=0
=> x[x(x-5)-6(x-5)]=0
=> x(x-5)(x-6)=0
=>\(\hept{\begin{cases}x=0\\x-5=0\\x-6=0\end{cases}}\)=>\(\hept{\begin{cases}x=0\\x=5\\x=6\end{cases}}\)
Vậy x=0 hoặc x=5 hoặc x=6
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a) (x-3)(x+3)-(x-1)^2=0
=> (x^2-9)-(x^2-2x+1)=0
=>x^2-9-x^2+2x-1=0
=>(x^2-x^2)-9-1+2x=0
=>-10+2x=0
=>-2.(-5-x)=0
=>-5-x=0
=>-x=0+5
=>x=-5
vậy x=-5
b) x^3-3x^2+3x-1=0
=>(x-1)^3=0
=>x-1=0
=>x=0+1
=>x=1
vậy x=1
c) 4x^2-28x=0
=>4x.(x-7)=0
=> 2 TH
* 4x=0=>x=0
*x-7=0=>x=0+7=>x=7
vậy x=0 hoặc x=7
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( x + 1 ) ( x - 1 )
= ..................
= x2 -1
100% là đúng
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a)\(x\left(x+2\right)-3x-6=0\)
=>\(x\left(x+2\right)-3\left(x+2\right)=0\)
=>\(\left(x-3\right)\left(x+2\right)=0\)
=>\(\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)
b)\(x^3+3x^2+3x-1-3x^2-3x=0\)
=>\(x^3-1=0\)
=>x3=1
=>x=1
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a) (3x + 4)2 - (3x - 1).(3x + 1) = 49
=> (3x + 4).3x + (3x + 4).4 - (9x2 - 1) = 49
=> 9x2 + 12x + 12x + 16 - 9x2 + 1 = 49
=> 24x + 17 = 49
=> 24x = 49 - 17
=> 24x = 32
=> \(x=\frac{32}{24}=\frac{4}{3}\)
Vậy \(x=\frac{4}{3}\)
b) (2x + 1)2 - (x - 1)2 = 0
=> (2x + 1 - x + 1).(2x + 1 + x - 1) = 0
=> (x + 2).3x = 0
=> (x + 2).x = 0
\(\Rightarrow\left[\begin{array}{nghiempt}x+2=0\\x=0\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=-2\\x=0\end{array}\right.\)
Vậy \(\left[\begin{array}{nghiempt}x=-2\\x=0\end{array}\right.\)
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a) 3x^3-12x=0
3x(x^2-4)=0
3x(x-2)(x+2)=0
suy ra 3x=0 suy ra x=0
x-2=0 x=2
x+2=0 x= -2
b) (x-3)^2-(x-3)(3-x)^2=0
(x-3)^2-(x-3)(x-3)^2=0
(x-3)^2(1-x+3)=0
(x-3)^2(4-x)=0
suy ra x-3=0 suy ra x=3
4-x=0 x=4
a) và b) đã nhé bạn
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1.1 / 3x(x-2005)-x+2005=0
<=>3x(x-2005)-(x-2005)=0
<=>(x-2005)(3x-1)=0
<=>x-2005=0 hoặc 3x-1=0
<=>x=2005 hoặc x=1/3
1.2/ x+1 =(x+1)2
<=>(x+1) - (x+1)2=0
<=>(x+1) (1-x+1)=0
<=> (x+1) (2-x) =0
<=>x+1=0 hoặc 2-x =0
<=> x=-1 hoặc x=2
1.3/x3=5x
<=>x3-5x=0
<=>x(x2-5)=0
<=>x=0 hoặc x2-5=0
<=>x=0 hoặc x2=5
<=>x=0 hoặc x=\(\sqrt{5}\)và \(-\sqrt{5}\)
1.4/x2(x2 -2)-4(2-x2)=0
<=>x2(x2-2) +4(x2-2)=0
<=> (x2 -2)(x2+4)=0
<=>x2-2=0 hoặc x2+4=0
<=>x2=2 hoặc x2=-4(vô lí)
<=>x=\(\sqrt{2}\)hoặc \(-\sqrt{2}\)
\(3x\left(x-1\right)+2\left(x-1\right)=0\)
\(\left(x-1\right)\left(3x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\3x+2=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{-2}{3}\end{cases}}\)
Vậy,............
3x là dấu 3.[x-1] hay là x thôi??