3x3-75x=0
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Ta có 3x3 - 75x=0
x.3.x2-75.x=0
x(3x2-75)=0
=>hoặc x=0
hoặc 3x2 - 75=0 => 3x2 = 75 => x2 = 25 => hoặc x = 25
hoặc x = -25
Vậy x thuộc {0;25;-25}
Tick nha
0,75x:2/3=3/5:4/15
0,75x:2/3=3/5*15/4
0,75x:2/3=9/4
0,75x=9/4*2/3
0,75x=3/2
x=3/2 :0,75 =2
vậy x=2
\(x^3+3x^2+2x=0\)
\(\Leftrightarrow x\left(x^2+3x+2\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)
\(x^4+3x^3-x-3=0\)
\(\Leftrightarrow x^3\left(x+3\right)-\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^3-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^3-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)
Vậy \(x\in\left\{-3;1\right\}\)
Ta có: \(x^5-x^4+3x^3+3x^2-x+1=0\)
\(\Leftrightarrow x^5+x^4-2x^4-2x^3+5x^3+5x^2-2x^2-2x+x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^4-2x^3+5x^2-2x+1\right)=0\)
\(\Leftrightarrow x+1=0\)
hay x=-1
Ta có : x4+3x3+4x2+3x+1=0
⇔ ( x4 + x3 ) + ( 2x3 + 2x2 ) + ( 2x2 + 2x ) + ( x + 1 ) = 0
⇔ x3 ( x + 1 ) + 2x2 ( x + 1 ) + 2x ( x+1 ) + ( x + 1 ) =0
⇔ ( x + 1 ) ( x3 + 2x2 + 2x + 1 ) = 0
⇔ ( x + 1 ) [ ( x3 + 1 ) + ( 2x2 + 2x ) ] = 0
⇔ ( x + 1 ) [ (x + 1 ) ( x2 - x +1 ) + 2x ( x + 1 ) ] =0
⇔ ( x +1 ) ( x + 1 ) ( x2 + x +1 ) =0
⇒ \(\left[{}\begin{matrix}x+1=0\\x^{2^{ }}+x+1=0\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}x=-1\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(VoLy\right)\end{matrix}\right.\)
Vậy x = -1
x4+3x3+4x2+3x+1=0
⇔(x4+2x3+x2)+(x3+2x2+1)+(x2+2x+1)=0
⇔x2(x2+2x+1)+x(x2+2x+1)+(x2+2x+1)=0
⇔x2(x+1)2+x(x+1)2+(x+1)2=0
⇔(x+1)2(x2+x+1)=0
Vì x2+x+1=x2+x+\(\dfrac{1}{4}\)+\(\dfrac{3}{4}\)=(x+\(\dfrac{1}{2}\))2+\(\dfrac{3}{4}\)>0 nên phương trình đã cho tương đương:
(x+1)2=0 ⇔(x+1)(x+1)=0 ⇔x=-1.
\(x=0\)
\(3x^3-75x=0\Leftrightarrow3x\left(x^2-25\right)=0\Leftrightarrow3x\left(x-5\right)\left(x+5\right)=0\Leftrightarrow x=0;x=-5;x=5\)