c)x3+8=0
d)4x3-16x=0
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N
3
24 tháng 10 2021
a) \(\Rightarrow\left(x-1\right)^3=0\Rightarrow x=1\)
b) \(\Rightarrow\left(x^3-1\right)\left(x^3+1\right)=0\Rightarrow\left(x-1\right)\left(x+1\right)\left(x^2-x+1\right)\left(x^2+x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)(do \(\left\{{}\begin{matrix}x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\\x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\end{matrix}\right.\))
c) \(\Rightarrow4x\left(x^2-9\right)=0\Rightarrow4x\left(x-3\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
d) \(\Rightarrow\left(x-2\right)^3=0\Rightarrow x=2\)
24 tháng 10 2021
a) \(x^3-3x^2+3x-1=0\Rightarrow\left(x-1\right)^3=0\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
b) \(x^6-1=0\Rightarrow\left(x^3\right)^2-1=0\Rightarrow\left(x^3-1\right)\left(x^3+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^3-1=0\\x^3+1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
c) \(4x^3-36x=0\Rightarrow4x\left(x^2-36\right)=0\Rightarrow4x\left(x-6\right)\left(x+6\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}4x=0\\x-6=0\\x+6=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\\x=-6\end{matrix}\right.\)
d) \(x^3-6x^2+12x-8=0\) (đề bài như vậy mới làm đc, nếu là +8 thì mình xin bó tay nhé)
\(\Rightarrow x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3=0\)
\(\Rightarrow\left(x-2\right)^3=0\Rightarrow x-2=0\Rightarrow x=2\)
LL
1
27 tháng 10 2021
b: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\\x=-1\end{matrix}\right.\)
c: \(\Leftrightarrow\left(x-1\right)\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\\x=-5\end{matrix}\right.\)
C
1
C
1
HN
2
27 tháng 10 2021
\(a,\Leftrightarrow x\left(x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\\ b,\Leftrightarrow\left(x+4-4\right)\left(x+4+4\right)=0\\ \Leftrightarrow x\left(x+8\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\\ c,\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\\ d,\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x=5\)
27 tháng 10 2021
a) \(\Leftrightarrow x\left(x+9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-9\end{matrix}\right.\)
b) \(\Leftrightarrow x\left(x+8\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)
c) \(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
d) \(\Leftrightarrow\left(x-5\right)^2=0\\ \Leftrightarrow x=5\)
15 tháng 10 2021
c: \(x^4+x^3-4x^2+x+1\)
\(=x^4-x^3+2x^3-2x^2-2x^2+2x-x+1\)
\(=\left(x-1\right)\left(x^3+2x^2-2x-1\right)\)
\(=\left(x-1\right)\left[\left(x-1\right)\left(x^2+x+1\right)+2x\left(x-1\right)\right]\)
\(=\left(x-1\right)^2\cdot\left(x^2+3x+1\right)\)
H9
HT.Phong (9A5)
CTVHS
11 tháng 8 2023
a) \(x^3-x^2+3x-3>0\)
\(\Leftrightarrow x^2\left(x-1\right)+3\left(x-1\right)>0\)
\(\Leftrightarrow\left(x^2+3\right)\left(x-1\right)>0\)
Mà: \(x^2+3>0\forall x\)
\(\Leftrightarrow x-1>0\)
\(\Leftrightarrow x>1\)
b) \(x^3+x^2+9x+9< 0\)
\(\Leftrightarrow x^2\left(x+1\right)+9\left(x+1\right)< 0\)
\(\Leftrightarrow\left(x^2+9\right)\left(x+1\right)< 0\)
Mà: \(x^2+9>0\forall x\)
\(\Leftrightarrow x+1< 0\)
\(\Leftrightarrow x< -1\)
d) \(4x^3-14x^2+6x-21< 0\)
\(\Leftrightarrow2x^2\left(2x-7\right)+3\left(2x-7\right)< 0\)
\(\Leftrightarrow\left(2x^2+3\right)\left(2x-7\right)< 0\)
Mà: \(2x^2+3>0\forall x\)
\(\Leftrightarrow2x-7< 0\)
\(\Leftrightarrow2x< 7\)
\(\Leftrightarrow x< \dfrac{7}{2}\)
d) \(x^2\left(2x^2+3\right)+2x^2>-3\)
\(\Leftrightarrow2x^4+3x^2+2x^2+3>0\)
\(\Leftrightarrow2x^4+5x^2+3>0\)
\(\Leftrightarrow\left(x^2+1\right)\left(2x^2+3\right)>0\)
Mà:
\(x^2+1>0\forall x\)
\(2x^2+3>0\forall x\)
\(\Rightarrow x\in R\)
11 tháng 8 2023
a: =>x^2(x-1)+3(x-1)>0
=>(x-1)(x^2+3)>0
=>x-1>0
=>x>1
b: =>x^2(x+1)+9(x+1)<0
=>(x+1)(x^2+9)<0
=>x+1<0
=>x<-1
c: 4x^3-14x^2+6x-21<0
=>2x^2(2x-7)+3(2x-7)<0
=>2x-7<0
=>x<7/2
d: =>x^2(2x^2+3)+2x^2+3>0
=>(2x^2+3)(x^2+1)>0(luôn đúng)
T
1
18 tháng 8 2023
a: =>2x^2+9x-6x-27=0
=>x(2x+9)-3(2x+9)=0
=>(2x+9)(x-3)=0
=>x=3 hoặc x=-9/2
b: =>-10x^2+6x-5x+3=0
=>-2x(5x-3)-(5x-3)=0
=>(5x-3)(-2x-1)=0
=>x=-1/2 hoặc x=5/3
c: =>-x^3+2x^2-x^2+4=0
=>-x^2(x-2)-(x-2)(x+2)=0
=>(x-2)(-x^2-x-2)=0
=>x-2=0
=>x=2
d: =>(x^3+8)-4x(x+2)=0
=>(x+2)(x^2-2x+4)-4x(x+2)=0
=>(x+2)(x^2-6x+4)=0
=>x=-2 hoặc \(x=3\pm\sqrt{5}\)
C
1
