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3 tháng 2 2019
\(2x^3+7x^2+7x+2=0\)
\(\Leftrightarrow\left(2x^3+4x^2\right)+\left(3x^2+6x\right)+\left(x+2\right)=0\)
\(\Leftrightarrow2x^2\left(x+2\right)+3x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(2x^2+3x+1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left[2x\left(x+1\right)+\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+1\right)\left(2x+1\right)=0\)
.......................................................................................
\(x^3-8x^2-8x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x+1\right)-8x\left(x+1\right)=0\)
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TM
20 tháng 6 2017
b)\(3x^3+6x^2-75x-150=0\Leftrightarrow3\left(x^3+2x^2-25x-50\right)=0\Leftrightarrow x^3+2x^2-25x-50=0\)
<=>\(x^2\left(x+2\right)-25\left(x+2\right)=0\Leftrightarrow\left(x^2-25\right)\left(x+2\right)=0\Leftrightarrow\left(x-5\right)\left(x+5\right)\left(x+2\right)=0\)
<=>x-5=0 hoặc x+5=0 hoặc x+2=0<=>x=5 hoặc x=-5 hoặc x=-2
c)\(2x^5-3x^4+6x^3-8x^2+3=0\Leftrightarrow2x^5+x^4-4x^4-2x^3+8x^3+4x^2-12x^2+3=0\)
<=>\(x^4\left(2x+1\right)-2x^3\left(2x+1\right)+4x^2\left(2x+1\right)-3\left(4x^2-1\right)=0\)
<=>\(x^4\left(2x+1\right)-2x^3\left(2x+1\right)+4x^2\left(2x+1\right)-3\left(2x-1\right)\left(2x+1\right)=0\)
<=>\(\left(2x+1\right)\left(x^4-2x^3+4x^2-6x+3\right)=0\)
<=>\(\left(2x+1\right)\left(x^4-2x^3+x^2+3x^2-6x+3\right)=0\)
<=>\(\left(2x+1\right)\left[x^2\left(x^2-2x+1\right)+3\left(x^2-2x+1\right)\right]=0\)
<=>\(\left(2x+1\right)\left(x^2+3\right)\left(x^2-2x+1\right)=0\Leftrightarrow\left(2x+1\right)\left(x^2+3\right)\left(x-1\right)^2=0\)
Vì \(x^2\ge0\Rightarrow x^2+3\ge3>0\Rightarrow\orbr{\begin{cases}2x+1=0\\\left(x-1\right)^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)
20 tháng 6 2017
a) 2x3 - x2 - 8x + 4 = 0
x2.(2x - 1) - 4.(2x - 1) = 0
(x2 - 4)(2x - 1) = 0
\(\Rightarrow\orbr{\begin{cases}x^2-4=0\\2x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x^2=4\\x=\frac{1}{2}\end{cases}}\)
Với x2 = 4
=> x = 2 hoặc x = -2
=> x = {-2 ; 2 ; \(\frac{1}{2}\))
ND
1
12 tháng 4 2023
a:=>6x^2-8x+4x-6x^2<-4
=>-4x<-4
=>x>1
b: =>6x+8x^2-8x^2-24x>5
=>-18x>5
=>x<-5/18
12 tháng 4 2023
a)\(6x^2-8x+2x\left(2-3x\right)< -4\)
\(\Leftrightarrow6x^2-8x+4x-6x^2< -4\)
\(\Leftrightarrow-4x< -4\)
\(\Leftrightarrow-4x.\dfrac{-1}{4}>-4\cdot\dfrac{-1}{4}\)
\(\Leftrightarrow x>1\)
Vậy bất phương trình có nghiệm là \(S=\left\{xIx>1\right\}\)
b)\(2\left(3x+4x^2\right)-8x\left(x+3\right)>5\)
\(\Leftrightarrow6x+8x^2-8x^2-24x>5\)
\(\Leftrightarrow-18x>5\)
\(\Leftrightarrow-18x\cdot\dfrac{-1}{18}< 5\cdot\dfrac{-1}{18}\)
\(\Leftrightarrow x< -\dfrac{5}{18}\)
Vậy bất phương trình có nghiệm là \(S=\left\{xIx< -\dfrac{5}{18}\right\}\)
12 tháng 5 2022
*vn:vô nghiệm.
a. \(\left(x^2-2\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2=0\\x^2+x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\pm\sqrt{2}\)
-Vậy \(S=\left\{\pm\sqrt{2}\right\}\).
b. \(16x^2-8x+5=0\)
\(\Leftrightarrow16x^2-8x+1+4=0\)
\(\Leftrightarrow\left(4x-1\right)^2+4=0\) (vô lí)
-Vậy S=∅.
c. \(2x^3-x^2-8x+4=0\)
\(\Leftrightarrow x^2\left(2x-1\right)-4\left(2x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(2x-1\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\pm2\end{matrix}\right.\)
-Vậy \(S=\left\{\dfrac{1}{2};\pm2\right\}\).
d. \(3x^3+6x^2-75x-150=0\)
\(\Leftrightarrow3x^2\left(x+2\right)-75\left(x+2\right)=0\)
\(\Leftrightarrow3\left(x+2\right)\left(x^2-25\right)=0\)
\(\Leftrightarrow3\left(x+2\right)\left(x+5\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\pm5\end{matrix}\right.\)
-Vậy \(S=\left\{-2;\pm5\right\}\)
1 tháng 3 2019
1) \(x^4-6x^3-x^2+54x-72=0\)
\(\Leftrightarrow x^3\left(x-2\right)-4x^2\left(x-2\right)-9x\left(x-2\right)+36\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-4x^2-9x+36\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x-4\right)-9\left(x-4\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-4\right)\left(x-3\right)\left(x+3\right)=0\)
Tự làm nốt...
2) \(x^4-5x^2+4=0\)
\(\Leftrightarrow x^2\left(x^2-1\right)-4\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)=0\)
Tự làm nốt...
1 tháng 3 2019
\(x^4-2x^3-6x^2+8x+8=0\)
\(\Leftrightarrow x^3\left(x-2\right)-6x\left(x-2\right)-4\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3-6x-4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x^2\left(x+2\right)-2x\left(x+2\right)-2\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x^2-2x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left[\left(x-1\right)^2-\left(\sqrt{3}\right)^2\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-1-\sqrt{3}\right)\left(x-1+\sqrt{3}\right)=0\)
...
\(2x^4-13x^3+20x^2-3x-2=0\)
\(\Leftrightarrow2x^3\left(x-2\right)-9x^2\left(x-2\right)+2x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^3-9x^2+2x+1\right)=0\)
Bí
Ta có :
\(3x^3-8x^2-2x+4=\left(3x-2\right)\left(x^2-2x-2\right)\)
\(\Leftrightarrow\left(3x-2\right)\left(x^2-2x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\x^2-2x-2=0\end{cases}}\)
Th1 : \(3x-2=0\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\)
Th2: \(x^2-2x-2=0\)
\(\Leftrightarrow x^2-2x+1=3\)
\(\Leftrightarrow\left(x-1\right)^2=3\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=\sqrt{3}\\x-1=-\sqrt{3}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1+\sqrt{3}\\x=1-\sqrt{3}\end{cases}}}\)
Vậy phương trình có 3 nghiệm : \(x=1\), \(x=1\pm\sqrt{3}\)
\(3x^3-8x^2-2x+4=0\)
\(\Leftrightarrow\left(3x-2\right)\left(x^2-2x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\x^2-2x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=1\pm\sqrt{3}\end{cases}}\)
Vậy tập nghiệm của phương trình \(S=\left\{\frac{2}{3};1\pm\sqrt{3}\right\}\)