-3x^3+8x^2+8x-24=0
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1 tháng 9 2023
c) \(x^2-9=2\cdot\left(x+3\right)^2\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)-2\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(x+3\right)\left[x-3-2\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-3-2x-6\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(-x-9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-9\end{matrix}\right.\)
b) \(x^3-3x^2+3x-1=0\)
\(\Leftrightarrow x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
d) \(x^2-8x+3x-24=0\)
\(\Leftrightarrow\left(x^2-8x\right)+\left(3x-24\right)=0\)
\(\Leftrightarrow x\left(x-8\right)+3\left(x-8\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=8\end{matrix}\right.\)
1 tháng 9 2023
a) \(x^2-9=2\left(x+3\right)^2\)
\(\Leftrightarrow\left(x+3\right)\left(x-3\right)=2\left(x+3\right)^2\)
\(\Leftrightarrow2\left(x+3\right)^2-\left(x+3\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left[2\left(x+3\right)-\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left[2x+6-x+3\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+9\right)=0\)
\(\)\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+9=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-9\end{matrix}\right.\)
b) \(x^2-8x+3x-24=0\)
\(\Leftrightarrow\left(x-8\right)x+3\left(x-8\right)=0\)
\(\Leftrightarrow\left(x-8\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-3\end{matrix}\right.\)
c) \(x^3-3x^2+3x-1=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
4 tháng 3 2019
Với dạng bài này ta chỉ việc chia hoocne là ra nhé!
\(C1:x^4+x^3-8x^2-9x-9=0\\ \Leftrightarrow\left(x-3\right)\left(x^3+4x^2+4x+3\right)\\ \Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x^2+x+1\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+3=0\\x^2+x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x^2+x+1=0\left(VN\right)\end{matrix}\right.\)
\(C2:x^4+2x^3-3x^2-8x-4=0\\ \Leftrightarrow\left(x+1\right)\left(x^3+x^2-4x-4\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x+1\right)\left(x^2-4\right)=0\\ \Leftrightarrow\left(x+1\right)^2\left(x^2-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)^2=0\\x^2-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-2\end{matrix}\right.\)
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 \(S=\left\{-2;\pm5\right\}\)
6 tháng 9 2017
a) x2 + 10x - 2x - 20 = 0
=> x(x + 10) - 2(x + 10) = 0
=> (x - 2)(x + 10) = 0
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+10=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-10\end{cases}}\)
b) \(x^2-5x-24=0\)
\(\Rightarrow x^2-5x+\frac{25}{4}-\frac{121}{4}=0\)
\(\Rightarrow\left(x-\frac{5}{2}\right)^2=\frac{121}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-\frac{5}{2}\right)^2=\left(-\frac{11}{2}\right)^2\\\left(x-\frac{5}{2}\right)^2=\left(\frac{11}{2}\right)^2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{5}{2}=\left(-\frac{11}{2}\right)\\x-\frac{5}{2}=\frac{11}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{6}{2}=3\\x=\frac{16}{2}=8\end{cases}}\)
c) x2 - 8x + 3x - 24 = 0
=> x(x - 8) + 3(x - 8) = 0
=> (x + 3)(x - 8) = 0
\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-8=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-3\\x=8\end{cases}}\)
NQ
0
PD
3
9 tháng 8 2021
\(\left(x^2-8x\right)-3x+24=0\)
\(\Leftrightarrow x\left(x-8\right)-3\left(x-8\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-8\right)=0\Leftrightarrow x=3;x=8\)
9 tháng 8 2021
bài 59; tìm x biết
2, ( x mũ 2 - 8x ) - 3x + 24 = 0
x= 3 ; x = 8
MH
17 tháng 2 2021
1/ \(x^4+x^2-2=0\)
\(\Leftrightarrow\left(x^2\right)^2-x^2+2x^2-2=0\\ \Leftrightarrow x^2\left(x^2-1\right)+2\left(x^2-1\right)=0\\ \Leftrightarrow\left(x^2+2\right)\left(x^2-1\right)=0\\ \Leftrightarrow\left(x^2+2\right)\left(x-1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+2=0\\x+1=0\\x-1-0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
2/ \(x^3+3x^2+6x+4=0\)
\(\Leftrightarrow\left(x^3+x^2\right)+\left(2x^2+2x\right)+\left(4x+4\right)=0\\ \Leftrightarrow x^2\left(x+1\right)+2x\left(x+1\right)+4\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x^2+2x+4\right)=0\)
\(\Leftrightarrow x+1=0\) (do \(x^2+2x+4=\left(x+1\right)^2+3>0,\forall x\))
\(\Leftrightarrow x=-1\).
3/ \(x^3-6x^2+8x=0\)
\(\Leftrightarrow x\left(x^2-6x+8\right)=0\\ \Leftrightarrow x\left[\left(x^2-2x\right)-\left(4x-8\right)\right]=0\\ \Leftrightarrow x\left[x\left(x-2\right)-4\left(x-2\right)\right]=0\\ \Leftrightarrow x\left(x-2\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=4\end{matrix}\right.\)
4/ \(x^4-8x^3-9x^2=0\)
\(\Leftrightarrow x^2\left(x^2-8x-9\right)=0\\ \Leftrightarrow x^2\left(x^2-9x+x-9\right)=0\\ \Leftrightarrow x^2\left(x\left(x-9\right)+\left(x-9\right)\right)=0\\ \Leftrightarrow x^2\left(x+1\right)\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=0\\x+1=0\\x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=9\end{matrix}\right.\)
TH
1
19 tháng 7 2019
a) x^2 +3x-2x-6=0
x^2 + x = 6
x^2 + 0.5x + 0.5x = 6
x (x + 0.5) + 0.5 (x + 0.5) =5.75
(x+0.5)^2 = 5.75
13 tháng 8 2018
a) \(\left(x+\frac{1}{x}\right)^2+2\left(x+\frac{1}{x}\right)-8=0\)
\(\Leftrightarrow x^2+2x+\frac{1}{x^2}+\frac{2}{x}-6=0\)
\(\Leftrightarrow x^2x^2+2xx^2+\frac{1}{x^2}x^2+\frac{2}{x^2}x^2-6x^2=0.x^2\)
\(\Leftrightarrow x^4+2x^3+1+2x-6x^2=0\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=\sqrt{3}\end{cases}}\)
b) \(x^3-8x^2-8x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-9x+1\right)=0\)
\(\Rightarrow x=-1\)
c) \(x^4-3x^3+4x^2-3x+1=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x^2-x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)
\(-3x^3+8x^2+8x-24=0\)
\(\Leftrightarrow-3x^3+6x^2+2x^2-4x+12x-24=0\)
\(\Leftrightarrow-3x^2\left(x-2\right)+2x\left(x-2\right)+12\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(-3x^2+2x+12\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\-3x^2+2x+12=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\-3x^2+2x+12=0\left(^∗\right)\end{cases}}\)
\(\left(^∗\right)\Leftrightarrow-3x^2+2x+12=0\)
\(\Leftrightarrow-3\left(x^2-\frac{2}{3}x-4\right)=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\frac{1}{3}+\frac{1}{9}-\frac{37}{9}=0\)
\(\Leftrightarrow\left(x-\frac{1}{3}\right)^2=\frac{37}{9}=\left(\frac{\pm\sqrt{37}}{3}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{\sqrt{37}}{3}\\x-\frac{1}{3}=\frac{-\sqrt{37}}{3}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1+\sqrt{37}}{3}\\x=\frac{1-\sqrt{37}}{3}\end{cases}}}\)
Vậy....