x3-7x-6=0
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25 tháng 10 2021
\(a,\Leftrightarrow\left(4x-8\right)\left(x+1\right)=0\\ \Leftrightarrow4\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ b,\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2=-1\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=-1\\ c,\Leftrightarrow x^2-2x-4x+8=0\\ \Leftrightarrow\left(x-2\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\\ d,\Leftrightarrow x^3-3x^2+3x-9x+2x-6=0\\ \Leftrightarrow\left(x-3\right)\left(x^2+3x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x^2+x+2x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\\x=-2\end{matrix}\right.\)
25 tháng 10 2021
a) \(\Rightarrow4\left(x+1\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
b) \(\Rightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x^2+1\right)=0\)
\(\Rightarrow x=-1\left(do.x^2+1\ge1>0\right)\)
c) \(\Rightarrow x\left(x-4\right)-2\left(x-4\right)=0\)
\(\Rightarrow\left(x-4\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
d) \(\Rightarrow x^2\left(x-3\right)+3x\left(x-3\right)+2\left(x-3\right)\)
\(\Rightarrow\left(x-3\right)\left(x^2+3x+2\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=-1\end{matrix}\right.\)
15 tháng 12 2023
(x-4)(2x+6)=0
=>\(\left[{}\begin{matrix}x-4=0\\2x+6=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=4\\2x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)
SK
1
5 tháng 1 2022
a: \(\Leftrightarrow8x^2+16x+14x+7=0\)
=>(2x+1)(8x+7)=0
=>x=-1/2 hoặc x=-7/8
b: \(=x^3-x-6x-6\)
\(=x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-6\right)=\left(x+1\right)\left(x-3\right)\left(x+2\right)\)
5 tháng 1 2022
\(a,\Rightarrow8x^2+2x+28x+7=0\\ \Rightarrow2x\left(4x+1\right)+7\left(4x+1\right)=0\\ \Rightarrow\left(2x+7\right)\left(4x+1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=-\dfrac{1}{4}\end{matrix}\right.\\ b,Sửa:x^3-7x-6=0\\ \Rightarrow x^3-x-6x-6=0\\ \Rightarrow x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)=0\\ \Rightarrow\left(x+1\right)\left(x^2-x-6\right)=0\\ \Rightarrow\left(x+1\right)\left(x^2-3x+2x-6\right)=0\\ \Rightarrow\left(x+1\right)\left(x-3\right)\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=3\\x=-2\end{matrix}\right.\)
26 tháng 10 2021
a: \(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)
b: \(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
21 tháng 4 2022
a. 3/5 + x = 5/7
x = 5/7 - 3/5
x = 4/35
b. x x 4/11= 1/3
x = 1/3 : 4/11
x = 11/12
NT
3
M
20 tháng 11 2017
\(\left(3x-7x+3-2x\right):\left(x-3\right)=\left(3x-2x-7x+3\right):\left(x-3\right)\)
\(\left(3-6x\right):\left(x-3\right)-\left(3-6x\right):\left(x-3\right)=0\)
\(0=0\)
vậy \(x\in\varnothing\)
H9
HT.Phong (9A5)
CTVHS
2 tháng 5 2023
a) Thu gọn và sắp xếp:
\(P\left(x\right)=2x^3-9x^2+5-4x^3+7x\)
\(P\left(x\right)=\left(2x^3-4x^3\right)-\left(9x^2+2x^2\right)+7x+5\)
\(P\left(x\right)=-2x^3-11x^2+7x+5\)
b) Thay x=1 vào đa thức P(x) ta được:
\(P\left(x\right)=\left(-1\right)^4-\left(-1\right)^3-\left(-1\right)-2=1\)
8 tháng 8 2021
Ta có: \(x^3-7x-6=0\)
\(\Leftrightarrow x^3-x-6x-6=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-x-6\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\\x=-2\end{matrix}\right.\)
8 tháng 8 2021
\(x^3-7x-6=0\)
\(\Leftrightarrow x^3-x-6x-6=0\)
\(\Leftrightarrow x\left(x^2-1\right)-6\left(x+1\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[x\left(x-1\right)-6\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+2x-3x-6\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[x\left(x+2\right)-3\left(x+2\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=-2\\x=3\end{matrix}\right.\)
Vậy...
\(x^3-7x-6=0\)
\(\Leftrightarrow x^3+x^2-x^2-6x-x-6=0\)
\(\Leftrightarrow\left(x^3+x^2\right)-\left(x^2+x\right)-\left(6x+6\right)=0\)
\(\Leftrightarrow x^2\left(x+1\right)-x\left(x+1\right)-6\left(x+1\right)=0\)
\(\Leftrightarrow\left(x^2-x-6\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[x\left(x-3\right)+2\left(x-3\right)\right]\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)\left(x+1\right)=0\)
\(\left[\begin{array}{nghiempt}x+2=0\\x-3=0\\x+1=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=-2\\x=3\\x=-1\end{array}\right.\)
Vây ...................