Tìm x \(\in\) Z biết:
a) x2 = 16
b) x3 = -8
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10 tháng 10 2021
c: Ta có: \(x^3-12x^2+48x-64=0\)
\(\Leftrightarrow x-4=0\)
hay x=4
10 tháng 10 2021
c: Ta có: \(x^3-12x^2+48x-64=0\)
\(\Leftrightarrow x-4=0\)
hay x=4
25 tháng 8 2021
a: Ta có: \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=16\)
\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=16\)
\(\Leftrightarrow9x+7=16\)
\(\Leftrightarrow9x=9\)
hay x=1
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) \(\Rightarrow x^2\left(x-3\right)+3x\left(x-3\right)+2\left(x-3\right)\)
\(\Rightarrow\left(x-3\right)\left(x^2+3x+2\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=-1\end{matrix}\right.\)
NM
2
15 tháng 11 2021
a) \(\left(2x-3\right)\left(x+2\right)-\left(4x-2\right)\left(x-5\right)=-16\)
\(\Rightarrow2x^2+x-6-4x^2+22x-10=-16\)
\(\Rightarrow2x^2-23x=0\Rightarrow x\left(2x-23\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{23}{2}\end{matrix}\right.\)
b) \(7x^2-7=x^2-2x+1\)
\(\Rightarrow7\left(x^2-1\right)-\left(x^2-2x+1\right)=0\)
\(\Rightarrow7\left(x-1\right)\left(x+1\right)-\left(x-1\right)^2=0\)
\(\Rightarrow\left(x-1\right)\left(7x+7-x+1\right)=0\Rightarrow2\left(x-1\right)\left(3x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{4}{3}\end{matrix}\right.\)
MH
15 tháng 11 2021
a) \(\left(2x-3\right)\left(x+2\right)-\left(4x-2\right)\left(x-5\right)=-16\)
\(2x^2+x-6-4x^2+22x-10=-16\)
\(-2x^2+23x-16=-16\)
\(23x-2x^2=0\)
\(x\left(23-2x\right)=0\)
⇔ \(\left[{}\begin{matrix}x=0\\x=\dfrac{23}{2}\end{matrix}\right.\)
b) \(7x^2-7=x^2-2x+1\)
\(7\left(x^2-1\right)=\left(x-1\right)^2\)
\(7\left(x-1\right)\left(x+1\right)-\left(x-1\right)^2=0\)
\(\left(7x+7\right)\left(x-1\right)-\left(x-1\right)^2=0\)
\(\left(x-1\right)\left(7x+7-x+1\right)=0\)
\(\left(x-1\right)\left(6x+8\right)=0\)
⇔ \(\left[{}\begin{matrix}x=1\\x=-\dfrac{4}{3}\end{matrix}\right.\)
23 tháng 10 2021
\(a,\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-2\right)^3=0\Leftrightarrow x-2=0\Leftrightarrow x=2\\ c,\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
14 tháng 8 2021
a: Ta có: \(\left(x^2+2\right)\left(x-4\right)-\left(x+2\right)^3=-16\)
\(\Leftrightarrow x^3-4x^2+2x-8-x^3-6x^2-12x-8=-16\)
\(\Leftrightarrow-10x^2-10x=0\)
\(\Leftrightarrow-10x\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
c: Ta có: \(x^3+3x^2+3x+28=0\)
\(\Leftrightarrow\left(x+1\right)^3=-27\)
\(\Leftrightarrow x+1=-3\)
hay x=-4
a) x2 = 16
x2 = 42 = (-4)2
=> \(x\in\left\{4;-4\right\}\)
Vậy \(x\in\left\{4;-4\right\}\)
b) x3 = -8
x3 = (-2)3
=> x = -2
Vậy x = -2
a )
\(x^2=16\)
\(x^2=4^2\)
\(\Rightarrow x=4\).
b )
\(x^3=-8\)
\(x^3=\left(-2\right)^3\)
\(\Rightarrow x=\left(-2\right)\).