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\(a,\Leftrightarrow\left(4x-8\right)\left(x+1\right)=0\\ \Leftrightarrow4\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ b,\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x^2=-1\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=-1\\ c,\Leftrightarrow x^2-2x-4x+8=0\\ \Leftrightarrow\left(x-2\right)\left(x-4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\\ d,\Leftrightarrow x^3-3x^2+3x-9x+2x-6=0\\ \Leftrightarrow\left(x-3\right)\left(x^2+3x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x^2+x+2x+2\right)=0\\ \Leftrightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\\x=-2\end{matrix}\right.\)
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) \(\Rightarrow x^2\left(x-3\right)+3x\left(x-3\right)+2\left(x-3\right)\)
\(\Rightarrow\left(x-3\right)\left(x^2+3x+2\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\\x=-1\end{matrix}\right.\)
2)
a) \(3x \left(x^2-4\right)=0 \)
\(\Leftrightarrow3x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-2=0\\x+2=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy x=0 ; x=2 ; x=-2
b) \(2x^2-x-6=0\)
\(\Leftrightarrow2x^2-4x+3x-6=0\)
\(\Leftrightarrow\left(2x^2-4x\right)+\left(3x-6\right)=0\)
\(\Leftrightarrow2x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-3}{2}\end{matrix}\right.\)
Vậy x=2 ; \(x=\dfrac{-3}{2}\)
Câu 1 .
a) x3 + x2 + x
= x( x2 + x + 1)
b) xy + y2 - x - y
= x( y - 1) + y( y - 1)
= ( y - 1)( x + y)
2)
a) \(3x\left(x^2-4\right)=0\)
\(\Leftrightarrow3x\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\x-2=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
Vậy \(x=0;x=2vàx=-2\)
b) \(2x^2-x-6=0\)
\(\Leftrightarrow2x^2-4x+3x-6=0\)
\(\Leftrightarrow\left(2x^2-4x\right)+\left(3x-6\right)=0\)
\(\Leftrightarrow2x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\2x=-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-3}{2}\end{matrix}\right.\)
Vậy \(x=2vàx=\dfrac{-3}{2}\)
\(\Leftrightarrow\left(x-2\right)^3=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)
a) x = -1. b) x = 4 hoặc x = 5.
c) x = ± 2 . d) x = 1 hoặc x = 2.
8 x 3 + 12 x 2 + 6x + 1 = 0
2 x 3 + 3. 2 x 2 .1 + 3.(2x). 1 2 + 1 3 = 0
2 x + 1 3 = 0
2x + 1 = 0
x = (-1)/2
Là x mũ 3 cộng 8 nha
\(\left(x-8\right).\left(x^3+8\right)=0\)
\(\Rightarrow x-8=0\left(x^3+8\ne0\right)\)
\(x=0+8\)
\(\Rightarrow x=8\)
Chúc bạn học tốt!