a) \(\Rightarrow x\left(x^2-0,25\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x^2=0,25\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\\x=0,5\\x=-0,5\end{matrix}\right.\)
b) Thiếu đề rồi bạn
c) \(\Rightarrow\left(x-1\right)\left(x^2+x+1\right)=0\Rightarrow x-1=0\Rightarrow x=1\)
(do \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\))
d) \(\Rightarrow2x\left(3x-2\right)-1\left(3x-2\right)=0\Rightarrow\left(3x-2\right)\left(2x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
a: \(\Leftrightarrow x\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)=0\)
hay \(x\in\left\{0;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)
b: \(=x^2\left(x+1\right)^2\)
c: =>x-1=0
hay x=1
d: \(\Leftrightarrow6x^2-3x-4x+2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x-2\right)=0\)
hay \(x\in\left\{\dfrac{1}{2};\dfrac{2}{3}\right\}\)
a)\(< =>x\left(x^2-0,25\right)=0\)
\(< =>\left[{}\begin{matrix}x=0\\x^2=0,25\end{matrix}\right.\)
\(< =>\left[{}\begin{matrix}x=0\\x=\pm0,5\end{matrix}\right.\)
b)\(< =>x^2\left(x^2+2x+1\right)=0\)
\(< =>x^2\left(x+1\right)^2=0\)
\(< =>\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\)
\(< =>\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
c)\(< =>x^3=1\)
\(< =>x=1\)
d)\(< =>\left(3x-2\right)\left(2x-1\right)=0\)
\(< =>\left[{}\begin{matrix}3x-2=0\\2x-1=0\end{matrix}\right.\)
\(< =>\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(a,x\left(x^2-0,25\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-0.25=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm\dfrac{1}{2}\end{matrix}\right.\)
\(b,\left(x^2+x\right)^2\)
\(\left(x\left(x+1\right)\right)^2\)
\(x^2\left(x+1\right)^2\)
\(c,x^3=1\)
\(x=1\)
\(d,\left(2x-1\right)\left(3x-2\right)=0\)
\(x=\left[{}\begin{matrix}\dfrac{1}{2}\\\dfrac{2}{3}\end{matrix}\right.\)
\(a,PT\Leftrightarrow x\left(x-0,5\right)\left(x+0,5\right)\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm0,5\end{matrix}\right.\)
Vậy: \(S=\left\{0;0,5\right\}\)
\(b,\) Sửa đề: \(x^4+2x^3+x^2=0\)
\(\Leftrightarrow x^2\left(x+1\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
Vậy: \(S=\left\{0;-1\right\}\)
\(c,PT\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=0\left(x^2+x+1>0\right)\)
\(\Rightarrow x=1\)
Vậy: \(S=\left\{1\right\}\)
\(d,PT\Leftrightarrow2x\left(3x-2\right)-\left(3x-2\right)=0\Leftrightarrow\left(3x-2\right)\left(2x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{2}{3};\dfrac{1}{2}\right\}\)
