Trả lời:
1, \(\left(2x-1\right)^2-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(2x-1-x-3\right)\left(2x-1+x+3\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(S=\left\{4;-\dfrac{2}{3}\right\}\)
2, \(x^3-\dfrac{1}{4}x=0\)
\(\Leftrightarrow x\left(x^2-\dfrac{1}{4}\right)=0\)
\(\Leftrightarrow x\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-\dfrac{1}{2}=0\\x+\dfrac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Vậy \(S=\left\{0;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)
3, \(x^3-0,25x=0\)
\(\Leftrightarrow x\left(x^2-0,25\right)=0\)
\(\Leftrightarrow x\left(x-0,5\right)\left(x+0,5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-0,5=0\\x+0,5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=0,5\\x=-0,5\end{matrix}\right.\)
Vậy \(S=\left\{0;0,5;-0,5\right\}\)
Trả lời:
4, \(x^2-x+\dfrac{1}{4}=0\)
\(\Leftrightarrow x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{1}{2}=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(S=\left\{\dfrac{1}{2}\right\}\)
5, \(x^2-10x=-25\)
\(\Leftrightarrow x^2-10x+25=0\)
\(\Leftrightarrow\left(x-5\right)^2=0\)
\(\Leftrightarrow x-5=0\)
\(\Leftrightarrow x=5\)
Vậy \(S=\left\{5\right\}\)
6, \(4x^2-4x=-1\)
\(\Leftrightarrow4x^2-4x+1=0\)
\(\Leftrightarrow\left(2x-1\right)^2=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow2x=1\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(S=\left\{\dfrac{1}{2}\right\}\)
Trả lời:
7, \(\left(2x-1\right)^2-25=0\)
\(\Leftrightarrow\left(2x-1-5\right)\left(2x-1+5\right)=0\)
\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\2x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy \(S=\left\{3;-2\right\}\)
8, \(27x^3+27x^2+9x+1=0\)
\(\Leftrightarrow\left(3x\right)^3+3.\left(3x\right)^2.1+3.3x.1^2+1^3=0\)
\(\Leftrightarrow\left(3x+1\right)^3=0\)
\(\Leftrightarrow3x+1=0\)
\(\Leftrightarrow x=-\dfrac{1}{3}\)
Vậy \(S=\left\{-\dfrac{1}{3}\right\}\)
9, \(9x^2\left(x+1\right)-4\left(x+1\right)=0\)
\(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\3x+2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{2}{3}\\x=-1\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{2}{3};-\dfrac{2}{3};-1\right\}\)
10, \(\left(x+1\right)^3-25\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[\left(x+1\right)^2-25\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+1-5\right)\left(x+1+5\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-4\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-4=0\\x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=4\\x=-6\end{matrix}\right.\)
Vậy \(S=\left\{-1;4;-6\right\}\)
1) Ta có: \(\left(2x-1\right)^2-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(2x-1-x-3\right)\left(2x-1+x+3\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)
2) Ta có: \(x^3-\dfrac{1}{4}x=0\)
\(\Leftrightarrow x\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{matrix}\right.\)
3) Ta có: \(x^3-0.25x=0\)
\(\Leftrightarrow x\left(x-0.5\right)\left(x+0.5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=0.5\\x=-0.5\end{matrix}\right.\)
4) Ta có: \(x^2-x+\dfrac{1}{4}=0\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\dfrac{1}{2}=0\)
hay \(x=\dfrac{1}{2}\)
5) Ta có: \(x^2-10x=-25\)
\(\Leftrightarrow x^2-10x+25=0\)
\(\Leftrightarrow\left(x-5\right)^2=0\)
hay x=5
6) Ta có: \(4x^2-4x=-1\)
\(\Leftrightarrow4x^2+4x+1=0\)
\(\Leftrightarrow\left(2x+1\right)^2=0\)
hay \(x=-\dfrac{1}{2}\)
7) Ta có: \(\left(2x-1\right)^2-25=0\)
\(\Leftrightarrow\left(2x-1-5\right)\left(2x-1+5\right)=0\)
\(\Leftrightarrow\left(2x-6\right)\left(2x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
8) Ta có: \(27x^3+27x^2+9x+1=0\)
\(\Leftrightarrow\left(3x+1\right)^3=0\)
\(\Leftrightarrow3x+1=0\)
hay \(x=-\dfrac{1}{3}\)
9) Ta có: \(9x^2\left(x+1\right)-4\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-2\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{2}{3}\\x=\dfrac{-2}{3}\end{matrix}\right.\)
10) Ta có: \(\left(x+1\right)^3-25\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+1-5\right)\left(x+1+5\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-4\right)\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=4\\x=-6\end{matrix}\right.\)
