a. \(\left(2x-1\right)^2-\left(4x^2-1\right)=0\Leftrightarrow\left(2x-1\right)^2-\left[\left(2x\right)^2-1^2\right]=0\Leftrightarrow\left(2x-1\right)^2-\left(2x-1\right)\left(2x+1\right)=0\Leftrightarrow\left(2x-1\right)\left(2x-1-2x-1\right)=0\Leftrightarrow-2\left(2x-1\right)=0\Leftrightarrow2x-1=0\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\)
Vay \(x=\dfrac{1}{2}\)
b. \(x^2\left(x^2+4\right)-x^2-4=0\Leftrightarrow x^2\left(x^2+4\right)-\left(x^2+4\right)=0\Leftrightarrow\left(x^2+4\right)\left(x^2-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x^2+4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=-4\\x^2=1\end{matrix}\right.\Leftrightarrow x=\pm1\)
\(x^2=-4\) bị loại vì \(x^2\ge0\)
Vay \(x=\pm1\)
a, \(\left(2x-1\right)^2-\left(4x^2-1\right)=0\)
\(\Rightarrow4x^2-4x-1-4x^2+1=0\)
\(\Rightarrow-4x=0\Rightarrow x=0\)
b, \(x^2\left(x^2+4\right)-x^2-4=0\)
\(\Rightarrow x^4+4x^2-x^2-4=0\)
\(\Rightarrow x^4+3x^3-4=0\)
\(\Rightarrow x^4-x^2+4x^2-4=0\)
\(\Rightarrow x^2.\left(x^2-1\right)+4.\left(x^2-1\right)=0\)
\(\Rightarrow\left(x^2-1\right).\left(x^2+4\right)=0\)
Với mọi giá trị của \(x\in R\) ta có:
\(x^2+4\ge4>0\)
\(\Rightarrow x^2-1=0\Rightarrow x^2=1\Rightarrow x=\pm1\)
Chúc bạn học tốt!!!
\(a,\left(2x-1\right)^2-\left(4x^2-1\right)=0\)
\(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)
\(\Leftrightarrow2-4x=0\Leftrightarrow2\left(1-2x\right)=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy \(S=\left\{x=\dfrac{1}{2}\right\}\)
\(b,x^2\left(x^2+4\right)-x^2-4=0\)
\(\Leftrightarrow x^2\left(x^2+4\right)-\left(x^2+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x^2+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x^2=-4\left(\text{ vô lí }\right)\end{matrix}\right.\)
Vậy \(S=\left\{-1;1\right\}\)
a. (2x−1)2−(4x2−1)=0⇔(2x−1)2−[(2x)2−12]=0⇔(2x−1)2−(2x−1)(2x+1)=0⇔(2x−1)(2x−1−2x−1)=0⇔−2(2x−1)=0⇔2x−1=0⇔2x=1⇔x=12(2x−1)2−(4x2−1)=0⇔(2x−1)2−[(2x)2−12]=0⇔(2x−1)2−(2x−1)(2x+1)=0⇔(2x−1)(2x−1−2x−1)=0⇔−2(2x−1)=0⇔2x−1=0⇔2x=1⇔x=12
Vay x=12x=12
b. x2(x2+4)−x2−4=0⇔x2(x2+4)−(x2+4)=0⇔(x2+4)(x2−1)=0⇔[x2+4=0x2−1=0⇔[x2=−4x2=1⇔x=±1x2(x2+4)−x2−4=0⇔x2(x2+4)−(x2+4)=0⇔(x2+4)(x2−1)=0⇔[x2+4=0x2−1=0⇔[x2=−4x2=1⇔x=±1
x2=−4x2=−4 bị loại vì x2≥0
