a, \(\left(x-1\right)^2-\left(2x+5\right)^2=0\Leftrightarrow\left(-x-6\right)\left(3x+4\right)=0\Leftrightarrow x=-6;x=-\dfrac{4}{3}\)
b, \(x^2-\left(x-1\right)^2-\left(2x-1\right)^2=0\)
\(\Leftrightarrow x^2-x^2+2x-1-4x^2+4x-1=0\)
\(\Leftrightarrow-4x^2+6x-2=0\Leftrightarrow x=1;x=\dfrac{1}{2}\)
c, \(x^3+8=-2x\left(x+2\right)\Leftrightarrow\left(x+2\right)\left(x^2-2x+4\right)=-2x\left(x+2\right)\)
\(\Leftrightarrow\left(x+2\right)\left(x^2-2x+4+2x\right)=0\Leftrightarrow x=-2\)
vì x^2 + 4 > 0
d, \(4x^2+8x-5=0\Leftrightarrow\left(2x-1\right)\left(2x+5\right)=0\Leftrightarrow x=\dfrac{1}{2};x=-\dfrac{5}{2}\)
Bài 11:
\(a,\left(x-1\right)^2-\left(2x+5\right)^2=0\\ \Rightarrow\left(x-1-2x-5\right)\left(x-1+2x+5\right)=0\\ \Rightarrow\left(-x-6\right)\left(3x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-6\\x=-\dfrac{4}{3}\end{matrix}\right.\)
Vậy ...
\(b,x^2-\left(1-x\right)^2-\left(2x-1\right)^2=0\\ \Rightarrow x^2-\left(1-2x+x^2\right)-\left(4x^2-4x+1\right)=0\\ \Rightarrow x^2-1+2x-x^2-4x^2+4x-1=0\\ \Rightarrow-4x^2+6x-2=0\\ \Rightarrow2x^2-3x+1=0\\ \Rightarrow\left(2x^2-2x\right)-\left(x-1\right)=0\\ \Rightarrow2x\left(x-1\right)-x\left(x-1\right)=0\\ \Rightarrow\left(x-1\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)
Vậy ...
\(c,x^3+8=-2x\left(x+2\right)\\ \Rightarrow x^3+8=-2x^2-4x\\ \Rightarrow x^3+2x^2+4x+8=0\\ \Rightarrow x^2\left(x+2\right)+4\left(x+2\right)=0\\ \Rightarrow\left(x^2+4\right)\left(x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2=-4\left(vô.lí\right)\\x=-2\end{matrix}\right.\)
Vậy ...
\(d,4x^2+8x-5=0\\ \Rightarrow\left(4x^2-2x\right)+\left(10x-5\right)=0\\ \Rightarrow2x\left(2x-1\right)+5\left(2x-1\right)=0\\ \Rightarrow\left(2x-1\right)\left(2x+5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
Vậy ...
