\(a,\left(2x-5\right)^2=\left(x-2\right)^2\)
\(\Rightarrow\left(2x-5\right)^2-\left(x-2\right)^2=0\)
\(\Rightarrow\left(2x-5-x+2\right)\left(2x-5+x-2\right)=0\)
\(\Rightarrow\left(x-3\right)\left(3x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\3x-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{7}{3}\end{matrix}\right.\)
\(b,\left(x+1\right)\left(2-x\right)-\left(3x+5\right)\left(x+2\right)=-4x^2+1\)
\(\Rightarrow2x-x^2+2-x-3x^2-6x-5x-10=-4x^2+1\)
\(\Rightarrow-10x-4x^2-12=-4x^2+1\)
\(\Rightarrow-10x-4x^2-12+4x^2-1=0\)
\(\Rightarrow-10x-13=0\)
\(\Rightarrow x=-\dfrac{13}{10}\)
\(b,\left(x+1\right)\left(2-x\right)-\left(3x+5\right)\left(x+2\right)=-4x^2+1\) \(\Rightarrow2x-x^2+2-x-3x^2-6x-5x-10+4x^2-1=0\)
\(\Rightarrow-10x-9=0\)
\(\Rightarrow x=-\dfrac{9}{10}\)
Trần Quốc Lộc, @Trương Tú Nhi, Hung nguyen, Ribi Nkok Ngok, Toshiro Kiyoshi, @Aki Tsuki, ...
