a: Ta có: \(x\left(2-x\right)+\left(x^2+x\right)=7\)
\(\Leftrightarrow2x-x^2+x^2+x=7\)
\(\Leftrightarrow3x=7\)
hay \(x=\dfrac{7}{3}\)
b: Ta có: \(\left(2x+1\right)^2-x\left(4-5x\right)=17\)
\(\Leftrightarrow4x^2+4x+1-4x+5x^2=17\)
\(\Leftrightarrow9x^2=16\)
\(\Leftrightarrow x^2=\dfrac{16}{9}\)
hay \(x\in\left\{\dfrac{4}{3};-\dfrac{4}{3}\right\}\)
c: Ta có: \(\left(x-4\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left(x-4-2x-1\right)\left(x-4+2x+1\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=1\end{matrix}\right.\)
d: ta có: \(\dfrac{2x^3-8x^2+10x}{2x}=0\)
\(\Leftrightarrow x^2-4x+5=0\)
\(\Leftrightarrow\left(x-2\right)^2+1=0\)(vô lý)
