a: Ta có: \(2x+3=x+1\)
\(\Leftrightarrow2x-x=1-3\)
hay x=-2
b: Ta có: \(2x\left(2x-1\right)-\left(2x+3\right)^2=5\)
\(\Leftrightarrow4x^2-2x-4x^2-12x-9=5\)
\(\Leftrightarrow-14x=14\)
hay x=-1
c: Ta có: \(4x^2-25\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(2x-5x-10\right)\left(2x+5x+10\right)=0\)
\(\Leftrightarrow\left(-3x-10\right)\left(7x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-10}{3}\\x=-\dfrac{10}{7}\end{matrix}\right.\)
d: Ta có: \(2x^2+7x+5=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{5}{2}\end{matrix}\right.\)
e: Ta có: \(4x^2-4x=-1\)
\(\Leftrightarrow4x^2-4x+1=0\)
\(\Leftrightarrow2x-1=0\)
hay \(x=\dfrac{1}{2}\)
f: Ta có: \(\dfrac{1}{9}x^3-x=0\)
\(\Leftrightarrow x\left(\dfrac{1}{9}x^2-1\right)=0\)
\(\Leftrightarrow x\left(\dfrac{1}{3}x-1\right)\left(\dfrac{1}{3}x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=-3\end{matrix}\right.\)
g: Ta có: \(x^3+3x^2+3x=7\)
\(\Leftrightarrow\left(x+1\right)^3=8\)
\(\Leftrightarrow x+1=2\)
hay x=1
a, x = -2
b, x = -1
c, x = -10/3 hoặc -10/7