Giải thích các bước giải:
\(\begin{array}{l}
20.\left( {x + 3} \right)\left( {x--2} \right) = {\left( {x + 1} \right)^2}\\
\to {x^2} + x - 6 = {x^2} + 2x + 1\\
\to x = - 7\\
21.\left( {x + 7} \right)\left( {x--7} \right) + {x^2}--2 = 2\left( {{x^2} + 5} \right)\\
\to {x^2} - 49 + {x^2}--2 = 2{x^2} + 10\\
\to 0 = 61\left( {voli} \right)
\end{array}\)
⇒ Vô nghiệm
\(\begin{array}{l}
22){\left( {x--1} \right)^2} + {\left( {x + 3} \right)^2} = 2\left( {x--2} \right)\left( {x + 2} \right)\\
\to {x^2} - 2x + 1 + {x^2} + 6x + 9 = 2{x^2} - 8\\
\to 4x = - 18\\
\to x = - \frac{9}{2}\\
23.{\left( {x--5} \right)^2} = {\left( {x + 3} \right)^2} + 2\\
\to {x^2} - 10x + 25 = {x^2} + 6x + 9 + 2\\
\to 16x = 14\\
\to x = \frac{7}{8}\\
24.{\left( {3x + 2} \right)^2}--{\left( {3x--2} \right)^2} = 5x + 38\\
\to 9{x^2} + 12x + 4 - 9{x^2} + 12x - 4 = 5x + 38\\
\to 19x = 38\\
\to x = 2
\end{array}\)
