Question

(𝑥 + 3)^2 − (𝑥 − 4)(𝑥 + 8) = 1 tìm x

Answer 1

\(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\\ \Leftrightarrow\left(x^2+6x+9\right)-\left(x^2-4x+8x-32\right)=1\\ \Leftrightarrow x^2+6x+9-\left(x^2+4x-32\right)=1\\ \Leftrightarrow x^2+6x+9-x^2-4x+32-1=0\\ \Leftrightarrow2x+40=0\\ \Leftrightarrow2x=-40\\ \Leftrightarrow x=-20\)

Answer 2

\(\Leftrightarrow x^2+6x+9-x^2+4x-32=1\)

=>10x=22

hay x=11/5