Question

giải phương trình 

\(\left(x^2-3x+2\right)\left(x^2+15x+56\right)+8\)\(=0\)

Answer 1

\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+7\right)\left(x+8\right)+8=0.\)

\(\Leftrightarrow\left(x-1\right)\left(x+7\right)\left(x-2\right)\left(x+8\right)+8=0.\)

\(\Leftrightarrow\left(x^2+6x-7\right)\left(x^2+6x-16\right)+8=0.\)

đặt \(\left(x^2+6x-7\right)=a.\)

\(a\left(a-9\right)+8=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=1\\a=8\end{cases}}\)

thay ròi giả tiếp .

Answer 2

(x^2-x-2x+2)(x^2+8x+7x+56)+8=0

{x(x-1)-2(x-1)}{x(x+8)+7(x+8)}+8=0

(x-1)(x-2)(x+8)(x+7)+8=0

(x-1)(x-2)(x+8)(x+7)=-8

Answer 3

jjjgjjjvjvjijmkjiuenndmddnđndndndndnndndndnndndnndnndnd

Answer 4

Nguyễn Nho Dũng