Question

\(c.\left(n-2\right)^2-\left(n+3\right)\left(n-3\right)=4\left(n-1\right)\)a.)(n-2)(x+2)+6(n-1)=(x+1)²

b.)(x-3)(x²+3x+9)=x(x³+3)

\(d,2\left(3-x\right)-3\left(n-1\right)=4\left(n-3\right)\)

Mk đăng ngược ý các bn sắp xếp rồi giải giúp mk nha! CẢM ƠN NHIỀU Ạ

Answer 1

c: \(\left(n-2\right)^2-\left(n+3\right)\left(n-3\right)=4\left(n-1\right)\)

\(\Leftrightarrow n^2-4n+4-n^2+9=4n-4\)

=>-4n+13=4n-4

=>-8n=-17

hay n=17/8

a: \(\left(n-2\right)\left(n+2\right)+6\left(n-1\right)=\left(n+1\right)^2\)

\(\Leftrightarrow n^2-4+6n-6=n^2+2n+1\)

=>6n-10=2n+1

=>4n=11

hay n=11/4

d: \(2\left(3-x\right)-3\left(x-1\right)=4\left(x-3\right)\)

=>6-2x-3x+3=4x-12

=>-5x+9=4x-12

=>-9x=-21

hay x=7/3