Question

[Nấm lùn]

Tìm x biết :

a) x^2 - 3x + 2 (x-3) = 0

b) (x-1)(x+1) + x (x-9) = 2x^2 - 4

c) x (x-3) - 3x + 9 = 0

d) x (x+2) - (x-3)(x+3) = 5

đ) 2x (x+1) - (2x+1)(x-3) = 6

Answer 1

[Nguyệt]

\(x^2-3x+2.\left(x-3\right)=0\)

\(x.\left(x-3\right)+2.\left(x-3\right)=0\)

\(\left(x-3\right).\left(x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

\(x.\left(x-3\right)-3x+9=0\)

\(x.\left(x-3\right)-3.\left(x-3\right)=0\)

\(\left(x-3\right)^2=0=>x=3\)

Answer 2

[Nguyễn Xuân Anh]

a,\(x^2-3x+2\left(x-3\right)=0.\)

\(\Leftrightarrow x^2-3x+2x-6=0\)

\(\Leftrightarrow x^2+x-6=0\)

\(\Leftrightarrow\left(x^2-2x\right)+\left(3x-6\right)=0\)

\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)