Question

Giải phương trình: (x-2)^2=x(x-3)

Help me

Answer 1

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

\(\Leftrightarrow x^2-4x+4-x^2+3x=0\)

\(\Leftrightarrow-x+4=0\)

\(\Leftrightarrow x=4\)

\(S=\left\{4\right\}\)

Answer 2

\(\left(x-2\right)^2=x\left(x-3\right)\\ \Leftrightarrow\left(x-2\right)^2-x\left(x-3\right)=0\\ \Leftrightarrow x^2-4x+4-\left(x^2-3x\right)=0\\ \Leftrightarrow x^2-4x+4-x^2+3x=0\\ \Leftrightarrow-x=-4\\ \Leftrightarrow x=4\)

Answer 3

(x-2)²=x(x-3)

=> x² - 4x + 4 = x² - 3x

=> x = 4

Vậy, x = 4.

Answer 4

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

⇔\(x^2-4x+4=x^2-3x\)

\(\Leftrightarrow-x+4=0\)

\(\Leftrightarrow-x=-4\)

\(\Leftrightarrow x=4\)

Answer 5

Answer 6

Ta có: \(\left(x-2\right)^2=x\left(x-3\right)\)

\(\Leftrightarrow x^2-4x+4=x^2-3x\)

\(\Leftrightarrow x^2-4x+4-x^2+3x=0\)

\(\Leftrightarrow4-x=0\)

\(\Leftrightarrow x=4\)

Vậy: S={4}