Question

Tìm x, biết: \(x.\left(x-3\right)-2x.\left(x-1\right)=-x^2+5x\)

Answer 1

Bài làm:

Ta có: \(x\left(x-3\right)-2x\left(x-1\right)=-x^2+5x\)

\(\Leftrightarrow x^2-3x-2x^2+2x+x^2-5x=0\)

\(\Leftrightarrow-6x=0\)

\(\Rightarrow x=0\)

Answer 2

Ta có: \(x\left(x-3\right)-2x\left(x-1\right)=x^2-3x-2x^2+2x=-x^2-x\)

=> \(-x^2-x=-x^2+5x\)

=> \(-x^2+x^2=5x+x\)

=> 6x=0

=> x=0

Đúng hong tar? '-'

Answer 3

\(x\left(x-3\right)-2x.\left(x-1\right)=-x^2+5x\)

\(\Leftrightarrow x^2-3x-2x^2+2x=-x^2+5x\)

\(\Leftrightarrow6x=0\)

\(\Leftrightarrow x=0\)

Answer 4

x(x - 3) - 2x(x - 1) = - x² + 5x

=> x² - 3x - 2x² + 2x = - x² + 5x

=> x² - 3x - 2x² + 2x + x² - 5x = 0

=> (x² - 2x² + x²) - (3x - 2x + 5x) = 0

=> -6x = 0

=> x = 0

Vậy x = 0

Answer 5

x. ( x-3) - 2x . ( x-1) = -x² + 5x

\(\Leftrightarrow\) x² - 3x -  2x² + 2x +x² - 5x = 0

\(\Leftrightarrow\) -6x = 0

\(\Rightarrow\) x = 6

Vậy ,..

Answer 6

x( x - 3 ) - 2x( x - 1 ) = -x² + 5x

<=> x² - 3x - 2x² + 2x = -x² + 5x

<=> -x² - x = -x² + 5x

<=> -x² - x + x² - 5x = 0

<=> -6x = 0

<=> x = 0

Answer 7

\(x\left(x-3\right)-2x\left(x-1\right)=-x^2+5x\)

\(\Leftrightarrow x^2-3x-2x^2+2x=-x^2+5x\)

\(\Leftrightarrow-x^2-x=-x^2+5x\Leftrightarrow-6x=0\Leftrightarrow x=0\)