Question

(x2−3x)×(2x−1)−x×(2x2−7x)=12       giúp mik vs

Answer 1

\(\left(x^2-3x\right).\left(2x-1\right)-x.\left(2x^2-7x\right)=12\\ \Leftrightarrow2x^3-x^2-6x^2+3x-2x^3+7x^2=12\\ \Leftrightarrow3x=12\\ \Leftrightarrow x=4\)

Answer 2

\(\left(x^2-3x\right).\left(2x-1\right)-x.\left(2x^2-7x\right)=12\)

\(\Leftrightarrow2x^3-x^2-6x^2+3x-x.\left(2x^2-7x\right)=12\)

\(\Leftrightarrow2x^3-x^2-6x^2+3x-2x^3+7x^2=12\)

\(\Leftrightarrow-x^2-6x^2+3x+7x^2=12\)

\(\Leftrightarrow0+3x=12\)

\(\Leftrightarrow3x=12\)

\(\Leftrightarrow x=12:3\)

\(\Leftrightarrow x=4\)

Answer 3

(x2−3x).(2x−1)−x.(2x2−7x)=12⇔2x3−x2−6x2+3x−2x3+7x2=12⇔3x=12⇔x=4

Answer 4

$\left({}^{}-3x\right).\left(2x-1\right)-x.\left(2{}^{}-7x\right)=12$

$\iff 2{}^{}-{}^{}-6{}^{}+3x-x.\left(2{}^{}-7x\right)=12$

$\iff 2{}^{}-{}^{}-6{}^{}+3x-2{}^{}+7{}^{}=12$

$\iff -{}^{}-6{}^{}+3x+7{}^{}=12$

$\iff \mathrm{0...}$

Answer 5

(x2−3x).(2x−1)−x.(2x2−7x)=12⇔2x3−x2−6x2+3x−2x3+7x2=12⇔3x=12⇔x=4