Question

2x×(x+3)−(x−5)×(7+2x)=44         giúp mik vs

Answer 1

\(\Leftrightarrow2x^2+6x-\left(7x+2x^2-35-10x\right)=44\)

\(\Leftrightarrow2x^2+6x-2x^2+3x+35=44\)

=>9x=9

hay x=1

Answer 2

\(2x\left(x+3\right)-\left(x-5\right)\left(7+2x\right)=44\\ \Leftrightarrow2x^2+6x-7x-2x^2+35+10x-44=0\\ \Leftrightarrow9x=9\\ \Leftrightarrow x=1\)

Answer 3

\(2x.\left(x+3\right)-\left(x-5\right).\left(7+2x\right)=44\)

\(\Leftrightarrow2x^2+6x-\left(x-5\right).\left(7+2x\right)=44\)

\(\Leftrightarrow2x^2+6x-\left(7x+2x^2-35-10x\right)=44\)

\(\Leftrightarrow2x^2+6x-\left(-3x+2x^2-35\right)=44\)

\(\Leftrightarrow2x^2+6x+3x-2x^2+35=44\)

\(\Leftrightarrow6x+3x+35=44\)

\(\Leftrightarrow9x+35=44\)

\(\Leftrightarrow9x=44-35\)

\(\Leftrightarrow9x=9\)

\(\Leftrightarrow x=9:9\)

\(\Leftrightarrow x=1\)

Answer 4

$\iff 2{}^{}+6x-\left(7x+2{}^{}-35-10x\right)=44$

$\iff 2{}^{}+6x-2{}^{}+3x+35=44$

=>9x=9

hay x=1

Answer 5

$2x.\left(x+3\right)-\left(x-5\right).\left(7+2x\right)=44$

$\iff 2{}^{}+6x-\left(x-5\right).\left(7+2x\right)=44$

$\iff 2{}^{}+6x-\left(7x+2{}^{}-35-10x\right)=44$

$\iff 2{}^{}+6x-\left(-3x+2{}^{}-35\right)=44$

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

Answer 6

2x(x+3)−(x−5)(7+2x)=44⇔2x2+6x−7x−2x2+35+10x−44=0⇔9x=9⇔x=1