Question

(x+2)×(x2−4x+1)−(x3−2x2+2)=−18   giúp mik vs

Answer 1

\(\left(x+2\right).\left(x^2-4x+1\right)-\left(x^3-2x^2+2\right)=-18\\ \Leftrightarrow x^3-4x^2+x+2x^2-8x+2-x^3+2x^2-2=-18\\ \Leftrightarrow-7x=-18\\ \Leftrightarrow x=\dfrac{18}{7}\)

Answer 2

`(x+2)(x^2-4x+1)-(x^3-2x^2+2)=-18`

`<=>x^3-4x^2+x+2x^2-8x+2-x^3+2x^2-2=18`

`<=>-7x=-18`

`<=>x={18}/7`

Vậy `S={{18}/7}`

$#Notkiller$

Answer 3

\(\left(x+2\right).\left(x^2-4x+1\right)-\left(x^3-2x^2+2\right)=-18\)

\(\Leftrightarrow x^3-4x^2+x+2x^2-8x+2-\left(x^3-2x^2+2\right)=-18\)

\(\Leftrightarrow x^3-4x^2+x+2x^2-8x+2-x^3+2x^2-2=-18\)

\(\Leftrightarrow-4x^2+x+2x^2-8x+2+2x^2-2=-18\)

\(\Leftrightarrow-4x^2+x+2x^2-8x+2x^2=-18\)

\(\Leftrightarrow0+x-8x=-18\)

\(\Leftrightarrow x-7x=-18\)

\(\Leftrightarrow-7x=-18\)

\(\Leftrightarrow x=\dfrac{18}{7}\)

Answer 4

(x+2).(x2−4x+1)−(x3−2x2+2)=−18⇔x3−4x2+x+2x2−8x+2−x3+2x2−2=−18⇔−7x=−18⇔x=\(\dfrac{7}{18}\)

Answer 5

(x+2).(x2−4x+1)−(x3−2x2+2)=−18⇔x3−4x2+x+2x2−8x+2−x3+2x2−2=−18⇔−7x=−18⇔x=\(\dfrac{18}{7}\)

Answer 6

$(x+2)({}^{}-4x+1)-({}^{}-2{}^{}+2)=-18$

$\iff {}^{}-4{}^{}+x+2{}^{}-8x+2-{}^{}+2{}^{}-2=18$

$\iff -7x=-18$

$\iff x=\frac{18}{7}$

Vậy $S=\left\{\frac{18}{7}\right\}$

Answer 7

$\left(x+2\right).\left({}^{}-4x+1\right)-\left({}^{}-2{}^{}+2\right)=-18$

$\iff {}^{}-4{}^{}+x+2{}^{}-8x+2-\left({}^{}-2{}^{}+2\right)=-18$

$\iff {}^{}-4{}^{}+x+2{}^{}-8x+2-{}^{}+2{}^{}-2=-18$

$\iff -4{}^{}+x+$

Answer 8

(x+2).(x2−4x+1)−(x3−2x2+2)=−18⇔x3−4x2+x+2x2−8x+2−x3+2x2−2=−18⇔−7x=−18⇔x\(=\dfrac{18}{7}\)

Answer 9

$(x+2)({}^{}-4x+1)-({}^{}-2{}^{}+2)=-18$

$\iff {}^{}-4{}^{}+x+2{}^{}-8x+2-{}^{}+2{}^{}-2=18$

$\iff -7x=-18$

$\iff x=\frac{18}{7}$

Vậy $S=\left\{\frac{18}{7}\right\}$