Question

Tìm x 

^{x^2-3=0}

(x+2)×(x^2-2x+4)+x×(5-x)×(x+5)=-17

X^3-3x^2-3x+1=0

​

Answer 1

\(x^2-3=0\)

\(\Leftrightarrow x^2=3\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=3\\x=-3\end{array}\right.\)

\(\left(x+2\right).\left(x^2-2x+4\right)+x.\left(5-x\right).\left(x-5\right)=-17\)

\(\Leftrightarrow x^3-8+x.-\left(x-5\right).\left(x+5\right)=-17\)

\(\Leftrightarrow x^3-8-x.\left(x^2-25\right)=-17\)

\(\Leftrightarrow x^3-8-x^3+25x=-17\)

\(\Leftrightarrow-8+25x=-17\)

\(\Leftrightarrow25x=-9\)

\(\Leftrightarrow x=-\frac{9}{25}\)

\(x^3-3x^2-3x+1=0\)

\(\Leftrightarrow\left(x-1\right)^3=0\)

\(\Leftrightarrow x=1\)