\(a,\\ \left(x+2\right)^2-x.\left(x-1\right)=10\\ \Leftrightarrow x^2+4x+4-x^2+x=10\\ \Leftrightarrow\left(x^2-x^2\right)+4x+x=10-4\\ \Leftrightarrow5x=6\\ \Leftrightarrow x=\dfrac{6}{5}\\ b,\\ x^3-6x^2+9x=0\\ \Leftrightarrow x.\left(x^2-6x+9\right)=0\\ \Leftrightarrow x.\left(x-3\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
a) (x + 2)2 - x(x - 1) = 10
x2 + 4 - x2+ x - 10 = 0
x - 6 = 0
=> x = 6
Vậy x = {6}
b) x3 - 6x2 + 9x = 0
x(x2 - 6x + 9) = 0
x(x - 3)2 = 0
=> x = 0
=> x - 3 = 0 => x = 3
Vậy x ={0; 3}
a,
(x+2)
2
−x.(x−1)=10
⇔x
2
+4x+4−x
2
+x=10
⇔(x
2
−x
2
)+4x+x=10−4
⇔5x=6
⇔x=
5
6
b,
x
3
−6x
2
+9x=0
⇔x.(x
2
−6x+9)=0
⇔x.(x−3)
2
=0
⇔[
x=0
x−3=0
⇔[
x=0
x=3
a ( x + 2)^2 -x(x-1) =10
x^2 +4x + 4 -x^2 +x =10
5x + 4 =10
5x =14
x=15/4
