Bài 1 : Tìm x,biết :
a, x2(x + 5) - 9x = 45
⇔ x2(x + 5) - 9x - 45 = 0
⇔ x2(x + 5) - 9(x + 5) = 0
⇔ (x + 5)(x2 - 9) = 0
⇔ (x + 5)(x - 3)(x + 3) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\x-3=0\\x+3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=3\\x=-3\end{matrix}\right.\)
Vậy x ={-5; 3; -3}
b, 9(5 - x) + x2 - 10x = -25
⇔ 45 - 9x + x2 - 10x + 25 = 0
⇔ x2 - 19x + 70 = 0
⇔ x2 - 14x - 5x + 70 = 0
⇔ (x2 - 5x) - (14x - 70) = 0
⇔ x(x - 5) - 14(x - 5) = 0
⇔ (x - 5)(x - 14) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-14=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=14\end{matrix}\right.\)
Vậy x ={5; 14}
a, x2( x+5 ) - 9x = 45
x3 + 5x2 - 9x - 45 = 0
x2( x+5 ) - 9( x+5) = 0
(x2 - 9)(x + 5) = 0
(x + 3)(x - 3)(x + 5) = 0
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=3\\x=-5\end{matrix}\right.\)
b, 9( 5-x ) + x2 -10x = -25
45 - 9x + x2 - 10x + 25 = 0
x2 - 19x + 70 = 0
x2 - 14x - 5x + 70 = 0
x( x-14 ) - 5( x-14) = 0
(x - 5)(x - 14) = 0
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=14\end{matrix}\right.\)
