Question

Bài 1 : Tìm x,biết :
a, x^2 ( x + 5 ) - 9x = 45
b, 9 ( 5 -x ) + x^2 - 10x = -25
Giúp mk vs ạ mk đang cần gấp

Answer 1

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}

Answer 2

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

\(\Rightarrow\left[{}\begin{matrix}x-5=0\\x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=14\end{matrix}\right.\)