Question

Tìm x
(x+7)(3x-15)=0

4x(x+7)=2(x+7)

(x-3)² - x(x - 4) = 5

Answer 1

\(\left(x+7\right)\left(3x-15\right)=0\\ \Rightarrow3\left(x-5\right)\left(x+7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\\ 4x\left(x+7\right)=2\left(x+7\right)\\ \Rightarrow4x\left(x+7\right)-2\left(x+7\right)=0\\ \Rightarrow2\left(2x-1\right)\left(x+7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-7\end{matrix}\right.\\ \left(x-3\right)^2-x\left(x-4\right)=5\\ \Rightarrow x^2-6x+9-x^2+4x-5=0\\ \Rightarrow-2x+4=0\\ \Rightarrow-2x=-4\Rightarrow x=2\)

hưng phúc                                                          đầy đủ chưa bạn nhỉ?

Answer 2

1) \(\left(x+7\right)\left(3x-15\right)=0\)

⇔\(\left[{}\begin{matrix}x+7=0\\3x-15=0\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=-7\\x=5\end{matrix}\right.\)

2) \(4x\left(x+7\right)=2\left(x+7\right)\)

\(2\left(2x+1\right)\left(x+7\right)=0\)

⇔\(\left[{}\begin{matrix}2x+1=0\\x+7=0\end{matrix}\right.\)

⇔\(\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-7\end{matrix}\right.\)

Answer 3

c: Ta có: \(\left(x-3\right)^2-x\left(x-4\right)=5\)

\(\Leftrightarrow x^2-6x+9-x^2+4x=5\)

\(\Leftrightarrow-2x=-4\)

hay x=2