Question

tìm x€Z :

a) -12 (x -5)+7 (3- x) = 5

b) (x - 2)(x + 4) = 0

c) (x - 2)(x + 15) = 0

d) (7-x)(x + 19) = 0

Answer 1

a) \(-12\left(x-5\right)+7\left(3-x\right)=5\\ < =>-12x+60+21-7x=5\\ < =>-12x-7x=-60-21+5\\ < =>-19x=-76\\ =>x=4\)

Vậy: x=4

b) \(\left(x-2\right)\left(x+4\right)=0\\ < =>\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

Vậy: x=2 hoặc x=4

c) \(\left(x-2\right)\left(x+15\right)=0\\ < =>\left[{}\begin{matrix}x-2=0\\x+15=0\end{matrix}\right.=>\left[{}\begin{matrix}x=2\\x=-15\end{matrix}\right.\)

Vậy: x=2 hoặc x= -15

d) \(\left(7-x\right)\left(x+19\right)=0\\ < =>\left[{}\begin{matrix}7-x=0\\x+19=0\end{matrix}\right.=>\left[{}\begin{matrix}x=7\\x=-19\end{matrix}\right.\)

Vậy: x=7 hoặc x=-19

Answer 2

b) \(\left(x-2\right)\left(x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

c) \(\left(x-2\right)\left(x+15\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x+15=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-15\end{matrix}\right.\)

d) \(\left(7-x\right)\left(x+19\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}7-x=0\\x+19=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-19\end{matrix}\right.\)

Answer 3

b, \(\left(x-2\right)\left(x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

Vậy, \(x\in\left\{-4;2\right\}\)

c, \(\left(x-2\right)\left(x+15\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x+15=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-15\end{matrix}\right.\)

Vậy, \(x\in\left\{-15;2\right\}\)

d, \(\left(7-x\right)\left(x+19\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}7-x=0\\x+19=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-19\end{matrix}\right.\)

Vậy, \(x\in\left\{-19;7\right\}\)