Question

tìm x

a, (-2012).(x+3)=0

b, (2+x).(7-x)=0

c, 104.x.(6-x)

d, x^2-3x=0

Answer 1

a, ⇒x+ 3=0

⇒x = 0-3

vậy x = -3

b, ⇒TH1: 2 + x= 0

⇒x = 0 - 2

vậy x = -2

TH2: 7-x=0

x=7-0

x=7

vậy x ∈(-2;7)

Answer 2

a) (-2012).(x+3)=0

<=> x+3=0

<=> x=-3

b) \(\left(2+x\right).\left(7-x\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}2+x=0\\7-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2\\x=7\end{matrix}\right.\)

c) thiếu thì phải?

\(104.x.\left(6-x\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\6-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)

d) \(x^2-3x=0\)

\(\Leftrightarrow x\left(x-3\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

Answer 3

a) (-2012).(x+3)=0

<=>x+3=0

<=>x=-3

b) \(\left(2+x\right)\left(7-x\right)=0\)

\(< =>\left\{{}\begin{matrix}2+x=0\\7-x=0\end{matrix}\right.< =>\left\{{}\begin{matrix}x=-2\\x=7\end{matrix}\right.\)

d) \(x^2-3x=0\)

\(< =>x\left(x-3\right)=0\)

\(< =>\left\{{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.< =>\left\{{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)

Câu c bn viết thiếu đề rồi

Answer 4

a, suy ra x+3=0

x=0-3

x=-3

b, TH1 2+x=0

x=0-2

x=-2

TH2 7-x=0

x=7-0

x=7

vậy xϵ(7; -2)

d,suy ra x.x-3x=0

⇒x.(x-3)=0

TH1x=0

TH2 x-3=0

x=0+3

x=3

vậy x∈(3;0)