Question

tìm x thuộc Z

a)13+x=12-x

b)19-(x+5)=(x-2)-(-38)

 

 

Answer 1

a) \(13+x=12-x\) \(\Leftrightarrow x+x=12-13\)

\(\Leftrightarrow2x=-1\)\(\Leftrightarrow x=\frac{-1}{2}\)

Vậy \(x=\frac{-1}{2}\)

b) \(19-\left(x+5\right)=\left(x-2\right)-\left(-38\right)\)

\(\Leftrightarrow19-x-5=x-2+38\)\(\Leftrightarrow14-x=x+36\)

\(\Leftrightarrow x+x=14-36\)\(\Leftrightarrow2x=-22\)\(\Leftrightarrow x=-11\)

Vậy \(x=-11\)

Answer 2

      \(13+x=12-x\)

\(\Rightarrow13+x+x=12\)

\(\Rightarrow13+2x=12\)

\(\Rightarrow2x=12-13\)

\(\Rightarrow2x=-1\)

\(\Rightarrow x=-1\div2\)

\(\Rightarrow x=-0.5\)

Answer 3

a, \(13+x=12-x\)

=> \(13-12=-2x\)

=> \(-2x=1\)

=> \(x=\frac{-1}{2}\)  ( thỏa mãn \(x\in Z\))
Vậy \(x=\frac{-1}{2}\)

b, \(19-\left(x+5\right)=\left(x-2\right)-\left(-38\right)\)

=> \(19-x-5=x-2+38\)

=> \(14-x=x+36\)

=> \(14-36=x+x\)

=> \(2x=-22\)

=> \(x=-11\)  ( thỏa mãn \(x\in Z\))

Vậy \(x=-11\)

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