Question

Tìm x:

1) 1/3+3/35<x/210<4/7+1/3

2) 5/3+4/3<x<16/5+18/10

Answer 1

\(\dfrac{1}{3}+\dfrac{3}{35}< \dfrac{x}{210}< \dfrac{4}{7}+\dfrac{1}{3}\)

Ta có: \(\dfrac{1}{3}+\dfrac{3}{35}=\dfrac{35+9}{105}=\dfrac{44}{105}\)

và \(\dfrac{4}{7}+\dfrac{1}{3}=\dfrac{12+7}{21}=\dfrac{19}{21}\)

=> \(\dfrac{44}{105}=\dfrac{44.2}{105.2}=\dfrac{88}{210}\)

=> \(\dfrac{19}{21}=\dfrac{19.10}{21.10}=\dfrac{190}{210}\)

=> \(\dfrac{88}{201}< \dfrac{x}{210}< \dfrac{190}{210}\)

=> Vậy x ∈ {89; 90; 91; 92; ... ; 188; 189}

Answer 2

2: =>3<x<16/5+9/5=5

=>x=4

1: =>70/210+18/210<x/210<120/210+70/210

=>88<x<190

hay \(x\in\left\{89;90;...;189\right\}\)