Question

giả sử  x= a/m, y= b/m ( a,b ,m thuộc Z m>0)            và x<y hãy chứng tỏ rằng nếu chọn Z = a+b/2m thì ta có x<z<y                                          

Answer 1

\(\text{Cách 1}:\text{Vì }x< y\)

\(\Rightarrow\frac{a}{m}< \frac{b}{m}\)

\(\Rightarrow a< b\)

\(\text{Ta có : }x=\frac{a}{m}=\frac{2.a}{2.m}=\frac{a+a}{2m}< \frac{a+b}{2m}=z\)

\(\Rightarrow x< z\left(1\right)\)

\(\text{Lại có : }z=\frac{a+b}{2m}< \frac{b+b}{2m}=\frac{2b}{2m}=\frac{b}{m}\)

\(\Rightarrow z< y\left(2\right)\)

\(\text{Từ }\left(1\right)\text{và }\left(2\right)\Rightarrow x< z< y\left(\text{đpcm}\right)\)

\(\text{Cách }2:\text{Vì }x< y\)

\(\Rightarrow\frac{a}{m}< \frac{b}{m}\)

\(\Rightarrow\frac{a}{m}\times\frac{1}{2}< \frac{b}{m}\times\frac{1}{2}\)

\(\Rightarrow\frac{a}{2m}< \frac{b}{2m}\)

\(\Rightarrow\frac{a}{2m}+\frac{a}{2m}< \frac{b}{2m}+\frac{a}{2m}\)

\(\Rightarrow\frac{2a}{2m}< \frac{a+b}{2m}\)

\(\Rightarrow\frac{a}{m}< \frac{a+b}{2m}\)

\(\Rightarrow x< z\left(1\right)\)

\(\text{Lại có : }\)\(\frac{a}{2m}< \frac{b}{2m}\)

\(\Rightarrow\frac{a}{2m}+\frac{b}{2m}< \frac{b}{2m}+\frac{b}{2m}\)

\(\Rightarrow\frac{a+b}{2m}< \frac{2b}{2m}\)

\(\Rightarrow\frac{a+b}{2m}< \frac{b}{m}\)

\(\Rightarrow z< y\left(2\right)\)

\(\text{Từ }\left(1\right)\text{và}\left(2\right)\Rightarrow x< z< y\left(\text{đpcm}\right)\)