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Ta có : z = \(\frac{m}{n}\)= \(\frac{\frac{a+c}{2}}{\frac{b+d}{2}}=\frac{a+c}{b+d}=\frac{2m}{2n}\)
Nếu x < y thì \(\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)\(\Rightarrow\frac{a}{b}< \frac{2m}{2n}< \frac{c}{d}\)
\(\Rightarrow\frac{a}{b}< \frac{m}{n}< \frac{c}{d}\)\(\Rightarrow x< z< y\)
Nếu x > y thì : \(\frac{a}{b}>\frac{a+c}{b+d}>\frac{c}{d}\)\(\Rightarrow\frac{a}{b}>\frac{2m}{2n}>\frac{c}{d}\)
\(\Rightarrow\frac{a}{b}>\frac{m}{n}>\frac{c}{d}\)\(\Rightarrow x>z>y\)
Vậy ...
Z = a+c/2 :b+d/2 =a+c/2 ·2/b+d =a+c/b+d
X =a/b = a(b+d)/b(b+d) =ab+ad/b2+bd
Z= a+c/b+d =(a+c).b/(b+d).b =ab+ac/b2+bd
(+) Nếu a dương ; d< c => ad < ac => ab +ad < ab +ac => X < Z
(+) Nếu a âm ; d< c => ad > ac => ab + ad > ab + ac => X>Z
(+) nếu a dương ; d > c => ad > ac => ab + ad > ab + ac => X > Z
(+) ..................................... ........................................... Z >X
Z = a+c/2 :b+d/2 =a+c/2 ·2/b+d =a+c/b+d
X =a/b = a(b+d)/b(b+d) =ab+ad/b2+bd
Z= a+c/b+d =(a+c).b/(b+d).b =ab+ac/b2+bd
(+) Nếu a dương ; d< c => ad < ac => ab +ad < ab +ac => X < Z
(+) Nếu a âm ; d< c => ad > ac => ab + ad > ab + ac => X>Z
(+) nếu a dương ; d > c => ad > ac => ab + ad > ab + ac => X > Z
(+) ..................................... ........................................... Z >X
Theo đề ra, ta có:
\(x=\frac{a}{b};y=\frac{c}{d};z=\frac{m}{n}=\frac{\frac{a+c}{2}}{\frac{b+d}{2}}=\frac{a+c}{b+d}\)
*) Nếu \(\frac{a}{b}>\frac{c}{d}\) \(=>ad>bc=>ad+cd>bc+cd=>d\left(a+c\right)>c\left(b+d\right)=>\frac{a+c}{b+d}>\frac{c}{d}\)
và \(ad+ab>bc+ab=>a\left(d+b\right)>b\left(a+c\right)=>\frac{a}{b}>\frac{a+c}{b+d}\) => \(\frac{a}{b}>\frac{a+c}{b+d}>\frac{c}{d}=>x>z>y\)
*) Nếu \(\frac{a}{b}< \frac{c}{d}\) thì tương tự và được \(x< z< y\)
-Nếu x < y thì \(\frac{a}{b}\) < \(\frac{a+c}{b+d}\) < \(\frac{c}{d}\) hay \(\frac{a}{b}\) < \(\frac{2m}{2n}\) < \(\frac{c}{d}\)
Suy ra \(\frac{a}{b}\) < \(\frac{m}{n}\) < \(\frac{c}{d}\)
hay x < z < y
- Nếu x > y thì \(\frac{a}{b}\) > \(\frac{a+c}{b+d}\) > \(\frac{c}{d}\) hay \(\frac{a}{b}\) > \(\frac{2m}{2n}\) > \(\frac{c}{d}\)
Suy ra \(\frac{a}{b}\) > \(\frac{m}{n}\) > \(\frac{c}{d}\)
hay x > z > y