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\(a>b\Rightarrow a+2016>b+2016\)
\(\Rightarrow\frac{a}{b}=\frac{b+a-b}{b}\)
\(\Rightarrow\frac{a+2016}{b+2016}=\frac{b+2016+a+2016-b+2016}{b+2016}=\frac{b+a-a}{b+2016}\)
Vì: \(\frac{b+a-a}{b}>\frac{b+a-b}{b+2016}\)
\(\Rightarrow\frac{a}{b}>\frac{a+2016}{b+2016}\)
Ta có:
- \(\frac{a}{b}=\frac{a\left(b+2016\right)}{b\left(b+2016\right)}\)
\(=\frac{ab+2016a}{b\left(b+2016\right)}\)
- \(\frac{a+2016}{b+2016}=\frac{b\left(a+2016\right)}{b\left(b+2016\right)}\)
\(=\frac{ab+2016b}{b\left(b+2016\right)}\)
Vì \(a>b\Rightarrow2016a>2016b\)
\(\Rightarrow ab+2016a>ab+2016b\)
\(\Rightarrow\frac{ab+2016a}{b\left(b+2016\right)}>\frac{ab+2016b}{b\left(b+2016\right)}\)
\(\Rightarrow\frac{a}{b}>\frac{a+2016}{b+2016}\)
1.
Ta có: \(\frac{a}{b}< \frac{c}{d}\Leftrightarrow ad< bc\Leftrightarrow ab+ad< ad+bc\Leftrightarrow a\left(b+d\right)< b\left(a+c\right)\Leftrightarrow\frac{a}{b}< \frac{a+c}{b+d}\) (1)
Lại có: \(\frac{a}{b}< \frac{c}{d}\Leftrightarrow bc>ad\Leftrightarrow bc+cd>ad+cd\Leftrightarrow c\left(b+d\right)>d\left(a+c\right)\Leftrightarrow\frac{c}{d}>\frac{a+c}{b+d}\) (2)
Từ (1) và (2) suy ra \(\frac{a}{b}< \frac{a+c}{b+d}< \frac{c}{d}\)
2.
Ta có: a(b + n) = ab + an (1)
b(a + n) = ab + bn (2)
Trường hợp 1: nếu a < b mà n > 0 thì an < bn (3)
Từ (1),(2),(3) suy ra a(b + n) < b(a + n) => \(\frac{a}{n}< \frac{a+n}{b+n}\)
Trường hợp 2: nếu a > b mà n > 0 thì an > bn (4)
Từ (1),(2),(4) suy ra a(b + n) > b(a + n) => \(\frac{a}{b}>\frac{a+n}{b+n}\)
Trường hợp 3: nếu a = b thì \(\frac{a}{b}=\frac{a+n}{b+n}=1\)
a) Ta có: a < b => a + 1 < b + 1
b) Ta có: a < b => a - 2 < b - 2
Xét 3 trường hợp :
+) Nếu b > a thì \(\frac{a}{b}=\frac{b-m}{b}=\frac{b}{b}-\frac{m}{b}=1-\frac{m}{b}\)
\(\frac{a+1}{b+1}=\frac{b-m+1}{b+1}=\frac{b+1-m}{b+1}=\frac{b+1}{b+1}-\frac{m}{b+1}=1-\frac{m}{b+1}\)
Vì \(\frac{m}{b}>\frac{m}{b+1}\)nên \(1-\frac{m}{b}< 1-\frac{m}{b+1}\)hay \(\frac{a}{b}< \frac{a+1}{b+1}\)
+) Nếu a = b thì \(\frac{a}{b}=1\)
\(\frac{a+1}{b+1}=1\)nên\(\frac{a}{b}=\frac{a+1}{b+1}\)
+) Nếu a > b thì \(\frac{a}{b}=\frac{b+m}{b}=\frac{b}{b}+\frac{m}{b}=1+\frac{m}{b}\)
\(\frac{a+1}{b+1}=\frac{b+m+1}{b+1}=\frac{b+1}{b+1}+\frac{m}{b+1}=1+\frac{m}{b+1}\)
Vì \(\frac{m}{b}>\frac{m}{b+1}\)nên \(1+\frac{m}{b}>1+\frac{m}{b+1}\)hay \(\frac{a}{b}>\frac{a+1}{b+1}\)
Ta có :
\(\frac{a}{b}=\frac{a\left(b+1\right)}{b\left(b+1\right)}=\frac{ab+a}{b^2+b}\)
\(\frac{a+1}{b+1}=\frac{b\left(a+1\right)}{b\left(b+1\right)}=\frac{ab+b}{b^2+b}\)
Từ 2 ý trên , ta xét từng trường hợp sau :
a < b thì \(\frac{a}{b}< \frac{a+1}{b+1}\)
a > b thì \(\frac{a}{b}>\frac{a+1}{b+1}\)
a = b thì \(\frac{a}{b}=\frac{a+1}{b+1}\)
ta có \(a< a+100\)(với a >0)
\(b< b+100\)(với b >0)
\(\Rightarrow\frac{a}{b}< \frac{a+100}{b+100}\)
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cùng mẫu mới so sánh đc bạn