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\(\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{a\left(b+2017\right)}{b\left(b+2017\right)}\\\dfrac{a+2017}{b+2017}=\dfrac{b\left(a+2017\right)}{b\left(b+2017\right)}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{b}=\dfrac{ab+2017a}{b^2+2017b}\\\dfrac{a+2017}{b+2017}=\dfrac{ab+2017b}{b^2+2017b}\end{matrix}\right.\)
Ta cần so sánh:
\(ab+2017a\) với \(ab+2017b\)
Cần so sánh \(a\) với \(b\)
Nếu \(a>b\Leftrightarrow\dfrac{a}{b}>\dfrac{a+2017}{b+2017}\)
Nếu \(a< b\Leftrightarrow\dfrac{a}{b}< \dfrac{a+2017}{b+2017}\)
Nếu \(a=b\Leftrightarrow\dfrac{a}{b}=\dfrac{a+2017}{b+2017}\)
Mấy câu sau dễ tương tự
1) Nếu a/b>1 thì a/b>b/b<=>a>b
2)Nếu a>b thì a.z>b.z=>a/b>z/z<=>a/b>1
3)Nếu a/b<1 thì a/b<b/b<=>a<b
4)Nếu a<b=>a.z<b.z=>a/b<z/z<=>a/b<1
+) Nếu \(a>b\Leftrightarrow\dfrac{a}{b}>1\Leftrightarrow\dfrac{a}{b}>\dfrac{a+n}{b+n}\)
+) Nếu \(a=b\Leftrightarrow\dfrac{a}{b}=1\Leftrightarrow\dfrac{a}{b}=\dfrac{a+n}{b+n}\)
+) Nếu \(a< b\Leftrightarrow\dfrac{a}{b}< 1\Leftrightarrow\dfrac{a}{b}< \dfrac{a+n}{b+n}\)
\(a>b\)
\(\Rightarrow\dfrac{a}{b}>1\Rightarrow\dfrac{a+n}{b+n}>1\Rightarrow\dfrac{a}{b}>\dfrac{a+n}{b+n}\)
\(a< b\)
\(\Rightarrow\dfrac{a}{b}< 1\Rightarrow\dfrac{a+n}{b+n}< 1\Rightarrow\dfrac{a}{b}< \dfrac{a+n}{b+n}\)
\(a=b\)
\(\Rightarrow\dfrac{a}{b}=1\Rightarrow\dfrac{a+n}{b+n}=1\Rightarrow\dfrac{a}{b}=\dfrac{a+n}{b+n}\)
Chúc bạn học tốt!
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 \(\frac{a}{b}=\frac{a+n}{b+n}=1\)
1
a) Vì \(\dfrac{a}{b}< \dfrac{c}{d}\)
\(\Rightarrow\dfrac{ad}{bd}< \dfrac{bc}{bd}\)
\(\Rightarrow ad< bc\)
2
b) Ta có : \(\dfrac{-1}{3}=\dfrac{-16}{48};\dfrac{-1}{4}=\dfrac{-12}{48}\)
Ta có dãy sau : \(\dfrac{-16}{48};\dfrac{-15}{48};\dfrac{-14}{48};\dfrac{-13}{48};\dfrac{-12}{48}\)
Vậy 3 số hữu tỉ xen giữa \(\dfrac{-1}{3}\) và \(\dfrac{-1}{4}\) là :\(\dfrac{-15}{48};\dfrac{-14}{48};\dfrac{-13}{48}\)
1a ) Ta có : \(\dfrac{a}{b}\) < \(\dfrac{c}{d}\)
\(\Leftrightarrow\) \(\dfrac{ad}{bd}\) < \(\dfrac{bc}{bd}\) \(\Rightarrow\) ad < bc
1b ) Như trên
2b) \(\dfrac{-1}{3}\) = \(\dfrac{-16}{48}\) ; \(\dfrac{-1}{4}\) = \(\dfrac{-12}{48}\)
\(\dfrac{-16}{48}\) < \(\dfrac{-15}{48}\) <\(\dfrac{-14}{48}\) < \(\dfrac{-13}{48}\) < \(\dfrac{-12}{48}\)
Vậy 3 số hữu tỉ xen giữa là.................
Ta có: \(\dfrac{a}{b}< \dfrac{a+n}{b+n}\Leftrightarrow a\left(b+n\right)< b\left(a+n\right)\)\(\Leftrightarrow ab+an< ab+bn\)\(\Leftrightarrow a< b\) (vì \(n>0\)).
Vậy \(\dfrac{a}{b}< \dfrac{a+n}{b+n}\Leftrightarrow a< b.\)
Tương tự
\(\dfrac{a}{b}>\dfrac{a+n}{b+n}\Leftrightarrow a>b\) ;
\(\dfrac{a}{b}=\dfrac{a+n}{b+n}\Leftrightarrow a=b\).
Bài 1:
Ta có:
\(\dfrac{a}{b}>\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{a.d}{b.d}>\dfrac{b.c}{b.d}\left(b;d>0\right)\)
\(\Leftrightarrow ad>bc\)
Vậy ...
Bài 2:
Ta có:
\(0< a< 5< b\)
\(\Leftrightarrow a;b>0\)
\(\Leftrightarrow\dfrac{b}{a}>0\)
Mà \(a< 5< b\)
\(\Leftrightarrow a< b\)
\(\Leftrightarrow\dfrac{b}{a}>1\)
Vậy ...
Giải:
Ta có:
\(\dfrac{a}{b}< \dfrac{a+n}{b+n}\) \(\Leftrightarrow a\left(b+n\right)< b\left(a+n\right)\)
\(\Leftrightarrow ab+an< ab+bn\Leftrightarrow a< b\) (Vì \(n>0\))
Vậy \(\dfrac{a}{b}< \dfrac{a+n}{b+n}\Leftrightarrow a< b\)
Tương tự ta cũng có:
\(\dfrac{a}{b}>\dfrac{a+n}{b+n}\Leftrightarrow a>b\)
\(\dfrac{a}{b}=\dfrac{a+n}{b+n}\Leftrightarrow a=b\)