Bài 6: So sánh phân số

A=-2015/2015x2016

A=-1/2016

B=-2014/2014x2015

B=-1/2015

vi 2016>2015,-1/2016>-1/2015

vay A>B

b) Ta có: \(A=\dfrac{10^{2009}+1}{10^{2010}+1}\)

\(\Leftrightarrow10A=\dfrac{10^{2010}+10}{10^{2010}+1}=1+\dfrac{9}{10^{2010}+1}\)

Ta có: \(B=\dfrac{10^{2010}+1}{10^{2011}+1}\)

\(\Leftrightarrow10B=\dfrac{10^{2011}+10}{10^{2011}+1}=1+\dfrac{9}{10^{2011}+1}\)

Ta có: \(10^{2010}+1< 10^{2011}+1\)

\(\Leftrightarrow\dfrac{9}{10^{2010}+1}>\dfrac{9}{10^{2011}+1}\)

\(\Leftrightarrow\dfrac{9}{10^{2010}+1}+1>\dfrac{9}{10^{2011}+1}+1\)

\(\Leftrightarrow10A>10B\)

hay A>B

Đặt $A=\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{100}}$

Ý bạn là muốn CMR $A<B$?

----------------------------

Lời giải:

$2A=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{99}}$

Trừ theo vế:

$2A-A=1-\frac{1}{2^{100}}< 1<2$

$\Leftrightarrow A< 2$ hay $A< B$

Lời giải:
a) 

$\frac{a}{b}< \frac{c}{d}\Leftrightarrow \frac{ad}{bd}< \frac{bc}{bd}$

$\Leftrightarrow \frac{ad-bc}{bd}< 0$

Vì $bd>0$ với mọi $b,d>0$ nên $ad-bc< 0\Leftrightarrow ad< bc$

b) Từ phần a suy ra $bc-ad>0$

$\frac{a+c}{b+d}-\frac{a}{b}=\frac{b(a+c)-a(b+d)}{b(b+d)}=\frac{bc-ad}{b(b+d)}>0$ do $bc-ad>0$ và $b(b+d)>0$ với mọi $b,d>0$)

$\Rightarrow \frac{a+c}{b+d}>\frac{a}{b}$

Lại có:
$\frac{a+c}{b+d}-\frac{c}{d}=\frac{d(a+c)-c(b+d)}{d(b+d)}=\frac{ad-bc}{d(b+d)}<0$ do $ad-bc<0$ và $d(b+d)>0$ với mọi $b,d>0$

$\Rightarrow \frac{a+c}{b+d}< \frac{c}{d}$ 

Ta có đpcm.

a) Gọi phân số cần tìm có dạng là \(\dfrac{a}{12}\)

Theo đề, ta có: \(\dfrac{-2}{3}< \dfrac{a}{12}< \dfrac{-1}{4}\)

\(\Leftrightarrow\dfrac{-8}{12}< \dfrac{a}{12}< \dfrac{-3}{12}\)

\(\Leftrightarrow-8< a< -3\)

\(\Leftrightarrow a\in\left\{-7;-6;-5;-4\right\}\)

Vậy: Các phân số cần tìm là \(\dfrac{-7}{12};\dfrac{-6}{12};\dfrac{-5}{12};\dfrac{-4}{12}\)

b) Gọi phân số cần tìm có dạng là \(\dfrac{15}{a}\left(a\ne0\right)\)

Theo đề, ta có: \(\dfrac{3}{7}< \dfrac{15}{a}< \dfrac{5}{8}\)

\(\Leftrightarrow\dfrac{15}{35}< \dfrac{15}{a}< \dfrac{15}{24}\)

Vậy: Các phân số cần tìm là \(\dfrac{15}{34};\dfrac{15}{33};...;\dfrac{15}{25}\)

a) \(\dfrac{-5}{6}=\dfrac{-340}{408}\);\(\dfrac{7}{8}=\dfrac{357}{408}\);\(\dfrac{7}{24}=\dfrac{119}{408}\)

\(\dfrac{16}{17}=\dfrac{384}{408}\); \(\dfrac{-3}{4}=\dfrac{-306}{408}\); \(\dfrac{2}{3}=\dfrac{272}{408}\)

Do đó: \(\dfrac{-5}{6}< \dfrac{-3}{4}< \dfrac{7}{24}< \dfrac{2}{3}< \dfrac{7}{8}< \dfrac{16}{17}\)

\(\dfrac{1}{2020}-\dfrac{1}{2021}=\dfrac{2021}{2020.2021}-\dfrac{2020}{2020.2021}=\dfrac{2021-2020}{2020.2021}=\dfrac{1}{2020.2021}\)

\(\dfrac{1}{2020\cdot2021}=\dfrac{2021-2020}{2020\cdot2021}=\dfrac{1}{2020}-\dfrac{1}{2021}\)(đpcm)

\(\dfrac{1}{13}-\dfrac{1}{18}=\dfrac{18}{13.18}-\dfrac{13}{13.18}=\dfrac{18-13}{13.18}=\dfrac{5}{13.18}\)

\(\dfrac{5}{13\cdot18}=\dfrac{18}{13\cdot18}-\dfrac{13}{13\cdot18}=\dfrac{1}{13}-\dfrac{1}{18}\)(đpcm)