Mục lục
PHƯƠNG PHÁP GIẢI
Để so sánh các phân số không cùng mẫu số, ta có các cách như sau:
– Cách 1. Quy đồng mẫu (hoặc tử).
– Cách 2. So sánh phần bù (hoặc phần hơn) với 1.
– Cách 3. Dùng số trung gian.
BÀI TẬP MINH HỌA
2A. So sánh hai phân số bằng cách quy đồng mẫu:
a) $ \displaystyle \dfrac{1}{3}v\grave{a}\dfrac{5}{6};\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}b)\dfrac{4}{5}v\grave{a}\dfrac{3}{7};$
c) $ \displaystyle \dfrac{{-3}}{{11}}v\grave{a}\dfrac{{-4}}{{13}};\begin{array}{*{20}{c}} {} & {} & {} \end{array}d)\dfrac{{-5}}{6}v\grave{a}\dfrac{{63}}{{-70}};$
2B. So sánh hai phân số bằng cách quy đồng mẫu:
$ \displaystyle \begin{array}{l}a)\dfrac{1}{2}v\grave{a}\dfrac{5}{6};\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}b)\dfrac{4}{7}v\grave{a}\dfrac{5}{9};\\\\c)\dfrac{{-3}}{7}v\grave{a}\dfrac{{-4}}{9};\begin{array}{*{20}{c}} {} & {} & {} \end{array}d)\dfrac{{-7}}{8}v\grave{a}\dfrac{{35}}{{-42}};\end{array}$
3A. So sánh hai phân số bằng cách quy đồng tử
$ \displaystyle \begin{array}{l}a)\dfrac{3}{4}v\grave{a}\dfrac{6}{7};\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}b)\dfrac{{17}}{{-21}}v\grave{a}\dfrac{{51}}{{-31}};\\\\c)\dfrac{{-4}}{9}v\grave{a}\dfrac{{-3}}{{13}};\begin{array}{*{20}{c}} {} & {} & {} \end{array}d)\dfrac{{-4}}{{-11}}v\grave{a}\dfrac{{-6}}{{-19}};\end{array}$
3B. So sánh hai phân số bằng cách quy đồng tử
$ \displaystyle \begin{array}{l}a)\dfrac{2}{3}v\grave{a}\dfrac{4}{5};\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}b)\dfrac{{13}}{{-27}}v\grave{a}\dfrac{{39}}{{-37}};\\\\c)\dfrac{{-3}}{7}v\grave{a}\dfrac{{-2}}{9};\begin{array}{*{20}{c}} {} & {} & {} \end{array}d)\dfrac{{-2}}{{-7}}v\grave{a}\dfrac{{-5}}{{-17}};\end{array}$
4A. So sánh hai phân số bằng cách so sánh phần bù (hoặc phần hơn) với 1:
$ \displaystyle \begin{array}{l}a)\dfrac{{26}}{{27}}v\grave{a}\dfrac{{96}}{{97}};\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}b)\dfrac{{102}}{{103}}v\grave{a}\dfrac{{103}}{{105}};\\\\c)\dfrac{{2017}}{{2016}}v\grave{a}\dfrac{{2019}}{{2018}};\begin{array}{*{20}{c}} {} & {} & {} \end{array}d)\dfrac{{73}}{{64}}v\grave{a}\dfrac{{51}}{{45}};\end{array}$
4B. So sánh hai phân số bằng cách so sánh phần bù (hoặc phần hơn) với 1:
$ \displaystyle \begin{array}{l}a)\dfrac{{22}}{{23}}v\grave{a}\dfrac{{16}}{{17}};\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}b)\dfrac{{64}}{{65}}v\grave{a}\dfrac{{45}}{{47}};\\\\c)\dfrac{{199}}{{198}}v\grave{a}\dfrac{{200}}{{199}};\begin{array}{*{20}{c}} {} & {} & {} \end{array}d)\dfrac{{61}}{{58}}v\grave{a}\dfrac{{73}}{{72}};\end{array}$
5A. So sánh hai phân số bằng cách dùng số trung gian:
$ \displaystyle \begin{array}{l}a)\dfrac{{16}}{{-19}}v\grave{a}\dfrac{{15}}{{17}};\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}b)\dfrac{{419}}{{-723}}v\grave{a}\dfrac{{-697}}{{-313}};\begin{array}{*{20}{c}} {} & {} & {} \end{array}c)\dfrac{{311}}{{256}}v\grave{a}\dfrac{{199}}{{203}};\\\\d)\dfrac{{30}}{{235}}v\grave{a}\dfrac{{168}}{{1323}};\begin{array}{*{20}{c}} {} & {} & {} \end{array}e)\dfrac{{19}}{{60}}v\grave{a}\dfrac{{31}}{{90}};\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}f)\dfrac{{15}}{{23}}v\grave{a}\dfrac{{70}}{{117}};\end{array}$
5B. So sánh hai phân số bằng cách dùng số trung gian:
$ \displaystyle \begin{array}{l}a)\dfrac{5}{{-17}}v\grave{a}\dfrac{2}{7};\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}b)\dfrac{{41}}{{-73}}v\grave{a}\dfrac{{-67}}{{-33}};\begin{array}{*{20}{c}} {} & {} & {} \end{array}c)\dfrac{{23}}{{21}}v\grave{a}\dfrac{{21}}{{23}};\\\\d)\dfrac{{19}}{{26}}v\grave{a}\dfrac{{21}}{{25}};\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}e)\dfrac{{19}}{{40}}v\grave{a}\dfrac{{41}}{{80}};\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}f)\dfrac{9}{{23}}v\grave{a}\dfrac{{34}}{{97}};\end{array}$
6A. a) Cho phân số $ \displaystyle \dfrac{a}{b}(a,b\in \mathbb{N},b\ne 0)$.Giả sử $ \displaystyle \dfrac{a}{b}$<1 và $ \displaystyle m\in \mathbb{N},m\ne 0$. Chứng tỏ rằng $ \displaystyle \dfrac{a}{b}<\dfrac{{a+m}}{{b+m}}$.
b) Áp dụng so sánh: $ \displaystyle \dfrac{{437}}{{564}}v\grave{a}\dfrac{{446}}{{573}}$.
6B. a) Cho phân số $ \displaystyle \dfrac{a}{b}(a,b\in \mathbb{N},b\ne 0)$.Giả sử $ \displaystyle \dfrac{a}{b}$>1 và $ \displaystyle m\in \mathbb{N},m\ne 0$. Chứng tỏ rằng $ \displaystyle \dfrac{a}{b}>\dfrac{{a+m}}{{b+m}}$.
b) Áp dụng so sánh: $ \displaystyle \dfrac{{237}}{{142}}v\grave{a}\dfrac{{246}}{{151}}$.
7A. So sánh:
$ \displaystyle a)\dfrac{{510}}{{714}}v\grave{a}\dfrac{{1717}}{{3535}};\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}b)\dfrac{{-292929}}{{333333}}v\grave{a}\dfrac{{-16665}}{{17776}};$
7B. So sánh:
$ \displaystyle a)\dfrac{{1734}}{{2346}}v\grave{a}\dfrac{{1919}}{{2323}};\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}b)\dfrac{{-15151515}}{{23232323}}v\grave{a}\dfrac{{-188887}}{{211109}};$
8A*. So sánh:
$ \displaystyle \begin{array}{l}a)A=\dfrac{{{98}^{{99}}+1}}{{{98}^{{89}}+1}}v\grave{a}\text{ }B=\dfrac{{{98}^{{98}}+1}}{{{98}^{{88}}+1}};\\\\b)C=\dfrac{{{100}^{{2008}}+1}}{{{100}^{{2018}}+1}}v\grave{a}\text{ D}=\dfrac{{{100}^{{2007}}+1}}{{{100}^{{2017}}+1}};\end{array}$
8B*. So sánh:
$ \displaystyle \begin{array}{l}a)A=\dfrac{{{17}^{{18}}+1}}{{{17}^{{19}}+1}}v\grave{a}\text{ }B=\dfrac{{{17}^{{17}}+1}}{{{17}^{{18}}+1}};\\\\b)C=\dfrac{{{100}^{{100}}+1}}{{{100}^{{90}}+1}}v\grave{a}\text{ D}=\dfrac{{{100}^{{99}}+1}}{{{100}^{{89}}+1}};\end{array}$
9A*. So sánh hai phân số : $ \displaystyle \left( {\dfrac{1}{{243}}} \right)^{9}v\grave{a}\text{ }\left( {\dfrac{1}{{83}}} \right)^{{13}}.$
9B*. So sánh hai phân số : $ \displaystyle \left( {\dfrac{1}{{32}}} \right)^{7}v\grave{a}\text{ }\left( {\dfrac{1}{{16}}} \right)^{9}.$
HƯỚNG DẪN GIẢI
2A. a) Ta có $ \displaystyle \dfrac{1}{3}=\dfrac{2}{6};\dfrac{2}{6}<\dfrac{5}{6}=>\dfrac{1}{3}<\dfrac{5}{6}$
Tương tự. $ \displaystyle b)\dfrac{4}{5}>\dfrac{3}{7}\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}c)\dfrac{{-3}}{{11}}>\dfrac{{-4}}{{13}}$
b) Ta có $ \displaystyle \dfrac{{-63}}{{70}}=\dfrac{{-9}}{{10}}$;
Qui đồng ta được :$ \displaystyle \dfrac{{-9}}{{10}}=\dfrac{{-27}}{{30}};\dfrac{{-5}}{6}=\dfrac{{-25}}{{30}}=>\dfrac{{-5}}{6}>\dfrac{{63}}{{-70}}$
2B. $ \displaystyle a)\dfrac{1}{2}<\dfrac{5}{6}\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}b)\dfrac{4}{7}>\dfrac{5}{9};\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}c)\dfrac{{-3}}{7}>\dfrac{{-4}}{9}\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}d)\dfrac{{-7}}{8}<\dfrac{{35}}{{-42}}$
3A. $ \displaystyle a)\dfrac{3}{4}=\dfrac{6}{8}<\dfrac{6}{7}\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}b)\dfrac{{17}}{{-21}}=\dfrac{{51}}{{-63}}>\dfrac{{51}}{{-31}}$
$ \displaystyle c)\dfrac{{-4}}{9}=\dfrac{{12}}{{27}}<\dfrac{{-12}}{{52}}=\dfrac{{-3}}{{13}}\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}d)\dfrac{{-4}}{{-11}}=\dfrac{{12}}{{33}}>\dfrac{{12}}{{38}}=\dfrac{{-6}}{{-19}}$
3B. Tương tự 3A.
$ \displaystyle a)\dfrac{2}{3}<\dfrac{4}{5}\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}b)\dfrac{{13}}{{-27}}>\dfrac{{39}}{{-37}};\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}c)\dfrac{{-3}}{7}<\dfrac{{-2}}{9}\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}d)\dfrac{{-2}}{{-7}}<\dfrac{{-5}}{{-17}}$
4A. a) Ta có: $ \displaystyle 1-\dfrac{{26}}{{27}}=\dfrac{1}{{27}};1-\dfrac{{96}}{{97}}=\dfrac{1}{{97}}$ . Vì $ \displaystyle \dfrac{1}{{27}}>\dfrac{1}{{97}}$ nên $ \displaystyle \dfrac{{26}}{{27}}<\dfrac{{96}}{{97}}$
b) Ta có: $ \displaystyle 1-\dfrac{{102}}{{103}}=\dfrac{1}{{103}};1-\dfrac{{103}}{{105}}=\dfrac{2}{{105}}$. Vì $ \displaystyle \dfrac{1}{{103}}=\dfrac{2}{{206}}<\dfrac{2}{{105}}n\hat{e}n\dfrac{{102}}{{103}}>\dfrac{{103}}{{105}}$
c) Ta có :$ \displaystyle \begin{array}{l}\dfrac{{2017}}{{2016}}=1+\dfrac{1}{{2016}};\dfrac{{2019}}{{2018}}=1+\dfrac{1}{{2018}}~.~~V\grave{i}~~\dfrac{1}{{2016}}>\dfrac{1}{{2018}}~~n\hat{e}n~~\dfrac{{2017}}{{2016}}>\dfrac{{2019}}{{2018}}\\\end{array}$
d) $ \displaystyle \dfrac{{73}}{{64}}>\dfrac{{51}}{{45}}.~~Ta\text{ }c\acute{o}:~~\dfrac{{73}}{{64}}=1+\dfrac{9}{{64}}~;\dfrac{{51}}{{45}}=1+\dfrac{6}{{45}}.~$$ \displaystyle \begin{array}{l}~V\grave{i}~~\dfrac{9}{{64}}=\dfrac{{18}}{{128}}~>\dfrac{6}{{45}}=\dfrac{{18}}{{135}}~n\hat{e}n~~\dfrac{{73}}{{64}}>\dfrac{{51}}{{45}}\\\end{array}$
4B. Tương tự 4A.
$ \displaystyle a)\dfrac{{22}}{{23}}>\dfrac{{16}}{{77}};\begin{array}{*{20}{c}} {} & {} & {} \end{array}b)\dfrac{{64}}{{65}}>\dfrac{{45}}{{47}};\begin{array}{*{20}{c}} {} & {} & {} \end{array}c)\dfrac{{199}}{{198}}>\dfrac{{200}}{{199}};\begin{array}{*{20}{c}} {} & {} & {} \end{array}d)\dfrac{{61}}{{58}}>\dfrac{{73}}{{72}}.$
5A. $ \displaystyle a)\dfrac{{16}}{{-19}}<0<\dfrac{{15}}{{17}}\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}\begin{array}{*{20}{c}} {} & {} \end{array}b)\dfrac{{419}}{{-723}}<0<\dfrac{{-697}}{{-313}}$
$ \displaystyle \begin{array}{l}c)\dfrac{{311}}{{256}}>1>\dfrac{{199}}{{203}}\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}\begin{array}{*{20}{c}} {} & {} \end{array}d)\dfrac{{30}}{{235}}=\dfrac{6}{{47}}>\dfrac{6}{{48}}=\dfrac{8}{{64}}>\dfrac{8}{{63}}=\dfrac{{168}}{{1323}}\\\\e)\dfrac{{19}}{{60}}<\dfrac{{20}}{{60}}=\dfrac{{30}}{{90}}<\dfrac{{31}}{{90}}\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}f)\dfrac{{15}}{{23}}>\dfrac{{14}}{{23}}=\dfrac{{70}}{{115}}>\dfrac{{70}}{{117}}\end{array}$
5B. Tương tự 5A.
$ \displaystyle \begin{array}{l}a)\dfrac{5}{{-17}}<\dfrac{2}{7};\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}b)\dfrac{{41}}{{-73}}<\dfrac{{-67}}{{-33}};\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}c)\dfrac{{23}}{{21}}>\dfrac{{21}}{{23}}\\\\d)\dfrac{{19}}{{26}}<\dfrac{{21}}{{25}};\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}e)\dfrac{{19}}{{40}}<\dfrac{{41}}{{80}};\begin{array}{*{20}{c}} {} & {} \end{array}\begin{array}{*{20}{c}} {} & {} & {} \end{array}f)\dfrac{9}{{23}}>\dfrac{{34}}{{97}}\end{array}$
6A. a) Thực hiện quy đồng $ \displaystyle \dfrac{a}{b}=\dfrac{{a(b+m)}}{{b(b+m)}}=\dfrac{{ab+am}}{{b^{2}+bm}};$
$ \displaystyle \dfrac{{a+m}}{{b+m}}=\dfrac{{b(a+m)}}{{b(b+m)}}=\dfrac{{ab+bm}}{{b^{2}+bm}}.$ Vì $ \displaystyle \dfrac{a}{b}$ < 1=> a < b => ab +am < ab + bm
Từ đó thu được $ \displaystyle \dfrac{a}{b}$ < $ \displaystyle \dfrac{{a+m}}{{b+m}}$
b) $ \displaystyle \dfrac{{437}}{{564}}<\dfrac{{437+9}}{{564+9}}=\dfrac{{446}}{{573}}.$
6B. a) Tương tự 6A
b) $ \displaystyle \dfrac{{237}}{{142}}<\dfrac{{237+9}}{{142+9}}=\dfrac{{246}}{{151}}.$
7A. a) $ \displaystyle \dfrac{{510}}{{714}}=\dfrac{5}{7}>\dfrac{{1717}}{{3535}}=\dfrac{{17}}{{35}}.$
b)$ \displaystyle \dfrac{{-292929}}{{333333}}=\dfrac{{-29}}{{33}}>\dfrac{{-29}}{{32}}>\dfrac{{-30}}{{32}}=\dfrac{{-15}}{{16}}=\dfrac{{-16665}}{{17776}}.$
7B. a)$ \displaystyle \dfrac{{1734}}{{2346}}=\dfrac{{17}}{{23}}<\dfrac{{19}}{{23}}=\dfrac{{1919}}{{2323}}$
b) $ \displaystyle \dfrac{{-15151515}}{{23232323}}=\dfrac{{-15}}{{23}}>\dfrac{{-188887}}{{211109}}=\dfrac{{-17}}{{19}}.$
8A. a) Do $ \displaystyle A=\dfrac{{{98}^{{99}}+1}}{{{98}^{{89}}+1}}>1$ nên
$ \displaystyle A=\dfrac{{{98}^{{99}}+1}}{{{98}^{{89}}+1}}>\dfrac{{{98}^{{99}}+1+97}}{{{98}^{{89}}+1+97}}=\dfrac{{98({98}^{{98}}+1)}}{{98({98}^{{88}}+1)}}=\dfrac{{{98}^{{98}}+1}}{{{98}^{{88}}+1}}=B$
Vậy A > B
b) Do C = $ \displaystyle \dfrac{{{100}^{{2008}}+1}}{{{100}^{{2018}}+1}}$ < 1 nên
C= $ \displaystyle \dfrac{{{100}^{{2008}}+1}}{{{100}^{{2018}}+1}}>\dfrac{{{100}^{{2008}}+1+99}}{{{100}^{{2018}}+1+99}}=\dfrac{{100({100}^{{2007}}+1)}}{{100({100}^{{2017}}+1)}}=\dfrac{{{100}^{{2007}}+1}}{{{100}^{{2017}}+1}}=D$
Vậy C > D.
8B. Tương tự 8A.
a) A < B b) C > D
9A. HS tự làm.
9B. HS tự làm