Cách tính nhanh tổng của nhiều phân số

PHƯƠNG PHÁP GIẢI

Để tính nhanh tổng của nhiều phân số ta làm như sau:

– Bước 1: Bỏ dấu ngoặc ( nếu cần)

– Bước 2: Sử dụng các tính chất cơ bản của phép cộng phân số để nhóm ghép một cách phù hợp.

– Bước 3: Tính tổng và rút gọn

BÀI TẬP MINH HỌA

1A. Tính nhanh:

$ \displaystyle \begin{array}{l}a)\dfrac{{-3}}{7}+\dfrac{5}{{13}}+\dfrac{3}{7};\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}\begin{array}{*{20}{c}} {} & {} \end{array}b)\dfrac{{-5}}{{21}}+\dfrac{{-2}}{{21}}+\dfrac{8}{{24}};\\\\c)\dfrac{{-5}}{{11}}+\left( {\dfrac{{-6}}{{11}}+2} \right);\begin{array}{*{20}{c}} {} & {} & {} \end{array}\begin{array}{*{20}{c}} {} & {} \end{array}d)\left( {\dfrac{{-1}}{{32}}+\dfrac{1}{2}} \right)+\dfrac{{-15}}{{32}}\end{array}$

1B. Tính nhanh:

$ \displaystyle \begin{array}{l}a)\dfrac{2}{3}+\dfrac{5}{7}+\dfrac{{-2}}{3};\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}\begin{array}{*{20}{c}} {} & {} \end{array}b)\dfrac{{-1}}{4}+\dfrac{5}{8}+\dfrac{{-3}}{8};\\\\c)\dfrac{{-6}}{{13}}+\left( {\dfrac{{-7}}{{13}}+2} \right);\begin{array}{*{20}{c}} {} & {} & {} \end{array}\begin{array}{*{20}{c}} {} & {} \end{array}d)\left( {\dfrac{{-5}}{{27}}+\dfrac{1}{3}} \right)+\dfrac{{-4}}{{27}}\end{array}$

2A. Tính nhanh:

$ \displaystyle \begin{array}{l}a)\dfrac{4}{7}+\dfrac{3}{4}+\dfrac{2}{7}+\dfrac{5}{4}+\dfrac{1}{7};\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}\begin{array}{*{20}{c}} {} & {} \end{array}b)\dfrac{{-5}}{7}+\dfrac{3}{4}+\dfrac{{-1}}{5}+\dfrac{{-2}}{7}+\dfrac{1}{4};\\\\c)\dfrac{5}{{13}}+\dfrac{{-5}}{7}+\dfrac{{-20}}{{41}}+\dfrac{8}{{13}}+\dfrac{{-21}}{{41}};\begin{array}{*{20}{c}} {} & {} & {} \end{array}d)\dfrac{1}{{28}}+\dfrac{{-1}}{{14}}+\dfrac{3}{{28}}+\dfrac{{-1}}{7}+\dfrac{3}{{14}}\end{array}$

2B. Tính nhanh:

$ \displaystyle \begin{array}{l}a)\dfrac{4}{3}+\dfrac{3}{5}+\dfrac{7}{3}+\dfrac{2}{5}+\dfrac{1}{3};\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}\begin{array}{*{20}{c}} {} & {} \end{array}b)\dfrac{2}{7}+\dfrac{{-3}}{8}+\dfrac{4}{7}+\dfrac{1}{7}+\dfrac{{-5}}{8};\\\\c)\dfrac{{-5}}{9}+\dfrac{8}{{15}}+\dfrac{{-2}}{{11}}+\dfrac{4}{{-9}}+\dfrac{7}{{15}};\begin{array}{*{20}{c}} {} & {} & {} \end{array}d)\dfrac{4}{{45}}+\dfrac{{-1}}{{15}}+\dfrac{7}{{45}}+\dfrac{4}{{15}}+\dfrac{{-2}}{{45}}+\dfrac{{-1}}{5}\end{array}$

$ \displaystyle e)\dfrac{1}{2}+\dfrac{{-1}}{3}+\dfrac{1}{4}+\dfrac{{-1}}{5}+\dfrac{1}{6}+\dfrac{{-1}}{6}+\dfrac{1}{5}+\dfrac{{-1}}{4}+\dfrac{1}{3}+\dfrac{{-1}}{2}$

3A. Tính các tổng sau một cách hợp lí:

$ \displaystyle \begin{array}{l}a)\dfrac{1}{3}+\dfrac{{-2}}{6}+\dfrac{3}{9};\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}\begin{array}{*{20}{c}} {} & {} \end{array}\begin{array}{*{20}{c}} {} & {} \end{array}b)\dfrac{{-4}}{{12}}+\dfrac{{18}}{{45}}+\dfrac{{-6}}{9}+\dfrac{{24}}{{30}};\\\\c)\dfrac{7}{{23}}+\dfrac{{-10}}{{18}}+\dfrac{{-4}}{9}+\dfrac{{16}}{{23}}+\dfrac{{-5}}{8};\begin{array}{*{20}{c}} {} & {} & {} \end{array}d)\dfrac{1}{3}+\dfrac{{-3}}{4}+\dfrac{3}{5}+\dfrac{{-1}}{{36}}+\dfrac{1}{{15}}+\dfrac{{-2}}{9}\\\\e)\dfrac{{-1}}{2}+\dfrac{3}{5}+\dfrac{{-1}}{9}+\dfrac{{-7}}{{18}}+\dfrac{4}{{35}}+\dfrac{2}{7}\end{array}$

3B. Tính các tổng sau một cách hợp lí:

$ \displaystyle \begin{array}{l}a)\dfrac{1}{2}+\dfrac{{-2}}{4}+\dfrac{4}{8};\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}\begin{array}{*{20}{c}} {} & {} \end{array}\begin{array}{*{20}{c}} {} & {} \end{array}b)\dfrac{{-3}}{9}+\dfrac{{16}}{{40}}+\dfrac{{-8}}{{12}}+\dfrac{2}{{10}};\\\\c)\dfrac{{-1}}{8}+\dfrac{6}{7}+\dfrac{2}{{14}}+\dfrac{{-7}}{8}+\dfrac{7}{9};\begin{array}{*{20}{c}} {} & {} & {} & {} \end{array}d)\dfrac{1}{2}+\dfrac{{-1}}{5}+\dfrac{{-5}}{7}+\dfrac{1}{6}+\dfrac{{-3}}{{35}}+\dfrac{1}{3}\\\\e)\dfrac{{-1}}{2}+\dfrac{1}{3}+\dfrac{{-1}}{4}+\dfrac{{-2}}{8}+\dfrac{4}{{18}}+\dfrac{4}{9}\end{array}$

HƯỚNG DẪN GIẢI

1A.  $ \displaystyle a)\dfrac{{-3}}{7}+\dfrac{5}{{13}}+\dfrac{3}{7}=\left( {\dfrac{{-3}}{7}+\dfrac{3}{7}} \right)+\dfrac{5}{{13}}=\dfrac{5}{{13}}.$

$ \displaystyle \begin{array}{l}b)\dfrac{{-5}}{{21}}+\dfrac{{-2}}{{21}}+\dfrac{8}{{24}}=\left( {\dfrac{{-5}}{{21}}+\dfrac{{-2}}{{21}}} \right)+\dfrac{8}{{24}}=\dfrac{{-1}}{3}+\dfrac{1}{3}=0\\\\c)\dfrac{{-5}}{{11}}+\left( {\dfrac{{-6}}{{11}}+2} \right)=\left( {\dfrac{{-5}}{{11}}+\dfrac{{-6}}{{11}}} \right)+2=-1+2=1\\\\d)\dfrac{{-1}}{{32}}+\dfrac{1}{2}+\dfrac{{-15}}{{32}}=\dfrac{{-1}}{{32}}+\dfrac{{-15}}{{32}}+\dfrac{1}{2}=\dfrac{{-1}}{2}+\dfrac{1}{2}=0\end{array}$

1B.  a) $ \displaystyle \dfrac{5}{7}$.                          b) 0.                      c)   1.                          d)   0.

2A.  $ \displaystyle a)\left( {\dfrac{4}{7}+\dfrac{2}{7}+\dfrac{1}{7}} \right)+\left( {\dfrac{3}{4}+\dfrac{5}{4}} \right)=3\begin{array}{*{20}{c}} {} & {} & {} \end{array}b)\left( {\dfrac{{-5}}{7}+\dfrac{{-2}}{7}} \right)+\dfrac{{-1}}{5}+\left( {\dfrac{3}{4}+\dfrac{1}{4}} \right)=\dfrac{{-1}}{5}$

$ \displaystyle \begin{array}{l}c)\left( {\dfrac{5}{{13}}+\dfrac{8}{{13}}} \right)+\dfrac{{-5}}{7}+\left( {\dfrac{{-20}}{{41}}+\dfrac{{-21}}{{41}}} \right)=\dfrac{{-5}}{7}\\\\d)\left( {\dfrac{1}{{28}}+\dfrac{3}{{28}}} \right)+\left( {\dfrac{3}{{14}}+\dfrac{{-1}}{{14}}} \right)+\dfrac{{-1}}{7}=\dfrac{1}{7}\\\\e)\left( {\dfrac{1}{2}+\dfrac{{-1}}{2}} \right)+\left( {\dfrac{{-2}}{3}+\dfrac{2}{3}} \right)+\left( {\dfrac{3}{4}+\dfrac{{-3}}{4}} \right)+\left( {\dfrac{{-4}}{5}+\dfrac{4}{5}} \right)+\left( {\dfrac{5}{6}+\dfrac{{-5}}{6}} \right)=0\end{array}$

2B. Tương tự 2A

a) 5.

b)  $ \displaystyle c)\dfrac{{-2}}{{11}}\begin{array}{*{20}{c}} {} & {} \end{array}d)\dfrac{1}{5}$       

e)  0.

3A.  a) $ \displaystyle \dfrac{1}{3}+\dfrac{{-1}}{3}+\dfrac{1}{3}=\dfrac{1}{3}$

$ \displaystyle \begin{array}{l}b)\dfrac{{-1}}{3}+\dfrac{2}{5}+\dfrac{{-2}}{3}+\dfrac{4}{5}=\dfrac{1}{5}\\\\c)\dfrac{7}{{23}}+\dfrac{{16}}{{23}}+\dfrac{{-5}}{9}+\dfrac{{-4}}{9}+\dfrac{{-5}}{8}=\dfrac{{-5}}{8}\\\\d)\left( {\dfrac{{-2}}{9}+\dfrac{{-3}}{4}+\dfrac{{-1}}{{36}}} \right)+\dfrac{1}{3}+\left( {\dfrac{3}{5}+\dfrac{1}{5}} \right)=0\\\\e)\left( {\dfrac{{-1}}{2}+\dfrac{{-1}}{9}+\dfrac{{-7}}{{18}}} \right)+\left( {\dfrac{4}{{35}}+\dfrac{3}{5}+\dfrac{2}{7}} \right)=0\end{array}$

3B. Tương tự 3A

$ \displaystyle a)\dfrac{1}{2}.\begin{array}{*{20}{c}} {} & {} & {} \end{array}b)\dfrac{{-2}}{5}.\begin{array}{*{20}{c}} {} & {} & {} \end{array}c)\dfrac{{-7}}{8}.\begin{array}{*{20}{c}} {} & {} & {} \end{array}d)0.\begin{array}{*{20}{c}} {} & {} & {} \end{array}e)0.$