Giải Bài 11 trang 13 sách bài tập toán 7 tập 1 – Cánh diều

Đề bài

Tính:

a) \( – {\rm{ }}6,07 + 3,68\);                                            

b) \(\dfrac{3}{5} + ( – {\rm{ }}4)\);

c) \(1\dfrac{3}{5} – \dfrac{{ – 7}}{{11}}\);                                    

d) \(\dfrac{{ – 7}}{{13}} + 0,6\);

e) \( – {\rm{ 1,221}}{\rm{.}}\dfrac{2}{{ – 5}}\);                           

g) \(\dfrac{{ – 17}}{7}:0,25\).

Lời giải chi tiết

a) \( – {\rm{ }}6,07 + 3,68 =-(6,07-3,68)=  – {\rm{ 2,39}}\);                              

b) Do \(\dfrac{3}{5} = \dfrac{6}{{10}} = 0,6\) nên:

\(\dfrac{3}{5} + ( – {\rm{ }}4) = \dfrac{3}{5} – 4 = 0,6 – 4 =-(4-0,6)=  – 3,4\);

c) \(1\dfrac{3}{5} – \dfrac{{ – 7}}{{11}} = \dfrac{8}{5} + \dfrac{7}{{11}} = \dfrac{{88}}{{55}} + \dfrac{{35}}{{55}} = \dfrac{{123}}{{55}}\);                                

d) \(\dfrac{{ – 7}}{{13}} + 0,6 = \dfrac{{ – 7}}{{13}} + \dfrac{6}{{10}} = \dfrac{{ – 70}}{{130}} + \dfrac{{78}}{{130}} = \dfrac{8}{{130}} = \dfrac{4}{{65}}\);

e) Do \(\dfrac{2}{{ – 5}} = \dfrac{4}{{ – 10}} =- 0,4\) nên:

 \( – {\rm{ 1,221}}{\rm{.}}\dfrac{2}{{ – 5}} =  (- {\rm{ 1,221}}){\rm{.}}( – {\rm{ 0,4) = 0,4844}}\);       

g) \(\dfrac{{ – 17}}{7}:0,25 = \dfrac{{ – 17}}{7}:\dfrac{1}{4} = \dfrac{{ – 17}}{7}.4 = \dfrac{{ – 68}}{7}\).