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Bài 1:Tính:a,\(\frac{6}{7}\)+\(\frac{7}{8}\)=.................;                  b,\(\frac{4}{5}\)-\(\frac{2}{3}\)=...........c,\(\frac{2}{3}\)nhân \(\frac{4}{9}\)=.................;       d,\(\frac{1}{5}\):\(\frac{2}{7}\)=............Bài 2 :Điền dấu >;<;=a,\(13\frac{1}{100}\)..........\(12\frac{99}{100}\);b,3+\(\frac{24}{96}\)........\(3\frac{1}{4}\);c,\(4\frac{1}{2}\):\(2\frac{1}{2}\)...........\(1\frac{1}{2}\)+\(\frac{1}{2}\)Bài 3 :Một băng giấy có 3 màu...
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Bài 1:Tính:

a,

\(\frac{6}{7}\)+\(\frac{7}{8}\)=.................;                  b,\(\frac{4}{5}\)-\(\frac{2}{3}\)=...........

c,\(\frac{2}{3}\)nhân \(\frac{4}{9}\)=.................;       d,\(\frac{1}{5}\):\(\frac{2}{7}\)=............

Bài 2 :Điền dấu >;<;=

a,\(13\frac{1}{100}\)..........\(12\frac{99}{100}\);b,3+\(\frac{24}{96}\)........\(3\frac{1}{4}\);c,\(4\frac{1}{2}\):\(2\frac{1}{2}\)...........\(1\frac{1}{2}\)+\(\frac{1}{2}\)

Bài 3 :Một băng giấy có 3 màu vàng,xanh,đỏ.Trong đó\(\frac{1}{3}\)băng giấy tô màu vàng,\(\frac{1}{2}\)băng giấy tô màu xanh.Tìm phân số chỉ số phần băng giấy tô màu đỏ.

Bài 4 :Để lát nền căn phòng hcn,người ta dùng loại gạch men hình vuông có cạnh 30 cm.Hỏi cần bn viên gạch để lát kín nền căn phòng đó,biết chiều dài căn phòng là 9m,chiều rộng 6m?(Diện tích phần mạch vữa là k đáng kể)

mk cần gấp trg hnay ai trả lời trc thì mk sẽ tick nha

4
29 tháng 9 2018

mình làm bài này rồi

29 tháng 9 2018

chỉ mik cách để viết phân số r mik giải bài này cho

HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023

a)

\(\begin{array}{l}0,75 - \frac{5}{6} + 1\frac{1}{2} = \frac{3}{4} - \frac{5}{6} + \frac{3}{2}\\ = \frac{9}{{12}} - \frac{{10}}{{12}} + \frac{{18}}{{12}} = \frac{{17}}{{12}}\end{array}\)                                    

b)

\(\begin{array}{l}\frac{3}{7} + \frac{4}{{15}} + \left( {\frac{{ - 8}}{{21}}} \right) + \left( { - 0,4} \right) = \frac{3}{7} + \frac{4}{{15}} - \frac{8}{{21}} - \frac{2}{5}\\ = \left( {\frac{3}{7} - \frac{8}{{21}}} \right) + \left( {\frac{4}{{15}} - \frac{2}{5}} \right)\\ = \left( {\frac{9}{{21}} - \frac{8}{{21}}} \right) + \left( {\frac{4}{{15}} - \frac{6}{{15}}} \right)\\ = \frac{1}{{21}} + \left( {\frac{{ - 2}}{{15}}} \right)\\ = \frac{5}{{105}} - \frac{{14}}{{105}}\\ = \frac{{ - 9}}{{105}} = \frac{{ - 3}}{{35}}\end{array}\)

c)

\(\begin{array}{l}0,625 + \left( {\frac{{ - 2}}{7}} \right) + \frac{3}{8} + \left( {\frac{{ - 5}}{7}} \right) + 1\frac{2}{3}\\ = \frac{5}{8} + \left( {\frac{{ - 2}}{7}} \right) + \frac{3}{8} - \frac{5}{7} + \frac{5}{3}\\ = \left( {\frac{5}{8} + \frac{3}{8}} \right) + \left( {\frac{{ - 2}}{7} - \frac{5}{7}} \right) + \frac{5}{3}\\ = 1 - 1 + \frac{5}{3} = \frac{5}{3}\end{array}\)          

 d)

\(\begin{array}{l}\left( { - 3} \right).\left( {\frac{{ - 38}}{{21}}} \right).\left( {\frac{{ - 7}}{6}} \right).\left( { - \frac{3}{{19}}} \right)\\ = \frac{{ - 3.\left( { - 38} \right).\left( { - 7} \right).\left( { - 3} \right)}}{{21.6.19}}\\ = \frac{{3.38.7.3}}{{21.6.19}}\\ = \frac{{3.2.19.7.3}}{{3.7.3.2.19}}\\ = 1\end{array}\)

e)

 \(\begin{array}{l}\left( {\frac{{11}}{{18}}:\frac{{22}}{9}} \right).\frac{8}{5} = \left( {\frac{{11}}{{18}}.\frac{9}{{22}}} \right).\frac{8}{5}\\ = \frac{{11.9.4.2}}{{9.2.2.11.5}} = \frac{2}{5}\end{array}\)                                   

 g)

\(\left[ {\left( {\frac{{ - 4}}{5}} \right).\frac{5}{8}} \right]:\left( {\frac{{ - 25}}{{12}}} \right) = \frac{{ - 20}}{{40}}:\left( {\frac{{ - 25}}{{12}}} \right)\\ = \frac{{ - 1}}{2}.\frac{{ - 12}}{{25}} = \frac{6}{{25}}\)

HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023

a)

\(\begin{array}{l}\frac{3}{7}.\left( { - \frac{1}{9}} \right) + \frac{3}{7}.\left( { - \frac{2}{3}} \right)\\ = \frac{3}{7}.\left( { - \frac{1}{9} + \frac{-2}{3}} \right)\\ = \frac{3}{7}.\left( { - \frac{1}{9} - \frac{6}{9}} \right)\\ = \frac{3}{7}.\frac{{ - 7}}{9} = \frac{{ - 1}}{3}\end{array}\)                 

b)

\(\begin{array}{l}\left( {\frac{{ - 7}}{{13}}} \right).\frac{5}{{12}} + \left( {\frac{{ - 7}}{{13}}} \right).\frac{7}{{12}} + \left( {\frac{{ - 6}}{{13}}} \right)\\ = \frac{{ - 7}}{{13}}.\left( {\frac{5}{{12}} + \frac{7}{{12}}} \right) + \left( {\frac{{ - 6}}{{13}}} \right)\\ = \frac{{ - 7}}{{13}}.1 + \left( {\frac{{ - 6}}{{13}}} \right)\\ = \frac{{ - 7}}{{13}} + \left( {\frac{{ - 6}}{{13}}} \right)\\ = \frac{{ - 13}}{{13}}\\ = -1\end{array}\)

c)

\(\begin{array}{l}\left[ {\left( {\frac{{ - 2}}{3} + \frac{3}{7}} \right)} \right]:\frac{5}{9} + \left( {\frac{4}{7} - \frac{1}{3}} \right):\frac{5}{9}\\ = \left[ {\left( {\frac{{ - 2}}{3} + \frac{3}{7}} \right)} \right].\frac{9}{5} + \left( {\frac{4}{7} - \frac{1}{3}} \right).\frac{9}{5}\\ = \left( {\frac{{ - 2}}{3} + \frac{3}{7} + \frac{4}{7} - \frac{1}{3}} \right).\frac{9}{5}\\ = \left[ {\left( {\frac{{ - 2}}{3} - \frac{1}{3}} \right) + \left( {\frac{3}{7} + \frac{4}{7}} \right)} \right].\frac{9}{5}\\ = \left( { - 1 + 1} \right).\frac{9}{5}\\ = 0.\frac{9}{5} = 0\end{array}\)

d)

\(\begin{array}{l}\frac{5}{9}:\left( {\frac{1}{{11}} - \frac{5}{{22}}} \right) + \frac{5}{9}:\left( {\frac{1}{{15}} - \frac{2}{3}} \right)\\ = \frac{5}{9}:\left( {\frac{2}{{22}} - \frac{5}{{22}}} \right) + \frac{5}{9}:\left( {\frac{1}{{15}} - \frac{{10}}{{15}}} \right)\\ = \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{5}{9}:\frac{{ - 9}}{15}\\= \frac{5}{9}:\frac{{ - 3}}{{22}} + \frac{5}{9}:\frac{{ - 3}}{5}\\ = \frac{5}{9}.\frac{{ - 22}}{3} + \frac{5}{9}.\frac{{ - 5}}{3}\\ = \frac{5}{9}.\left( {\frac{{ - 22}}{3} - \frac{5}{3}} \right)\\ = \frac{5}{9}.\frac{-27}{3}= \frac{5}{9}.\left( { - 9} \right) =  - 5\end{array}\)

e)

\(\begin{array}{l}\frac{3}{5} + \frac{3}{{11}} - \left( {\frac{{ - 3}}{7}} \right) + \left( {\frac{{ - 2}}{{97}}} \right) - \frac{1}{{35}} - \frac{3}{4} + \left( {\frac{{ - 23}}{{44}}} \right)\\ = \frac{3}{5} + \frac{3}{{11}} + \frac{3}{7} - \frac{2}{{97}} - \frac{1}{{35}} - \frac{3}{4} - \frac{{23}}{{44}}\\ = \left( {\frac{3}{5} + \frac{3}{7} - \frac{1}{{35}}} \right) + \left( {\frac{3}{{11}} - \frac{3}{4} - \frac{{23}}{{44}}} \right) - \frac{2}{{97}}\\ = \left( {\frac{{21}}{{35}} + \frac{{15}}{{35}} - \frac{1}{{35}}} \right) + \left( {\frac{{12}}{{44}} - \frac{{33}}{{44}} - \frac{{23}}{{44}}} \right) - \frac{2}{{97}}\\ = \frac{35}{{35}}+ \frac{-44}{{44}}- \frac{2}{{97}}\\= 1 + \left( { - 1} \right) - \frac{2}{{97}}\\ =  - \frac{2}{{97}}\end{array}\)

HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023

\(\begin{array}{l}a)\left[ {{{\left( {\dfrac{3}{7}} \right)}^4}.{{\left( {\dfrac{3}{7}} \right)}^5}} \right]:{\left( {\dfrac{3}{7}} \right)^7}\\ = {\left( {\dfrac{3}{7}} \right)^{4 + 5}}:{\left( {\dfrac{3}{7}} \right)^7}\\ = {\left( {\dfrac{3}{7}} \right)^9}:{\left( {\dfrac{3}{7}} \right)^7}\\ = {\left( {\dfrac{3}{7}} \right)^{9-7}}\\= {\left( {\dfrac{3}{7}} \right)^2}\\b)\left[ {{{\left( {\dfrac{7}{8}} \right)}^5}:{{\left( {\dfrac{7}{8}} \right)}^4}} \right].\left( {\dfrac{7}{8}} \right)\\ = {\left( {\dfrac{7}{8}} \right)^{5 - 4}}.\left( {\dfrac{7}{8}} \right)\\ = \left( {\dfrac{7}{8}} \right).\left( {\dfrac{7}{8}} \right)\\ = {\left( {\dfrac{7}{8}} \right)^2}\\c)\left[ {{{\left( {0,6} \right)}^3}.{{\left( {0,6} \right)}^8}} \right]:\left[ {{{\left( {0,6} \right)}^7}.{{\left( {0,6} \right)}^2}} \right]\\ = {\left( {0,6} \right)^{3 + 8}}:{\left( {0,6} \right)^{7 + 2}}\\ = {\left( {0,6} \right)^{11}}:{\left( {0,6} \right)^9}\\ = {\left( {0,6} \right)^{11-9}}\\={\left( {0,6} \right)^2}.\end{array}\)

HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023

a)

\(\begin{array}{l}\left( {\frac{{ - 3}}{7}} \right) + \left( {\frac{5}{6} - \frac{4}{7}} \right)\\ = \left( {\frac{{ - 3}}{7}} \right) + \frac{5}{6} - \frac{4}{7}\\ = \left[ {\left( {\frac{{ - 3}}{7}} \right) - \frac{4}{7}} \right] + \frac{5}{6}\\ =\frac{-7}{7}+\frac{5}{6}\\=  - 1 + \frac{5}{6}\\ = \frac{{ - 1}}{6}\end{array}\)                          

b)

\(\begin{array}{l}\frac{3}{5} - \left( {\frac{2}{3} + \frac{1}{5}} \right)\\ = \frac{3}{5} - \frac{2}{3} - \frac{1}{5}\\ = (\frac{3}{5} - \frac{1}{5}) - \frac{2}{3}\\ = \frac{2}{5} - \frac{2}{3}\\ = \frac{6}{{15}} - \frac{{10}}{{15}}\\ = \frac{{ - 4}}{{15}}\end{array}\)

c)

\(\begin{array}{l}\left[ {\left( {\frac{{ - 1}}{3}} \right) + 1} \right] - \left( {\frac{2}{3} - \frac{1}{5}} \right)\\ = \left( {\frac{{ - 1}}{3}} \right) + 1 - \frac{2}{3} + \frac{1}{5}\\ = \left( {\frac{{ - 1}}{3} - \frac{2}{3}} \right) + 1 + \frac{1}{5}\\ = \frac{-3}{3}+1+\frac{1}{5}\\= - 1 + 1 + \frac{1}{5}\\ = \frac{1}{5}\end{array}\)                  

d)

\(\begin{array}{l}1\frac{1}{3} + \left( {\frac{2}{3} - \frac{3}{4}} \right) - \left( {0,8 + 1\frac{1}{5}} \right)\\ = 1 + \frac{1}{3} + \frac{2}{3} - \frac{3}{4} - \left( {\frac{4}{5} + 1 + \frac{1}{5}} \right)\\=1+\frac{3}{3}-\frac{3}{4}-(\frac{5}{5}+1)\\ = 1 + 1 - \frac{3}{4} - (1+1)\\ =  - \frac{3}{4}\end{array}\).

HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023

a)

\(\begin{array}{l}\left( {\frac{3}{4}:1\frac{1}{2}} \right) - \left( {\frac{5}{6}:\frac{1}{3}} \right)\\ = \left( {\frac{3}{4}:\frac{3}{2}} \right) - \left( {\frac{5}{6}.3} \right)\\ = \left( {\frac{3}{4}.\frac{2}{3}} \right) - \frac{5}{2}\\ = \frac{1}{2} - \frac{5}{2}\\ = \frac{-4}{2}\\= - 2.\end{array}\)                         

b)

\(\begin{array}{l}\left[ {\left( {\frac{{ - 1}}{5}} \right):\frac{1}{{10}}} \right] - \frac{5}{7}.\left( {\frac{2}{3} - \frac{1}{5}} \right)\\ = \left( {\frac{{ - 1}}{5}} \right).10 - \frac{5}{7}.\left( {\frac{{10}}{{15}} - \frac{3}{{15}}} \right)\\ =  - 2 - \frac{5}{7}.\frac{7}{{15}}\\ =  - 2 - \frac{1}{3}\\ = \frac{{ - 6}}{3} - \frac{1}{3}\\ = \frac{{ - 7}}{3}\end{array}\)

c)

\(\begin{array}{l}\left( { - 0,4} \right) + 2\frac{2}{5}.{\left[ {\left( {\frac{{ - 2}}{3}} \right) + \frac{1}{2}} \right]^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.{\left[ {\left( {\frac{{ - 4}}{6}} \right) + \frac{3}{6}} \right]^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.{\left( {\frac{{ - 1}}{6}} \right)^2}\\ = \left( { - \frac{2}{5}} \right) + \frac{{12}}{5}.\frac{1}{{36}}\\ = \left( { - \frac{2}{5}} \right) + \frac{1}{{15}}\\ = \left( { - \frac{6}{{15}}} \right) + \frac{1}{{15}}\\ = \frac{{ - 5}}{{15}}\\ = \frac{{ - 1}}{3}\end{array}\)             

d)

\(\begin{array}{l}\left\{ {\left[ {{{\left( {\frac{1}{{25}} - 0,6} \right)}^2}:\frac{{49}}{{125}}} \right].\frac{5}{6}} \right\} - \left[ {\left( {\frac{{ - 1}}{3}} \right) + \frac{1}{2}} \right]\\ = \left\{ {\left[ {{{\left( {\frac{1}{{25}} - \frac{3}{5}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \left[ {\left( {\frac{{ - 2}}{6}} \right) + \frac{3}{6}} \right]\\ = \left\{ {\left[ {{{\left( {\frac{{ 1}}{{25}}-\frac{15}{25}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left\{ {\left[ {{{\left( {\frac{{ - 14}}{{25}}} \right)}^2}.\frac{{125}}{{49}}} \right].\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left\{ {\frac{{196}}{{{{25}^2}}}.\frac{{25.5}}{{49}}.\frac{5}{6}} \right\} - \frac{1}{6}\\ = \left( {\frac{{4.49.25.5.5}}{{{{25}^2}.49.6}}} \right) - \frac{1}{6}\\ = \frac{4}{6} - \frac{1}{6}\\ = \frac{3}{6}\\ = \frac{1}{2}\end{array}\)

29 tháng 10 2023

5và 3/8-1 và 5/6

 

Kiểm tra bài : Nhân, chia số hữu tỉThực hiện phép tính...
Đọc tiếp

Kiểm tra bài : Nhân, chia số hữu tỉ

Thực hiện phép tính :

(1) \(-\frac{3}{2}.\frac{7}{10}=\frac{-3.7}{2.10}=\frac{-21}{20}\)

(2) \(\frac{-5}{3}.\frac{6}{11}=\frac{-5.6}{3.11}=\frac{-30}{33}\)

(3) \(2\frac{1}{3}.\left(-1\frac{2}{3}\right)=\frac{7}{3}.\left(-\frac{5}{3}\right)=\frac{7.\left(-5\right)}{3.3}=-\frac{35}{9}\)

(4) \(\frac{9}{10}:\left(-\frac{15}{11}\right)=\frac{9}{10}.\left(\frac{-11}{15}\right)=\frac{9.\left(-11\right)}{10.15}=-\frac{99}{150}=-\frac{33}{50}\)

(5) \(\left(-1\right):\frac{3}{8}=\frac{\left(-1\right).8}{3}=-\frac{8}{3}\)

(6) \(\frac{1}{2}.\left(-\frac{5}{4}\right).\frac{8}{7}=\frac{1.\left(-5\right)}{2.4}.\frac{8}{7}=-\frac{5}{8}.\frac{8}{7}=-\frac{5.8}{8.7}=-\frac{5}{7}\)

(7) \(\frac{-9}{2}.\frac{2}{18}.\frac{1}{7}=\left(-\frac{9}{2}.\frac{2}{18}\right).\frac{1}{7}=\left(-\frac{9.2}{2.18}\right).\frac{1}{7}=-\frac{18}{36}.\frac{1}{7}=-\frac{18.1}{36.7}=-\frac{1}{14}\)

(8) \(\left(\frac{9}{2}-\frac{1}{3}\right).\frac{6}{17}=\left(\frac{27}{6}-\frac{2}{6}\right).\frac{6}{17}=\frac{27-2}{6}.\frac{6}{17}=\frac{25}{6}.\frac{6}{17}=\frac{25.6}{6.17}=\frac{25}{17}\)

(9) \(\left(-\frac{12}{13}:\frac{36}{39}\right).\frac{3}{5}=\left(-\frac{12}{13}.\frac{39}{36}\right).\frac{3}{5}=\left(\frac{-12.39}{13.36}\right).\frac{3}{5}=-\frac{1.3}{5}=-\frac{3}{5}\)

(10) \(\left(-\frac{3}{7}+\frac{7}{9}\right):\frac{4}{7}+\left(-\frac{4}{7}+\frac{2}{9}\right):\frac{4}{7}=\left(\left(-\frac{3}{7}+\frac{7}{9}\right)+\left(-\frac{4}{7}+\frac{2}{9}\right)\right):\frac{4}{7}\)

\(=\left(\left(-\frac{27}{63}+\frac{49}{63}\right)+\left(-\frac{36}{63}+\frac{14}{63}\right)\right):\frac{4}{7}=\left(\left(-\frac{27+49}{63}\right)+\left(\frac{-36+14}{63}\right)\right):\frac{4}{7}\)

\(=\left(\left(\frac{22}{63}\right)+\left(-\frac{22}{63}\right)\right):\frac{4}{7}\)

\(=\frac{22+\left(-22\right)}{63}:\frac{4}{7}=\frac{0}{63}:\frac{4}{7}=0\)

Mình đăng các bài toán này lên thứ nhất là để kiểm tra năng lực thứ hai các bạn có thể xem đây và rút ra lời giải cho các bài khác và nếu mình sai chỗ nào các bạn chỉ mình sẽ chỉnh

0
16 tháng 4 2017

A. \(\frac{3}{4}\) x \(\frac{8}{9}\)\(\frac{15}{16}\)x .... x \(\frac{899}{900}\)

\(\frac{1.3}{2^2}\) x \(\frac{2.4}{3^3}\)\(\frac{3.5}{4^2}\)x ... x \(\frac{29.31}{30^2}\)

\(\left(\frac{1.2.3...29}{2.3.4...30}\right).\left(\frac{3.4.5...31}{2.3.4...30}\right)\)

\(\frac{1}{30}.\frac{31}{2}\)\(\frac{31}{60}\)

B. 

\(\frac{1}{3}+\frac{3}{8}-\frac{7}{12}=\frac{8}{24}+\frac{9}{24}-\frac{14}{24}=\frac{8+9-14}{24}=\frac{3}{24}=\frac{1}{8}\)

12 tháng 9 2016

giúp với ạ

13 tháng 9 2016

giải dc nhưng mà hoi lâu