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Câu 1 mình làm rồi
Câu 2
\(\frac{5\cdot6+5\cdot7}{5\cdot8+20}\) và \(\frac{8\cdot9-4\cdot15}{12\cdot7-180}\)
\(\frac{5\cdot6+5\cdot7}{5\cdot8+20}\)
= \(\frac{5\cdot6+5\cdot7}{5\cdot8+5\cdot4}\)
= \(\frac{5\cdot\left(6+7\right)}{5\cdot\left(8+4\right)}\)
= \(\frac{6+7}{8+4}\)
= \(\frac{13}{12}\)
\(\frac{8\cdot9-4\cdot15}{12\cdot7-180}\)
= \(\frac{4\cdot2\cdot9-4\cdot15}{4\cdot3\cdot7-4\cdot45}\)
= \(\frac{4\cdot\left(18-15\right)}{4\cdot\left(21-45\right)}\)
= \(\frac{-3}{13}\)
Quy đồng \(\frac{13}{12}\) và \(\frac{-3}{13}\)
Mẫu số chung: 156
TSP: 13; 12
\(\frac{13}{12}=\frac{13\cdot13}{12\cdot13}=\frac{169}{156}\)
\(\frac{-3}{13}=\frac{-3\cdot12}{13\cdot12}=\frac{-36}{156}\)
Chúc bn học tốt (Mình làm dài để dễ hiểu thôi, nếu bạn không thích có thể đoạn trung gian)
Bài 1:
\(C=\frac{3}{7}\cdot\frac{9}{26}-\frac{1}{14}\cdot\frac{1}{13}-\frac{1}{7}\)
\(C=\frac{1}{7}\cdot3\cdot\frac{9}{26}-\frac{1}{7}\cdot\frac{1}{2}\cdot\frac{1}{13}-\frac{1}{7}\cdot1\)
\(C=\frac{1}{7}\cdot\frac{27}{26}-\frac{1}{26}\cdot\frac{1}{7}-\frac{1}{7}\cdot1\)
\(C=\frac{1}{7}\cdot\left(\frac{27}{26}-\frac{1}{26}-1\right)\)
\(C=\frac{1}{7}\cdot0=0\)
Bài 2:
\(\frac{5\cdot6+5\cdot7}{5\cdot8+20}=\frac{5\cdot\left(6+7\right)}{5\cdot\left(8+4\right)}=\frac{13}{12}\) và \(\frac{8\cdot9-4\cdot15}{12\cdot7-180}=\frac{4\cdot2\cdot9-4\cdot15}{4\cdot3\cdot7-4\cdot45}=\frac{4\cdot18-4\cdot15}{4\cdot21-4\cdot45}=\frac{4\cdot\left(18-15\right)}{4\cdot\left(21-45\right)}=\frac{3}{-24}=-\frac{1}{8}\)
Quy đồng mẫu số các phân số:
\(\frac{13}{12}=\frac{13\cdot2}{12\cdot2}=\frac{26}{24};-\frac{1}{8}=\frac{-1\cdot3}{8\cdot3}=-\frac{3}{24}\)
Đặt mỗi số số hạng của M là A;B
Ta quy đồng tử và mẫu của A bằng cách nhân cả tử và mẫu với 3.7.13;rồi dùng tính chất phân phối bỏ ngoặc ra.Tính toán bình thường.Khi đó A=1/7
Tương tự với B ta quy đồng lên với 11.17.19;bỏ ngoặc đi và tính bình thường ;B=2/7
=>M=1/7+2/7-136/7=3/7-136/7=-19
a)
\(\begin{array}{l}\frac{2}{3} + \frac{{ - 2}}{5} + \frac{{ - 5}}{6} - \frac{{13}}{{10}}\\ = \frac{2}{3} + \frac{{ - 5}}{6} + \frac{{ - 2}}{5} - \frac{{13}}{{10}}\\ = \left( {\frac{2}{3} + \frac{{ - 5}}{6}} \right) + \left( {\frac{{ - 2}}{5} - \frac{{13}}{{10}}} \right)\\ = \left( {\frac{4}{6} + \frac{{ - 5}}{6}} \right) + \left( {\frac{{ - 4}}{{10}} - \frac{{13}}{{10}}} \right)\\ = \frac{{ - 1}}{6} + \frac{{ - 17}}{{10}}\\ = \frac{{ - 5}}{{30}} + \frac{{ - 51}}{{30}}\\ = \frac{{ - 56}}{{30}}\\ = \frac{{ - 28}}{{15}}\end{array}\)
b)
\(\begin{array}{l}\frac{{ - 3}}{7}.\frac{{ - 1}}{9} + \frac{7}{{ - 18}}.\frac{{ - 3}}{7} + \frac{5}{6}.\frac{{ - 3}}{7}\\ = \frac{{ - 3}}{7}.\left( {\frac{{ - 1}}{9} + \frac{7}{{ - 18}} + \frac{5}{6}} \right)\\ = \frac{{ - 3}}{7}.\left( {\frac{{ - 2}}{{18}} + \frac{{ - 7}}{{18}} + \frac{{15}}{{18}}} \right)\\ = \frac{{ - 3}}{7}.\frac{{ 6}}{{18}}\\ = \frac{-1}{7}\end{array}\).
\(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)
\(A=49\frac{8}{23}-5\frac{7}{32}+14\frac{8}{23}\)
\(A= \left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}\)
\(A=\left[\left(49-14\right)-\left(\frac{8}{23}-\frac{8}{23}\right)\right]-5\frac{7}{32}\)
\(A=\left[35-0\right]-5\frac{7}{32}\)
\(A=35-5\frac{7}{32}\)
\(A=\frac{953}{32}\)
\(B=71\frac{38}{45}-\left(43\frac{38}{45}-1\frac{17}{57}\right)\)
\(B=71\frac{38}{45}-\frac{36377}{855}\)
\(B=\frac{1670}{57}\)
\(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right):\frac{4}{5}\)
\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]:\frac{4}{5}\)
\(C=\left[\frac{51}{8}:\frac{7}{12}\right]:\frac{4}{5}\)
\(C=\frac{153}{14}:\frac{4}{5}\)
\(C=\frac{765}{56}\)
\(D=\left[\left(\frac{10}{15}-\frac{2}{3}\right):\frac{1}{7}\right]\cdot0,15-\frac{1}{4}\)
\(D=\left[0:\frac{1}{7}\right]\cdot\frac{3}{20}-\frac{1}{4}\)
\(D=0\cdot\frac{3}{20}-\frac{1}{4}\)
\(D=0-\frac{1}{4}\)
\(D=-\frac{1}{4}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot2\frac{1}{2}-\left[\left(\frac{1}{2}+\frac{1}{3}\right):\frac{53}{90}\right]:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\left[\frac{5}{6}:\frac{53}{90}\right]:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{75}{53}:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{14}{9}-\frac{3}{2}\)
\(\)\(E=\frac{22}{45}\)
CHUC BAN HOC TOT >.<
\(\begin{array}{l}\left( {\frac{3}{5} + \frac{{ - 2}}{7}} \right) + \frac{{ - 1}}{5} = \left( {\frac{3}{5} + \frac{{ - 1}}{5}} \right) + \frac{{ - 2}}{7}\\ = \frac{2}{5} + \frac{{ - 2}}{7} = \frac{{14}}{{35}} + \frac{{ - 10}}{{35}} = \frac{4}{{35}}\end{array}\).
a) \(\frac{{ - 3}}{7}.\frac{2}{5} + \frac{2}{5}.\left( { - \frac{5}{{14}}} \right) - \frac{{18}}{{35}}\)
\(\begin{array}{l} = \frac{2}{5}.\left( {\frac{{ - 3}}{7} + \frac{{ - 5}}{{14}}} \right) - \frac{{18}}{{35}}\\ = \frac{2}{5}.\left( {\frac{{ - 6}}{{14}} + \frac{{ - 5}}{{14}}} \right) - \frac{{18}}{{35}}\\ = \frac{2}{5}.\frac{{ - 11}}{{14}} - \frac{{18}}{{35}} = \frac{{ - 11}}{{35}} - \frac{{18}}{{35}} = \frac{{ -29}}{{35}}\end{array}\)
b) \(\left( {\frac{2}{3} - \frac{5}{{11}} + \frac{1}{4}} \right):\left( {1 + \frac{5}{{12}} - \frac{7}{{11}}} \right)\)
\(\begin{array}{l} = \left( {\frac{{2.11.4}}{{3.11.4}} - \frac{{5.3.4}}{{11.3.4}} + \frac{{1.3.11}}{{4.3.11}}} \right):\left( {\frac{11.12}{11.12} + \frac{{5.11}}{{12.11}} - \frac{{7.12}}{{11.12}}} \right)\\ = \left( {\frac{{88 - 60 + 33}}{{121}}} \right):\left( { \frac{{121+55 - 84}}{{121}}} \right)\\ = \frac{{61}}{{121}}:\frac{{92}}{{121}} = \frac{{61}}{{121}}.\frac{{121}}{{92}}= \frac{{61}}{{92}}\end{array}\)
c) \(\left( {13,6 - 37,8} \right).\left( { - 3,2} \right)\)
\( = \left( { - 24,2} \right).\left( { - 3,2} \right) = 77,44\)
d) \(\left( { - 25,4} \right).\left( {18,5 + 43,6 - 16,8} \right):12,7\)
\(\begin{array}{l} = \left( { - 25,4} \right).\left( {62,1 - 16,8} \right):12,7\\ = \left( { - 25,4} \right).45,3:12,7\\ = \left( { - 25,4} \right):12,7.45,3\\ = (- 2).45,3 = - 90,6\end{array}\)
a: \(=\dfrac{2}{5}\cdot\left(-\dfrac{3}{7}-\dfrac{5}{14}\right)-\dfrac{18}{35}\)
\(=\dfrac{2}{5}\cdot\dfrac{-6-5}{14}-\dfrac{18}{35}\)
\(=\dfrac{2}{5}\cdot\dfrac{-11}{14}-\dfrac{18}{35}=-\dfrac{22}{70}-\dfrac{18}{35}=\dfrac{-58}{70}=-\dfrac{29}{35}\)
b: \(=\dfrac{88-60+33}{132}:\dfrac{132+55-84}{132}\)
\(=\dfrac{61}{132}\cdot\dfrac{132}{103}=\dfrac{61}{103}\)
c: \(=-24.2\cdot\left(-3.2\right)=24.2\cdot3.2=77.44\)
d: \(=\dfrac{-25.4}{12.7}\cdot45.3=-2\cdot45.3=-90.6\)
C = \(\frac{3}{7}\cdot\frac{9}{26}-\frac{1}{14}\cdot\frac{1}{13}-\frac{1}{7}\)
C = \(\frac{1}{7}\cdot3\cdot\frac{9}{26}-\frac{1}{7}\cdot\frac{1}{2}\cdot\frac{1}{13}-\frac{1}{7}\)
C = \(\frac{1}{7}\cdot\frac{27}{26}-\frac{1}{7}\cdot\frac{1}{26}-\frac{1}{7}\)
C = \(\frac{1}{7}\left(\frac{27}{26}-\frac{1}{26}-1\right)\)
C = \(\frac{1}{7}\left(1-1\right)\)
C = 0
Chúc bn học tốt
0