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bài 1b)
\(8\frac{1}{14}-6\frac37\)
C1:\(\frac{113}{14}-\frac{45}{7}\) =\(\frac{113}{14}-\frac{90}{14}=\frac{23}{14}\)
C2:\(8\frac{1}{14}-6\frac37=\left(8-6\right)+\left(\frac{1}{14}-\frac37\right)=2+\left(\frac{1}{14}-\frac{6}{14}\right)\)
\(=2+\frac{-5}{14}=\frac{28}{14}-\frac{5}{14}=\frac{23}{14}\)
bài 1 c)\(7-3\frac67\)
C1:\(\) \(7-3\frac67=7-\frac{27}{7}=\frac{49}{7}-\frac{27}{7}=\frac{22}{7}\)
C2:\(7-3\frac67=\left(7-3\right)-\frac67=4-\frac67=\frac{28}{7}-\frac67=\frac{22}{7}\)
\(\frac27\times5\frac14-\frac27\times3\frac14\)
=\(\frac27\times\left(5\frac14-3\frac14\right)\)
=\(\frac27\times\left(\left(5-3\right)+\left(\frac14-\frac15\right)\right)\)
=\(\frac27\times\left(2+0\right)\)
=\(\frac27\times2\)
=\(\frac47\)
a)\(\frac{11^4.6-11^5}{11^4-11^5}:\frac{9^8.3-9^9}{9^8.5+9^8.7}\)
\(=1.6:\frac{9^8.3-9^8.9}{9^8.\left(5+7\right)}\)
\(=6:\frac{9^8.\left(3-9\right)}{9^8.12}\)
\(=6:\frac{9^8.\left(-6\right)}{9^8.12}\)
\(=6:\left(-\frac{6}{12}\right)\)
\(=6:\left(-\frac{1}{2}\right)\)
\(=-12\)
b) 3/5 : ( -1/5-1/6)+3/5:(-1/3-16/15) ( mình chuyển về ps luôn )
=3/5: (-11/30) + 3/5 : (-7/5)
=3/5:[-11/30+(-7/5)]
=3/5:53/30
=18/53
c) (1/2-13/14):5/7-(-2/21+1/7):5/7
= -3/7:5/7-1/21:5/7
=(-3/7-1/21):5/7
=-10/21:5/7
=-2/3
câu b vá c mình làm tắt nha. chúc bạn học tốt
\(\frac{\frac{3}{5}+\frac{3}{7}+\frac{3}{11}}{\frac{7}{5}+1+\frac{7}{17}}\div\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(=\frac{\frac{3}{5}+\frac{3}{7}+\frac{3}{11}}{\frac{7}{5}+\frac{7}{7}+\frac{7}{17}}\div\frac{\frac{2}{6}-\frac{2}{8}+\frac{2}{10}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(=\frac{3\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{11}\right)}{7\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{11}\right)}\div\frac{2\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\)
\(=\frac{3}{7}\div\frac{2}{7}\)
\(=\frac{3}{7}\times\frac{7}{2}=\frac{3}{2}\)
1: Ta có: \(\frac{12}{7}:\frac{16}{5}\)
\(=\frac{12}{7}\cdot\frac{5}{16}=\frac{60}{112}=\frac{15}{28}\)
2: Ta có: \(\frac{-2}{7}:\frac{6}{11}\)
\(=\frac{-2}{7}\cdot\frac{11}{6}=\frac{-22}{42}=\frac{-11}{21}\)
3: Ta có: \(\frac{-1}{7}:\frac{-1}{5}\)
\(=\frac{-1}{7}\cdot\frac{5}{-1}=\frac{5}{7}\)
4: Ta có: \(\left(-2\right):\frac{6}{11}\)
\(=\frac{-2\cdot11}{6}=\frac{-22}{6}=\frac{-11}{3}\)
5: Ta có: \(\frac{-6}{11}:\left(-3\right)\)
\(=\frac{-6}{11}\cdot\frac{1}{-3}=\frac{-6}{-33}=\frac{2}{11}\)
a,Ta có \(\frac{\frac{1}{2}-\frac{1}{3}-\frac{1}{4}}{1-\frac{2}{3}-\frac{1}{2}}-\frac{\frac{3}{5}-\frac{3}{7}-\frac{3}{11}}{\frac{6}{5}-\frac{6}{7}-\frac{6}{11}}\)
\(=\frac{\frac{1}{2}-\frac{1}{3}-\frac{1}{4}}{2.\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)}-\frac{3.\left(\frac{1}{5}-\frac{1}{7}-\frac{1}{11}\right)}{6.\left(\frac{1}{5}-\frac{1}{7}-\frac{1}{11}\right)}\)
=\(\frac{1}{2}-\frac{3}{6}=\frac{1}{2}-\frac{1}{2}=0\)
Vậy giá trị biểu thức bằng 0
b, Mình không hiểu cho lắm ạ , nếu ko phiền xin xem lại đầu bài ạ
\(\frac{-2}{3}+\frac{4}{15}=\frac{-10}{15}+\frac{4}{15}=\frac{\left(-10\right)+4}{15}=\frac{-6}{15}\)
\(\frac{-3}{7}\cdot\frac{5}{11}+\frac{-3}{7}\cdot\frac{6}{11}+2\frac{3}{7}=\frac{-3}{7}\cdot\left(\frac{5}{11}+\frac{6}{11}\right)+2\frac{3}{7}=\frac{-3}{7}\cdot1+2\frac{3}{7}=\frac{-3}{7}+2\frac{3}{7}=2\)
a ) \(\frac{-2}{3}+\frac{4}{15}\)
= \(\frac{-10}{15}+\frac{4}{15}\)
= \(\frac{-2}{5}\)
B1a)\(11\frac34-\left(6\frac56-4\frac12\right)+1\frac23\)
=\(11\frac34-6\frac56+4\frac12+1\frac23\)
=\(\left(11-6+4+1\right)+\left(\frac34-\frac56+\frac12+\frac23\right)\)
=\(10+\left(\frac{9}{12}-\frac{10}{12}+\frac{6}{12}+\frac{8}{12}\right)\)
=\(10+\left(-\frac{1}{12}+\frac{6}{12}+\frac{8}{12}\right)\)
=10+\(\frac{13}{12}\)
=\(\frac{120}{12}+\frac{13}{12}\)
=\(\frac{133}{12}\)
b)\(2\frac{17}{20}-1\frac{11}{5}+6\frac{9}{20}:3\)
= \(\frac{57}{20}-\frac{16}{5}+\frac{129}{20}\times\frac13\)
=\(\frac{57}{20}-\frac{16}{5}+\frac{129}{60}\)
=\(\frac{171}{60}-\frac{192}{60}+\frac{129}{60}\)
=\(\frac{108}{60}\)
=\(\frac95\)
\(\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{\frac{6}{5}+\frac{6}{7}-\frac{6}{11}}=\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{2.\left(\frac{3}{5}+\frac{3}{7}-\frac{3}{11}\right)}=\frac{1}{2}\)