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7: \(=\dfrac{-12}{7}\cdot15+\dfrac{2}{7}\cdot\left(-15\right)+\left(-105\right)\cdot\dfrac{70-84+15}{105}\)
\(=\dfrac{-12\cdot15+2\cdot\left(-15\right)}{7}-1\)
\(=\dfrac{-15\cdot14}{7}-1=-15\cdot2-1=-31\)
8: \(=\dfrac{13}{29}\cdot\dfrac{29}{5}-\dfrac{13}{29}\cdot\dfrac{45}{8}-\dfrac{45}{8}\cdot\dfrac{9}{8}+\dfrac{45}{8}\cdot\dfrac{13}{29}\)
\(=-\dfrac{1193}{320}\)
\(a,\left(\dfrac{7}{20}+\dfrac{11}{15}-\dfrac{15}{12}\right):\left(\dfrac{11}{20}-\dfrac{26}{45}\right).\)
\(=\left(\dfrac{21}{60}+\dfrac{44}{60}-\dfrac{75}{60}\right):\left(\dfrac{99}{180}-\dfrac{104}{180}\right).\)
\(=\left(\dfrac{65}{60}-\dfrac{75}{60}\right):\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{10}{60}:\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{1}{6}:\left(-\dfrac{1}{36}\right).\)
\(=-\dfrac{1}{6}.\left(-36\right).\)
\(=\dfrac{-1.\left(-36\right)}{6}=\dfrac{36}{6}=6.\)
Vậy......
\(b,\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}:\dfrac{15-\dfrac{15}{11}+\dfrac{15}{121}}{16-\dfrac{16}{11}+\dfrac{16}{121}}.\)
\(=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}:\dfrac{15\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}{16\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}.\)
\(=\dfrac{5}{8}:\dfrac{15}{16}.\)
\(=\dfrac{5}{8}.\dfrac{16}{15}=\dfrac{5.16}{8.15}=\dfrac{1.2}{1.3}=\dfrac{2}{3}.\)
Vậy......
c, (làm tương tự câu b).
~ Học tốt!!! ~
3)\(\dfrac{-41}{32}\left(\dfrac{15}{8}-\dfrac{16}{41}\right)+\dfrac{15}{8}\left(\dfrac{41}{32}-\dfrac{8}{3}\right)\)
=\(\dfrac{-41}{32}.\dfrac{15}{8}-\dfrac{-41}{32}.\dfrac{16}{41}+\dfrac{15}{8}.\dfrac{41}{32}-\dfrac{15}{8}.\dfrac{8}{3}\)
=\(\left(\dfrac{-41}{32}.\dfrac{15}{8}+\dfrac{15}{8}.\dfrac{41}{32}\right)+\dfrac{-16}{41}.\dfrac{-41}{32}-\dfrac{15}{8}.\dfrac{8}{3}\)
=\(0+\dfrac{1}{2}-5=\dfrac{-9}{2}\)
4)\(\dfrac{13}{29}\left(\dfrac{29}{5}-\dfrac{45}{8}\right)-\dfrac{45}{8}\left(\dfrac{9}{8}-\dfrac{13}{29}\right)\)
=\(\dfrac{13}{29}.\dfrac{29}{5}-\dfrac{45}{8}.\dfrac{13}{29}-\dfrac{45}{8}.\dfrac{9}{8}-\dfrac{45}{8}.\dfrac{13}{29}\)
=\(\left(\dfrac{45}{8}.\dfrac{13}{29}-\dfrac{45}{8}.\dfrac{13}{29}\right)-\dfrac{13}{29}.\dfrac{29}{5}-\dfrac{45}{8}.\dfrac{9}{8}\)
=\(0-\dfrac{13}{5}-\dfrac{405}{64}=\dfrac{-2857}{320}\)
b) \(\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}=\dfrac{5}{8}\)
Vì không có thời gian nên mình chỉ làm câu khó nhất thôi, tick mình nhé
a) \(\dfrac{1}{2}.\left(\dfrac{2}{9}+\dfrac{3}{7}-\dfrac{5}{27}\right)\)
\(=\dfrac{1}{2}.\left(\dfrac{41}{63}-\dfrac{5}{27}\right)\)
\(=\dfrac{1}{2}.\dfrac{88}{189}\)
\(=\dfrac{44}{189}\)
b) \(\left(\dfrac{-5}{28}+1,75+\dfrac{8}{35}\right):\left(-3\dfrac{9}{20}\right)\)
\(=\left(\dfrac{11}{7}+\dfrac{8}{35}\right):\left(-3\dfrac{9}{20}\right)\)
\(=\dfrac{9}{5}:\left(-3\dfrac{9}{20}\right)\)
\(=\dfrac{9}{5}:\dfrac{-69}{20}\)
\(=\dfrac{-12}{23}\)
c) \(\dfrac{1}{3}.\dfrac{5}{7}-\dfrac{7}{27}.\dfrac{36}{14}\)
\(=\dfrac{5}{21}-\dfrac{7}{27}.\dfrac{36}{14}\)
\(=\dfrac{5}{21}-\dfrac{2}{3}\)
\(=\dfrac{-3}{7}\)
d) \(70,5-528:\dfrac{15}{2}\)
\(=70,5-\dfrac{352}{5}\)
\(=\dfrac{1}{10}\)
em không trả lời được câu hỏi của chị nhưng chị có thể giúp em đăng bài toán lên bằng cách nào không
1) \(\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=\dfrac{-7}{4}+\dfrac{1}{4}:\dfrac{1}{8}\)
\(\Leftrightarrow\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=\dfrac{-7}{4}+2\)
\(\Leftrightarrow\left(-1\dfrac{1}{5}+x\right):\left(-3\dfrac{3}{5}\right)=\dfrac{1}{4}\)
\(\Leftrightarrow-1\dfrac{1}{5}+x=\dfrac{1}{4}.\left(-3\dfrac{3}{5}\right)\)
\(\Leftrightarrow-1\dfrac{1}{5}+x=\dfrac{1}{4}.\left(-\dfrac{18}{5}\right)\)
\(\Leftrightarrow-1\dfrac{1}{5}+x=-\dfrac{9}{10}\)
\(\Leftrightarrow x=\left(-\dfrac{9}{10}\right)-\left(-1\dfrac{1}{5}\right)\)
\(\Leftrightarrow x=\dfrac{3}{10}\)
a) \(2\dfrac{3}{4}.\left(-0,4\right)-1\dfrac{3}{5}.2,75+\left(-1,2\right):\dfrac{4}{11}\)
= \(2,75.\left(-0,4\right)-\left(1,6\right).\left(2,75\right)+\left(-1,2\right).\dfrac{11}{4}\)
= \(2,75.\left(-0,4\right)-\left(1,6\right).\left(2,75\right)+\left(-1,2\right).\left(2,75\right)\)
= \(2,75.\left\{\left(-0,4\right)-\left(1,6\right)+\left(-1,2\right)\right\}\)
= \(2,75.\left(-3,2\right)\)
= \(-8,8\)
b) \(1,4.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):2\dfrac{1}{5}\)
= \(\dfrac{7}{5}.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):\dfrac{11}{5}\)
= \(\dfrac{7}{5}.\dfrac{15}{49}-\dfrac{22}{15}.\dfrac{5}{11}\)
= \(\dfrac{3}{7}-\dfrac{2}{3}\)
= \(-\dfrac{5}{21}\)
c) \(\left(-3,2\right).\dfrac{15}{64}+\left(0,8-2\dfrac{4}{15}\right):3\dfrac{2}{3}\)
= \(-\dfrac{16}{5}.\dfrac{15}{64}+\left(\dfrac{4}{5}-2\dfrac{4}{15}\right):\dfrac{11}{3}\)
= \(-\dfrac{16}{5}.\dfrac{15}{64}+\left(-\dfrac{22}{15}\right).\dfrac{3}{11}\)
= \(\left(-\dfrac{3}{4}\right)+\left(-\dfrac{2}{5}\right)\)
= \(-\dfrac{23}{20}\)
d) \(0,02.\dfrac{-25}{2}+\dfrac{3}{8}+\left(-2\dfrac{9}{20}\right).\dfrac{2}{7}\)
= \(\dfrac{1}{50}.\dfrac{-25}{2}+\dfrac{3}{8}+\left(-\dfrac{49}{20}\right).\dfrac{2}{7}\)
=\(\left(-\dfrac{1}{4}\right)+\dfrac{3}{8}+\left(-\dfrac{7}{10}\right)\)
= \(\dfrac{1}{8}+\left(-\dfrac{7}{10}=\right)\)
= \(-\dfrac{23}{40}\)
e) \(34\%:\dfrac{51}{16}-3\dfrac{7}{9}.6,5-\left(0,4\right)^2\)
= \(\dfrac{17}{50}.\dfrac{16}{51}-\dfrac{34}{9}.\dfrac{13}{2}-\dfrac{4}{25}\)
= \(\dfrac{8}{75}-\dfrac{221}{9}-\dfrac{4}{15}\)
= \(-\dfrac{5501}{225}\)
Mấy bài này bạn tự làm đi, chuyển vế tìm x gần giống cấp I mà.
b)\(\dfrac{-3}{5}.x=\dfrac{1}{4}+0,75\)
=>\(\dfrac{-3}{5}.x=1\)
=>\(x=1:\dfrac{-3}{5}\)
=>\(x=\dfrac{-5}{3}\)
Vậy \(x=\dfrac{-5}{3}\)
a) \(\dfrac{-5}{9}+\dfrac{8}{15}+\dfrac{-2}{11}+\dfrac{4}{-9}+\dfrac{7}{15}\)
=\(\left(\dfrac{-5}{9}+\dfrac{-4}{9}\right)+\left(\dfrac{8}{15}+\dfrac{7}{15}\right)+\dfrac{-2}{11}\)
=\(\left(-1\right)+1+\dfrac{-2}{11}\)
=\(\dfrac{-2}{11}\)
b) \(\left(\dfrac{-5}{12}+\dfrac{6}{11}\right)+\left(\dfrac{7}{17}+\dfrac{5}{11}+\dfrac{5}{12}\right)\)
=\(\dfrac{-5}{12}+\dfrac{6}{11}+\dfrac{7}{17}+\dfrac{5}{11}+\dfrac{5}{12}\)
=\(\left(\dfrac{-5}{12}+\dfrac{5}{12}\right)+\left(\dfrac{6}{11}+\dfrac{5}{11}\right)+\dfrac{7}{17}\)
=\(0+0+\dfrac{7}{17}\)
=\(\dfrac{7}{17}\)
c) A= \(49\dfrac{8}{23}-\left(5\dfrac{7}{32}+14\dfrac{8}{23}\right)\)
A=\(49\dfrac{8}{23}-5\dfrac{7}{32}-14\dfrac{8}{23}\)
A=\(\left(49\dfrac{8}{23}-14\dfrac{8}{23}\right)-5\dfrac{7}{32}\)
A=\(35-5\dfrac{7}{32}\)
A=\(35-\dfrac{167}{32}=\dfrac{953}{32}\)
d) C=\(\dfrac{-3}{7}.\dfrac{5}{9}+\dfrac{4}{9}.\dfrac{-3}{7}+2\dfrac{3}{7}\)
C=\(\dfrac{-3}{7}.\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{17}{7}\)
C=\(\dfrac{-3}{7}.1+\dfrac{17}{7}\)
C=\(\dfrac{-3}{7}+\dfrac{17}{7}=2\)
a, `(-5)/9+8/15+(-2)/11+4/(-9)+7/15`
`=-5/9+8/15-2/11-4/9+7/15`
`=(-5/9-4/9)+(8/15+7/15)-2/11`
`=-9/9+15/15-2/11`
`=-1+1-2/11`
`=-2/11`
b, `((-5)/12+6/11)+(7/17+5/11+5/12)`
`=-5/12+6/11+7/17+5/11+5/12`
`=(-5/12+5/12)+(6/11+5/11)+7/17`
`=0+11/11+7/17`
`=1+7/17`
`=17/17+7/17`
`=24/17`
c, `A=49 8/23 - (5 7/32 + 14 8/23)`
`A=49 8/23 - 5 7/32 - 14 8/23`
`A=(49 8/23 - 14 8/23)-5 7/32`
`A=35 - 167/32`
`A=953/32`
d, `C=(-3)/7.5/9+4/9.(-3)/7+2 3/7`
`C=-3/7 . 5/9-4/9 . 3/7+17/7`
`C=-3/7.(5/9+4/9)+17/7`
`C=-3/7 . 1+17/7`
`C=2`
13)\(\dfrac{45}{8}\left(\dfrac{4}{15}-\dfrac{7}{8}\right)+\dfrac{45}{8}\left(\dfrac{11}{15}+\dfrac{9}{8}\right)\)
=\(\dfrac{45}{8}\left(\dfrac{4}{15}-\dfrac{7}{8}+\dfrac{11}{15}+\dfrac{9}{8}\right)\)
=\(\dfrac{45}{8}\left[\left(\dfrac{4}{15}+\dfrac{11}{15}\right)-\left(\dfrac{7}{8}-\dfrac{9}{8}\right)\right]\)
=\(\dfrac{45}{8}.\dfrac{5}{4}\)=\(\dfrac{225}{32}\)
14)\(\dfrac{15}{7}\left(\dfrac{49}{35}-\dfrac{27}{8}\right)-\left(\dfrac{2}{5}-\dfrac{27}{4}\right):\dfrac{7}{15}\)
=\(\dfrac{15}{7}\left(\dfrac{49}{35}-\dfrac{27}{8}\right)-\left(\dfrac{2}{5}-\dfrac{27}{4}\right).\dfrac{15}{7}\)
=\(\dfrac{15}{7}\left[\left(\dfrac{49}{35}-\dfrac{27}{8}\right)-\left(\dfrac{2}{5}-\dfrac{27}{4}\right)\right]\)
=\(\dfrac{15}{7}\left(\dfrac{49}{35}-\dfrac{27}{8}-\dfrac{2}{5}+\dfrac{27}{4}\right)\)
=\(\dfrac{15}{7}.\dfrac{35}{8}\)=\(\dfrac{75}{8}\)