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\(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)
\(A=49\frac{8}{23}-5\frac{7}{32}+14\frac{8}{23}\)
\(A= \left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}\)
\(A=\left[\left(49-14\right)-\left(\frac{8}{23}-\frac{8}{23}\right)\right]-5\frac{7}{32}\)
\(A=\left[35-0\right]-5\frac{7}{32}\)
\(A=35-5\frac{7}{32}\)
\(A=\frac{953}{32}\)
\(B=71\frac{38}{45}-\left(43\frac{38}{45}-1\frac{17}{57}\right)\)
\(B=71\frac{38}{45}-\frac{36377}{855}\)
\(B=\frac{1670}{57}\)
\(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right):\frac{4}{5}\)
\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]:\frac{4}{5}\)
\(C=\left[\frac{51}{8}:\frac{7}{12}\right]:\frac{4}{5}\)
\(C=\frac{153}{14}:\frac{4}{5}\)
\(C=\frac{765}{56}\)
\(D=\left[\left(\frac{10}{15}-\frac{2}{3}\right):\frac{1}{7}\right]\cdot0,15-\frac{1}{4}\)
\(D=\left[0:\frac{1}{7}\right]\cdot\frac{3}{20}-\frac{1}{4}\)
\(D=0\cdot\frac{3}{20}-\frac{1}{4}\)
\(D=0-\frac{1}{4}\)
\(D=-\frac{1}{4}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot2\frac{1}{2}-\left[\left(\frac{1}{2}+\frac{1}{3}\right):\frac{53}{90}\right]:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\left[\frac{5}{6}:\frac{53}{90}\right]:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{75}{53}:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{14}{9}-\frac{3}{2}\)
\(\)\(E=\frac{22}{45}\)
CHUC BAN HOC TOT >.<
\(A=\frac{\left(23\frac{11}{15}-26\frac{13}{20}\right)}{12^2+5^2}\cdot\frac{1-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}}{3^2.13.2-13.5}-\frac{19}{37}\)
\(A=\frac{\left(23+\frac{11}{15}-26+\frac{13}{20}\right)}{144+25}\cdot\frac{1-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}}{9.13.2-13.5}-\frac{19}{37}\)
\(A=\frac{\left(23+26+\frac{11}{15}-\frac{13}{20}\right)}{169}\cdot\frac{1-\left(\frac{1}{5}-\frac{1}{6}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-\left(\frac{1}{7}-\frac{1}{8}\right)}{13.\left(9.2-5\right)}-\frac{19}{37}\)
\(A=\frac{49+\frac{44}{60}-\frac{39}{60}}{169}\cdot\frac{1-\frac{1}{5}+\frac{1}{6}-\frac{1}{6}+\frac{1}{7}-\frac{1}{7}+\frac{1}{8}}{13.13}-\frac{19}{37}\)
\(A=\frac{49+\frac{1}{20}}{169}\cdot\frac{1-\frac{1}{5}+\frac{1}{8}}{169}-\frac{19}{37}\)
\(A=\frac{49\frac{1}{20}}{169}\cdot\frac{\frac{4}{5}+\frac{5}{40}}{169}-\frac{19}{37}\)
\(A=\frac{981}{169}\cdot\frac{\frac{32}{40}+\frac{5}{40}}{169}-\frac{19}{37}\)
\(A=\frac{981}{169}\cdot\frac{\frac{37}{40}}{169}-\frac{19}{37}\)
\(A=\frac{981.\frac{37}{40}}{169^2}-\frac{19}{37}\)
\(A=\frac{\frac{36297}{40}}{28561}-\frac{19}{37}\)
\(A=\frac{907,425}{28561}-\frac{19}{37}\)
\(A=\frac{33574,725}{1056757}-\frac{542659}{1056757}\)
\(A=\frac{-509084,275}{1056757}=-0,04604282...\)
Mik chỉ làm đc thế này thôi, ôn thi học kì II tốt nha bạn!
\(a.=\left(\frac{153}{37}-\frac{19}{5}+\frac{199}{23}\right)-\left(\frac{116}{37}-\frac{188}{29}\right)\)
\(=\frac{153}{37}-\frac{19}{5}+\frac{199}{23}-\frac{116}{37}+\frac{188}{29}\)
\(=\frac{37}{37}-\frac{19}{5}+\frac{199}{23}+\frac{188}{29}\)tự giải tiếp ^^
\(b.=\frac{8}{3}.\frac{-15}{4}.\frac{4}{5}\)
\(=\frac{8.\left(-15\right).4}{3.4.5}\)
\(=\frac{-480}{60}=-8\)
\(\left(\frac{-3}{42}+\frac{10}{-21}-\frac{-9}{14}\right).\left(\frac{-7}{5}\right).\)
\(=\left(\frac{-1}{14}-\frac{10}{21}+\frac{9}{14}\right).\left(\frac{-7}{5}\right)\)
\(\left(\frac{12}{21}-\frac{10}{21}\right).\left(\frac{-7}{5}\right)=\frac{2.\left(-7\right)}{21.5}=\frac{-2}{15}\)
\(\left(\frac{2}{9}+\frac{7}{9}\right)+\left(\frac{1}{5}+\frac{4}{5}\right)=1+1=2\)\(2\)
\(\frac{19}{37}+1-\frac{19}{37}=\left(\frac{19}{37}-\frac{19}{37}\right)+1=1\)
\(\left(4\frac{46}{65}+x\right).1\frac{1}{12}=5,75\)
\(\Rightarrow\frac{306}{65}+x.\frac{13}{12}=\frac{23}{4}\)
\(\Rightarrow\frac{51}{10}+\frac{13}{12}x=\frac{23}{4}\)
\(\Rightarrow306x=65x=345\)
\(\Rightarrow65x=39\)
\(\Rightarrow x=\frac{3}{5}\)
b, \(\frac{5}{4}-\left(\frac{3}{2}x+0,5\right)=1\frac{1}{4}\)
\(\Rightarrow\frac{5}{4}-\frac{3}{2}x-0,5=\frac{5}{4}\)
\(\Rightarrow\frac{5}{4}-\frac{3}{2}x-\frac{1}{2}=\frac{5}{4}\)
\(\Rightarrow\frac{3}{4}-\frac{3}{2}x=\frac{5}{4}\)
\(\Rightarrow3-6x=5\)
\(\Rightarrow-6x=2\)
\(\Rightarrow x=-\frac{1}{3}\)
Phần b) chị sai nhé ! Dấu [ ] là phần nguyên nâng cao của lớp 6 nhé.