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a) \(\left(\frac{-3}{4}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\)
= \(\left(-\frac{3}{4}+\frac{2}{5}+\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\)
= \(0:\frac{3}{7}\)
= \(0\)
b) \(\frac{2}{8}:\left(\frac{2}{9}-\frac{1}{18}\right)+\frac{7}{8}:\left(\frac{1}{36}-\frac{5}{12}\right)\)
= \(\frac{1}{4}:\frac{1}{6}+\frac{7}{8}:\frac{-7}{18}\)
=\(\frac{1}{4}.6+\frac{7}{8}.\frac{-18}{7}\)
= \(\frac{3}{2}-\frac{3}{4}\)
= \(\frac{3}{4}\)
Bài 2
a. \(-1\frac{2}{3}-|2x-1|:\frac{3}{5}=-2\)
\(|2x-1|:\frac{3}{5}=\frac{5}{3}-2\)
\(|2x-1|:\frac{3}{5}=-\frac{1}{3}\)
\(|2x-1|=-\frac{1}{5}\)
Vì giá trị tuyệt đối luôn \(\ge0\)với mọi x
mà \(-\frac{1}{5}< 0\)
=> \(x\in\varnothing\)
a) \(\left(\frac{-3}{2}+\frac{2}{5}\right):\frac{3}{7}+\left(\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\)
= \(\left(\frac{-3}{2}+\frac{2}{5}+\frac{3}{5}+\frac{-1}{4}\right):\frac{3}{7}\)
= \(\frac{-3}{4}:\frac{3}{7}\)
= \(\frac{-7}{4}\)
b) \(\frac{7}{8}:\left(\frac{2}{9}-\frac{1}{18}\right)+\frac{7}{8}:\left(\frac{1}{36}-\frac{5}{12}\right)\)
\(=\frac{7}{8}:\frac{1}{6}+\frac{7}{8}:\left(\frac{-7}{18}\right)\)
\(=\frac{7}{8}.6+\frac{7}{8}.\left(\frac{-18}{7}\right)\)
\(=\frac{7}{8}.\left(6+\frac{-18}{7}\right)\)
\(=\frac{7}{8}.\frac{24}{7}\)
\(=3\)
a. \(\left(\frac{2}{3}\right)^3-\left(\frac{3}{4}\right)^2.\left(-1\right)^5=\frac{8}{27}-\frac{9}{16}.\left(-1\right)=\frac{8}{27}+\frac{9}{16}=\frac{371}{432}\)
b. \(12:\left(\frac{3}{4}-\frac{5}{6}\right)^2=12:\left(-\frac{1}{12}\right)^2=12:\frac{1}{144}=12.144=1728\)
c. \(\frac{7}{22}:\frac{3}{11}+\frac{7}{22}:\frac{4}{11}=\frac{7}{22}.\frac{11}{3}+\frac{7}{22}.\frac{11}{4}=\frac{7}{22}\left(\frac{11}{3}+\frac{11}{4}\right)\)
\(=\frac{7}{22}.\frac{77}{12}=\frac{49}{24}\)
d. \(\frac{12}{35}\left(\frac{7}{4}+\frac{13}{4}\right)-\frac{1}{3}=\frac{12}{35}.5-\frac{1}{3}=\frac{12}{7}-\frac{1}{3}=\frac{29}{21}\)
\(A=49\frac{8}{23}-\left(5\frac{7}{32}+14\frac{8}{23}\right)\)
\(A=49\frac{8}{23}-5\frac{7}{32}+14\frac{8}{23}\)
\(A= \left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}\)
\(A=\left[\left(49-14\right)-\left(\frac{8}{23}-\frac{8}{23}\right)\right]-5\frac{7}{32}\)
\(A=\left[35-0\right]-5\frac{7}{32}\)
\(A=35-5\frac{7}{32}\)
\(A=\frac{953}{32}\)
\(B=71\frac{38}{45}-\left(43\frac{38}{45}-1\frac{17}{57}\right)\)
\(B=71\frac{38}{45}-\frac{36377}{855}\)
\(B=\frac{1670}{57}\)
\(C=\left(19\frac{5}{8}:\frac{7}{12}-13\frac{1}{4}:\frac{7}{12}\right):\frac{4}{5}\)
\(C=\left[\left(19\frac{5}{8}-13\frac{1}{4}\right):\frac{7}{12}\right]:\frac{4}{5}\)
\(C=\left[\frac{51}{8}:\frac{7}{12}\right]:\frac{4}{5}\)
\(C=\frac{153}{14}:\frac{4}{5}\)
\(C=\frac{765}{56}\)
\(D=\left[\left(\frac{10}{15}-\frac{2}{3}\right):\frac{1}{7}\right]\cdot0,15-\frac{1}{4}\)
\(D=\left[0:\frac{1}{7}\right]\cdot\frac{3}{20}-\frac{1}{4}\)
\(D=0\cdot\frac{3}{20}-\frac{1}{4}\)
\(D=0-\frac{1}{4}\)
\(D=-\frac{1}{4}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot2\frac{1}{2}-\left[\left(\frac{1}{2}+\frac{1}{3}\right):\frac{53}{90}\right]:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\left[\frac{5}{6}:\frac{53}{90}\right]:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{28}{45}\cdot\frac{5}{2}-\frac{75}{53}:\frac{50}{53}\)
\(E=\frac{13}{30}+\frac{14}{9}-\frac{3}{2}\)
\(\)\(E=\frac{22}{45}\)
CHUC BAN HOC TOT >.<
Vì A:B=\(\frac{7}{12}\)\(\frac{7}{12}\)
\(\Rightarrow\)\(\frac{A}{B}\)= \(\frac{7}{12}\)
\(\Rightarrow\)A=7;B=12