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Bài 1:
a) \(\left(\dfrac{3}{8}+\dfrac{-3}{4}+\dfrac{7}{12}\right):\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\left(\dfrac{9}{24}+\dfrac{-18}{24}+\dfrac{14}{24}\right):\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\dfrac{5}{24}:\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\dfrac{5}{24}.\dfrac{6}{5}+\dfrac{1}{2}\)
\(=\dfrac{1}{4}+\dfrac{1}{2}\)
\(=\dfrac{1}{4}+\dfrac{2}{4}\)
\(=\dfrac{3}{4}\)
b) \(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}-\dfrac{4}{5}\right)\)
\(=\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}+\dfrac{4}{5}\)
\(=\left(\dfrac{1}{2}+\dfrac{4}{5}\right)+\left(\dfrac{3}{4}-\dfrac{3}{4}\right)\)
\(=\dfrac{1}{2}+\dfrac{4}{5}\)
\(=\dfrac{5}{10}+\dfrac{8}{10}\)
\(=\dfrac{9}{5}\)
c) \(6\dfrac{5}{12}:2\dfrac{3}{4}+11\dfrac{1}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{77}{12}:\dfrac{11}{4}+\dfrac{42}{4}.\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{77}{12}.\dfrac{4}{11}+\dfrac{42}{4}.\left(\dfrac{5}{15}+\dfrac{3}{15}\right)\)
\(=\dfrac{7}{3}+\dfrac{42}{4}.\dfrac{8}{15}\)
\(=\dfrac{7}{3}+\dfrac{14.2}{1.3}\)
\(=\dfrac{7}{3}+\dfrac{28}{3}\)
\(=\dfrac{35}{3}\)
d) \(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).1\dfrac{1}{3}-\dfrac{2}{7}.\left(3,5\right)^2\)
\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right).\dfrac{4}{3}-\dfrac{2}{7}.12\dfrac{1}{4}\)
\(=\dfrac{1}{8}.\dfrac{4}{3}-\dfrac{2}{7}.\dfrac{49}{4}\)
\(=\dfrac{1}{6}-\dfrac{7}{2}\)
\(=\dfrac{1}{6}-\dfrac{21}{6}\)
\(=\dfrac{-10}{3}\)
e) \(\left(\dfrac{3}{5}+0,415-\dfrac{3}{200}\right).2\dfrac{2}{3}.0,25\)
\(=\left(\dfrac{3}{5}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)
\(=\left(\dfrac{120}{200}+\dfrac{83}{200}-\dfrac{3}{200}\right).\dfrac{8}{3}.\dfrac{1}{4}\)
\(=1.\dfrac{8}{3}.\dfrac{1}{4}\)
\(=\dfrac{2}{3}\)
f) \(\dfrac{5}{16}:0,125-\left(2\dfrac{1}{4}-0,6\right).\dfrac{10}{11}\)
\(=\dfrac{5}{16}:\dfrac{1}{8}-\left(\dfrac{9}{4}-\dfrac{3}{5}\right).\dfrac{10}{11}\)
\(=\dfrac{5}{16}.\dfrac{8}{1}-\left(\dfrac{45}{20}-\dfrac{12}{20}\right).\dfrac{10}{11}\)
\(=\dfrac{5}{2}-\dfrac{33}{20}.\dfrac{10}{11}\)
\(=\dfrac{5}{2}-\dfrac{3}{2}\)
\(=\dfrac{2}{2}=1\)
g) \(0,25:\left(10,3-9,8\right)-\dfrac{3}{4}\)
\(=\dfrac{1}{4}:\dfrac{1}{2}-\dfrac{3}{4}\)
\(=\dfrac{1}{4}.\dfrac{2}{1}-\dfrac{3}{4}\)
\(=\dfrac{1}{2}-\dfrac{3}{4}\)
\(=\dfrac{2}{4}-\dfrac{3}{4}\)
\(=\dfrac{-1}{4}\)
h) \(1\dfrac{13}{15}.0,75-\left(\dfrac{11}{20}+20\%\right):\dfrac{7}{3}\)
\(=\dfrac{28}{15}.\dfrac{3}{4}-\left(\dfrac{11}{20}+\dfrac{1}{5}\right):\dfrac{7}{3}\)
\(=\dfrac{7}{5}-\left(\dfrac{11}{20}+\dfrac{4}{20}\right):\dfrac{7}{3}\)
\(=\dfrac{7}{5}-\dfrac{3}{4}:\dfrac{7}{3}\)
\(=\dfrac{7}{5}-\dfrac{9}{28}\)
\(=\dfrac{196}{140}-\dfrac{45}{140}\)
\(=\dfrac{151}{140}\)
i) \(\dfrac{\left(\dfrac{1}{2-0,75}\right).\left(0,2-\dfrac{2}{5}\right)}{\dfrac{5}{9}-1\dfrac{1}{12}}\)
\(=\dfrac{\left(\dfrac{1}{1,25}\right).\left(\dfrac{1}{5}-\dfrac{2}{5}\right)}{\dfrac{5}{9}-\dfrac{13}{12}}\)
\(=\dfrac{\dfrac{1}{1,25}.\dfrac{-1}{5}}{\dfrac{20}{36}-\dfrac{39}{36}}\)
\(=\dfrac{\dfrac{-1}{6,25}}{\dfrac{-19}{36}}\)
k) \(\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{1}{14}}{-1-\dfrac{3}{7}+\dfrac{3}{28}}\)
\(=\dfrac{\dfrac{2}{3}+\dfrac{2}{7}-\dfrac{2}{28}}{-\dfrac{3}{3}-\dfrac{3}{7}+\dfrac{3}{28}}\)
\(=\dfrac{2\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}{\left(-3\right)\left(\dfrac{1}{3}+\dfrac{1}{7}-\dfrac{1}{28}\right)}\)
\(=-\dfrac{2}{3}\)
\(A=0,7.2\dfrac{2}{3}.20.0,375.\dfrac{5}{28}\)
\(A=\dfrac{7}{10}.\dfrac{8}{3}.20.\dfrac{3}{8}.\dfrac{5}{28}\)
\(A=\left(\dfrac{7}{10}.\dfrac{5}{28}\right).\left(\dfrac{8}{3}.\dfrac{3}{8}\right).20\)
\(A=\dfrac{1}{8}.1.20\)
\(A=\dfrac{20}{8}=\dfrac{5}{2}\)
\(B=\left(9\dfrac{30303}{80808}+7\dfrac{303030}{484848}\right)+4,03\)
\(B=\left(9\dfrac{3}{8}+7\dfrac{5}{8}\right)+4,03\)
\(B=\left[\left(9+7\right)+\left(\dfrac{3}{8}+\dfrac{5}{8}\right)\right]+4,03\)
\(B=\left(16+1\right)+4,03\)
\(B=17+4,03\)
\(B=21,03\)
\(C=\left(9,75.21\dfrac{3}{7}+\dfrac{39}{4}.18\dfrac{4}{7}\right).\dfrac{15}{78}\)
\(C=\left(\dfrac{39}{4}.\dfrac{150}{7}+\dfrac{39}{4}.\dfrac{130}{7}\right).\dfrac{15}{78}\)
\(C=\dfrac{39}{4}.\left(\dfrac{150}{7}+\dfrac{130}{7}\right).\dfrac{15}{78}\)
\(C=\dfrac{39}{4}.40.\dfrac{15}{78}\)
\(C=390.\dfrac{15}{78}\)
\(C=75\)
a. \(\dfrac{-3}{5}\)
b. \(\dfrac{-2}{3}\) c. \(\dfrac{4}{39}\) d. \(\dfrac{26}{45}\)
a: \(=\dfrac{5\cdot\left(8-6\right)}{10}=\dfrac{5\cdot2}{10}=1\)
b: \(\dfrac{\left(-4\right)^2}{5}=\dfrac{16}{5}\)
\(B=\dfrac{3}{7}-\dfrac{1}{5}-\dfrac{3}{7}=-\dfrac{1}{5}\)
c: \(C=\left(6-2.8\right)\cdot\dfrac{25}{8}-\dfrac{8}{5}\cdot4\)
\(=\dfrac{16}{5}\cdot\dfrac{25}{8}-\dfrac{32}{5}\)
\(=5\cdot2-\dfrac{32}{5}=10-\dfrac{32}{5}=\dfrac{18}{5}\)
d: \(D=\left(\dfrac{-5}{24}+\dfrac{18}{24}+\dfrac{14}{24}\right):\dfrac{-17}{8}\)
\(=\dfrac{27}{24}\cdot\dfrac{-8}{17}=\dfrac{-9}{8}\cdot\dfrac{8}{17}=\dfrac{-9}{17}\)
a) A = 3/7
b) B = 73/13
c) C = 37/7
d) D = 12
ba câu a) ,b) ,c) bn đổi ra hỗn số giúp mk nha
tick cho tớ nha
a. \(\dfrac{5}{7}+\dfrac{9}{23}+\dfrac{-12}{7}+\dfrac{14}{23}\)
= \(\left(\dfrac{5}{7}+\dfrac{-12}{7}\right)+\left(\dfrac{9}{23}+\dfrac{14}{23}\right)\)
= \(\left(-1\right)+1=0\)
b. \(\left(7-2\dfrac{3}{5}\right).2\dfrac{4}{33}-2\dfrac{3}{5}:\dfrac{1}{2}+\dfrac{7}{5}\)
= \(\dfrac{22}{5}.2\dfrac{4}{33}-\dfrac{26}{5}+\dfrac{7}{5}\)
= \(\dfrac{28}{3}-\dfrac{33}{5}=\dfrac{41}{15}\)
c\(\dfrac{-7}{9}:\dfrac{8}{15}+\dfrac{-7}{9}.\dfrac{7}{15}+5\dfrac{7}{9}\)
= \(\dfrac{-56}{135}+\dfrac{-49}{135}+5\dfrac{7}{9}\)
= \(\dfrac{-7}{9}+5\dfrac{7}{9}=5\)
d. \(\left(16\dfrac{3}{8}-19\dfrac{3}{4}\right)-\left(12\dfrac{3}{8}-17\dfrac{3}{4}\right)\)
= \(16\dfrac{3}{8}-19\dfrac{3}{4}-12\dfrac{3}{8}+17\dfrac{3}{4}\)
= \(\left(16\dfrac{3}{8}-12\dfrac{3}{8}\right)-\left(19\dfrac{3}{4}-17\dfrac{3}{4}\right)\)
=\(4+2=6\)
a)\(\dfrac{5}{7}+\dfrac{9}{23}+\dfrac{-12}{7}+\dfrac{14}{23}\)
=\(\left(\dfrac{5}{7}-\dfrac{12}{7}\right)+\left(\dfrac{9}{23}+\dfrac{14}{23}\right)=-1+1=0\)
b)\(\left(7-2\dfrac{3}{5}\right).2\dfrac{4}{33}-2\dfrac{3}{5}:\dfrac{1}{2}+\dfrac{7}{5}\)
=\(\dfrac{22}{5}.\dfrac{70}{33}-\dfrac{13}{5}.2+\dfrac{7}{5}\)
=\(\dfrac{28}{3}-\dfrac{26}{5}+\dfrac{7}{5}=\dfrac{83}{15}\)
Các câu sau tương tự
d, Vì B=10^1993+1/10^1992+1 > 1 =>10^1993+1/10^1992+1>10^1993+1+9/10^1992+1+9 = 10^1993+10/10^1992+10= 10. (10^1992+1)/10. (10^1991+1) = 10^1992+1/10^1991+1=A Vậy A=B
cau d B>1 ta co tinh chat (\(\dfrac{a}{b}>\dfrac{a+m}{b+m}\) ) B> \(\dfrac{10^{1993}+1+9}{10^{1992}+1+9}\)\(=\dfrac{10^{1993}+10}{10^{1992}+10}\)=\(\dfrac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}\)=\(\dfrac{10^{1992}+1}{10^{1991}+1}\)=A
Suy ra B>A(chuc ban hoc goi nhe)
Bài 1 : Rút gọn các phân số sau đến tối giản :
a) \(\dfrac{3.21}{14.15}=\dfrac{3.3.7}{2.7.3.5}=\dfrac{1.3.1}{2.1.1.5}=\dfrac{3}{10}\)
b) \(\dfrac{49+49.7}{49}=\dfrac{49\left(1+7\right)}{49}=\dfrac{49.8}{49}=\dfrac{1.8}{1}=\dfrac{8}{1}=8\)
\(a)\dfrac{3}{4}-\dfrac{-5}{2}-\dfrac{7}{-24}\)
\(=\dfrac{13}{4}-\dfrac{7}{-24}\)
\(=\dfrac{85}{24}\)
\(b)\dfrac{4}{7}+\dfrac{-5}{8}-\dfrac{3}{28}\)
\(=\dfrac{-3}{56}-\dfrac{3}{28}\)
\(=\dfrac{-9}{56}\)
\(c)\dfrac{7}{36}-\dfrac{8}{-9}+\dfrac{-2}{3}\)
\(=\dfrac{13}{12}\)\(+\dfrac{-2}{3}\)
\(=\dfrac{5}{12}\)
\(d)\dfrac{-1}{2}+\dfrac{3}{7}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}\)
\(=\dfrac{-1}{14}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}\)
\(=\dfrac{-23}{126}+\dfrac{-7}{18}+\dfrac{4}{7}\)
\(=\dfrac{-4}{7}+\dfrac{4}{7}\)
\(=0\)
\(e)\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)
\(=\dfrac{-5}{56}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)
\(=\dfrac{83}{56}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)
\(=\dfrac{305}{168}+\dfrac{1}{7}+\dfrac{5}{-8}\)
\(=\dfrac{47}{24}+\dfrac{5}{-8}\)
\(=\dfrac{4}{3}\)
Bài 2 : Tính
a) \(\dfrac{3}{4}-\dfrac{-5}{2}-\dfrac{7}{-24}\)
\(=\dfrac{18}{24}-\dfrac{-60}{24}-\dfrac{-4}{24}\)
\(=\dfrac{18-\left(-60\right)-\left(-7\right)}{24}\)
\(=\dfrac{85}{24}\)
b) \(\dfrac{4}{7}+\dfrac{-5}{8}-\dfrac{3}{28}\)
\(=\dfrac{32}{56}+\dfrac{-35}{56}-\dfrac{6}{56}\)
\(=\dfrac{32+\left(-35\right)-6}{56}\)
\(=\dfrac{-9}{56}\)
c) \(\dfrac{7}{36}-\dfrac{8}{9}+\dfrac{-2}{3}\)
\(=\dfrac{7}{36}-\dfrac{32}{36}+\dfrac{-24}{36}\)
\(=\dfrac{7-32+\left(-24\right)}{36}\)
\(=\dfrac{-49}{36}\)
d) \(\dfrac{-1}{2}+\dfrac{3}{7}-\dfrac{1}{9}+\dfrac{-7}{18}+\dfrac{4}{7}\)
\(=\dfrac{-9}{18}+\dfrac{3}{7}-\dfrac{2}{18}+\dfrac{-7}{18}+\dfrac{4}{7}\)
\(=\left(\dfrac{-9}{18}+\dfrac{-7}{18}-\dfrac{2}{18}\right)+\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\)
\(=\left(-1\right)+1\)
\(=0\)
e) \(\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-8}\)
\(=\left(\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{11}{7}\right)+\left(\dfrac{-3}{8}+\dfrac{-5}{8}\right)+\dfrac{1}{3}\)
\(=2+\left(-1\right)+\dfrac{1}{3}\)
\(=1+\dfrac{1}{3}\)
\(=\dfrac{4}{3}\)
\(\dfrac{4\cdot5\cdot36}{35\cdot9\cdot2}=\dfrac{2^2\cdot5\cdot2^2\cdot3^2}{5\cdot7\cdot3^2\cdot2}=\dfrac{2^3}{7}=\dfrac{8}{7}\left(C\right)\)
C