Tính giá trị biểu thức :
B = \(1\dfrac{13}{15}\). (0,5)2 . 3+ \(\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):1\dfrac{23}{24}\)
Tính hợp lí giá trị biểu thức :
C = \(49\dfrac{8}{23}-\left(5\dfrac{7}{32}+14\dfrac{8}{23}\right)\)
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a: \(A=49+\dfrac{8}{23}-14-\dfrac{8}{23}-5-\dfrac{7}{32}=30-\dfrac{7}{32}=\dfrac{953}{32}\)
b:
Sửa đề: \(B=71\dfrac{38}{45}-\left(43\dfrac{8}{45}-1\dfrac{17}{51}\right)\)
\(B=71+\dfrac{38}{45}-43-\dfrac{8}{45}+1+\dfrac{17}{51}\)
\(=71-43+1+1\)
=28+2=30
e: \(=\left(\dfrac{18}{37}+\dfrac{19}{37}\right)+\left(\dfrac{8}{24}+\dfrac{2}{3}\right)-\dfrac{47}{24}=2-\dfrac{47}{24}=\dfrac{1}{24}\)
f: \(=-8\cdot\dfrac{1}{2}:\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=-4:\dfrac{13}{12}=\dfrac{-48}{13}\)
g: \(=\dfrac{4}{25}+\dfrac{11}{2}\cdot\dfrac{5}{2}-\dfrac{8}{4}=\dfrac{4}{25}+\dfrac{55}{4}-2=\dfrac{1191}{100}\)
Gợi ý: Sử dụng tính chất phân phối của phép nhân đối với phép cộng để nhóm thừa số chung ra ngoài.
\(A=\left(3-\dfrac{1}{4}+\dfrac{3}{2}\right)-\left(5+\dfrac{1}{3}-\dfrac{5}{6}\right)-\left(6-\dfrac{7}{4}+\dfrac{2}{3}\right)\\ \Rightarrow A=3-\dfrac{1}{4}+\dfrac{3}{2}-5-\dfrac{1}{3}+\dfrac{5}{6}-6+\dfrac{7}{4}-\dfrac{2}{3}\\ \Rightarrow A=\left(3-5-6\right)-\left(\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{3}{2}+\dfrac{5}{6}-\dfrac{2}{3}\right)\\ \Rightarrow A=-8-\dfrac{3}{2}+\dfrac{5}{3}\\ =-\dfrac{47}{6}.\\ B=0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{41}\)
\(\Rightarrow B=\left(0,5+0,4\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{5}{7}-\dfrac{4}{35}\right)+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{9}{10}+\dfrac{1}{2}+\dfrac{3}{5}+\dfrac{1}{41}\\ \Rightarrow B=2+\dfrac{1}{41}\\ \Rightarrow B=\dfrac{83}{41}.\)
a) \(=\dfrac{157}{8}.\dfrac{12}{7}-\dfrac{61}{4}.\dfrac{12}{7}=\dfrac{12}{7}\left(\dfrac{157}{8}-\dfrac{61}{4}\right)=\dfrac{12}{7}.\dfrac{35}{8}=\dfrac{15}{2}\)
b) \(\dfrac{2}{5}.\dfrac{1}{3}-\dfrac{2}{15}\div\dfrac{1}{5}+\dfrac{3}{5}.\dfrac{1}{3}=\dfrac{1}{3}\left(\dfrac{2}{5}+\dfrac{3}{5}\right)-\dfrac{2}{15}.5=\dfrac{1}{3}.1-\dfrac{2}{3}=\dfrac{1}{3}-\dfrac{2}{3}=-\dfrac{1}{3}\)
c) \(=-\dfrac{80}{9}\)
a: \(=\dfrac{99}{100}:\left(\dfrac{3}{12}-\dfrac{1}{12}+\dfrac{4}{12}\right)-\dfrac{49}{25}\)
\(=\dfrac{99}{100}:\dfrac{1}{2}-\dfrac{49}{25}\)
\(=\dfrac{99}{50}-\dfrac{98}{50}=\dfrac{1}{50}\)
b: \(=\dfrac{13}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{32}{60}-1-\dfrac{19}{60}\right):\dfrac{47}{24}\)
\(=\dfrac{39}{60}+\dfrac{-19}{60}\cdot\dfrac{24}{47}\)
=459/940
\(1\dfrac{13}{15}.\left(0,5\right)^2.3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right):1\dfrac{23}{24}\\ =\dfrac{28}{15}.0,25.3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right):\dfrac{47}{24}\\ =\dfrac{7}{15}.3+\dfrac{-47}{60}:\dfrac{47}{24}\\ =\dfrac{7}{5}+\dfrac{-2}{5}\\ =\dfrac{5}{5}=1\)
a)
\(\dfrac{4}{9}+\dfrac{2}{9}-\dfrac{5}{18}\\ =\dfrac{6}{9}-\dfrac{5}{18}\\ =\dfrac{6\times2}{9\times2}-\dfrac{5}{18}\\ =\dfrac{12}{18}-\dfrac{5}{18}\\ =\dfrac{7}{18}\)
b)
\(2-\dfrac{3}{5}+\dfrac{8}{15}\\ =\dfrac{2\times15}{15}-\dfrac{3\times3}{5\times3}+\dfrac{8}{15}\\ =\dfrac{30-9+8}{15}\\ =\dfrac{29}{15}\)
c)
\(\dfrac{9}{8}-\left(\dfrac{11}{8}-\dfrac{19}{32}\right)\\ =\dfrac{9}{8}-\left(\dfrac{11\times4}{8\times4}-\dfrac{19}{32}\right)\\ =\dfrac{9}{8}-\left(\dfrac{44}{32}-\dfrac{19}{32}\right)\\ =\dfrac{9}{8}-\dfrac{25}{32}\\ =\dfrac{9\times4}{8\times4}-\dfrac{25}{32}\\ =\dfrac{36-25}{32}\\ =\dfrac{11}{32}\)
a) \(\dfrac{4}{9}+\dfrac{2}{9}-\dfrac{5}{18}=\dfrac{8}{18}+\dfrac{4}{18}-\dfrac{5}{18}=\dfrac{8+4-5}{18}=\dfrac{7}{18}\)
b) \(2-\dfrac{3}{5}+\dfrac{8}{15}=\dfrac{30}{15}-\dfrac{9}{15}+\dfrac{8}{15}=\dfrac{30-9+8}{15}=\dfrac{29}{15}\)
c) \(\dfrac{9}{8}-\left(\dfrac{11}{8}-\dfrac{19}{32}\right)=9-\left(\dfrac{44}{32}-\dfrac{19}{32}\right)=9-\dfrac{25}{32}=\dfrac{288}{32}-\dfrac{25}{32}=\dfrac{288-25}{32}=\dfrac{263}{32}\)
\(C = 49\dfrac{8}{23} - (5\dfrac{7}{32} + 14\dfrac{8}{23} )\)
\(C = 49\dfrac{8}{23} - 5\dfrac{7}{32} - 14\dfrac{8}{23}\)
\(C =( 49\dfrac{8}{23} - 4\dfrac{8}{23}) - 5\dfrac{7}{32}\)
\(C = 45 - 5\dfrac{7}{32}\)
\(C = \dfrac{1273}{32}\)