.

Áp dụng tính chất phân phối, rồi tính giá trị biểu thức.

Chẳng hạn,

Với , thì

ĐS. ; C = 0.

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Lời giải

Gợi ý: Sử dụng tính chất phân phối của phép nhân đối với phép cộng để nhóm các thừa số chung a; b; c ra ngoài, sau đó tính phép tính trong ngoặc rồi thay giá trị a; b; c đã cho vào.

a: \(A=\dfrac{19}{9}+\dfrac{4}{11}+\dfrac{2}{3}=\dfrac{209}{99}+\dfrac{44}{99}+\dfrac{66}{99}=\dfrac{319}{99}\)

b: \(B=\dfrac{-50}{60}+\dfrac{-35}{60}+\dfrac{12}{60}=\dfrac{-73}{60}\)

c: \(C=\dfrac{-27}{36}+\dfrac{132}{36}+\dfrac{10}{36}=\dfrac{115}{36}\)

d: \(D=\dfrac{-19}{3}+\dfrac{2}{3}-\dfrac{4}{5}=\dfrac{-17}{3}-\dfrac{4}{5}=\dfrac{-85-12}{15}=-\dfrac{97}{15}\)

a) Ta có: \(a\left(-\dfrac{3}{2}\right)+a\cdot\dfrac{1}{4}-a\cdot\dfrac{5}{6}\)

\(=a\left(-\dfrac{3}{2}+\dfrac{1}{4}-\dfrac{5}{6}\right)\)

\(=a\left(\dfrac{-18}{12}+\dfrac{3}{12}-\dfrac{10}{12}\right)\)

\(=a\cdot\dfrac{-25}{12}\)(1)

Thay \(a=\dfrac{3}{5}\) vào biểu thức (1), ta được:

\(\dfrac{3}{5}\cdot\dfrac{-25}{12}=\dfrac{-75}{60}=\dfrac{-5}{4}\)

35⋅−2512=−7560=−54

1. Tính hợp lí

a) \(0,7+\dfrac{-7}{19}-\left(-0,3\right)\)

\(=\dfrac{7}{10}+\dfrac{-7}{19}+\dfrac{3}{10}\)

\(=\left(\dfrac{7}{10}+\dfrac{3}{10}\right)+\dfrac{-7}{19}\)

\(=1+\dfrac{-7}{19}\)

\(=\dfrac{12}{19}\)

b) \(\dfrac{5}{3}.\left(-2,5\right):\dfrac{5}{6}\)

\(=\dfrac{5}{3}.\dfrac{-5}{2}.\dfrac{6}{5}\)

\(=\left(\dfrac{5}{3}.\dfrac{6}{5}\right).\dfrac{-5}{2}\)

\(=2.\dfrac{-5}{2}\)

\(=-5\)

c) \(0,6.\dfrac{-5}{17}-\dfrac{3}{5}.\dfrac{12}{17}\)

\(=\dfrac{3}{5}.\dfrac{-5}{17}-\dfrac{3}{5}.\dfrac{12}{17}\)

\(=\dfrac{3}{5}.\left(\dfrac{-5}{17}-\dfrac{12}{17}\right)\)

\(=\dfrac{3}{5}.-1\)

\(=\dfrac{-3}{5}\)

d) \(\dfrac{7}{4}.\dfrac{5}{2}-\dfrac{7}{4}.\dfrac{3}{2}\)

\(=\dfrac{7}{4}.\left(\dfrac{5}{2}-\dfrac{3}{2}\right)\)

\(=\dfrac{7}{4}.1\)

\(=\dfrac{7}{4}\)

Chúc bạn học tốt

a) 0,7+−719−(−0,3)0,7+19−7​−(−0,3)

=710+−719+310=107​+19−7​+103​

=(710+310)+−719=(107​+103​)+19−7​

=1+−719=1+19−7​

=1219=1912​

b) 53.(−2,5):5635​.(−2,5):65​

=53.−52.65=35​.2−5​.56​

=(53.65).−52=(35​.56​).2−5​

=2.−52=2.2−5​

$=-5$

c) 0,6.−517−35.12170,6.17−5​−53​.1712​

=35.−517−35.1217=53​.17−5​−53​.1712​

=35.(−517−1217)=53​.(17−5​−1712​)

=35.−1=53​.−1

=−35=5−3​

d) 74.52−74.3247​.25​−47​.23​

=74.(52−32)=47​.(25​−23​)

=74.1=47​.1

=74=47​

mình giúp rùi đó nhớ tick mình nha

Ta có:B=1\(\dfrac{6}{41}\)( \(\dfrac{12+\dfrac{12}{19}-\dfrac{12}{37}-\dfrac{12}{53}}{3+\dfrac{1}{3}-\dfrac{3}{37}-\dfrac{3}{53}}:\dfrac{4+\dfrac{4}{17}+\dfrac{4}{19}+\dfrac{4}{2006}}{5+\dfrac{5}{17}+\dfrac{5}{19}+\dfrac{5}{2006}}\) )

B=\(\dfrac{47}{41}\) [\(\dfrac{12\left(1+\dfrac{1}{19}-\dfrac{1}{37}-\dfrac{1}{53}\right)}{3\left(1+\dfrac{1}{3}-\dfrac{1}{37}-\dfrac{1}{53}\right)}:\dfrac{4\left(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2006}\right)}{5\left(1+\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{2006}\right)}\) B = \(\dfrac{47}{41}\) [ \(\dfrac{12}{3}:\dfrac{4}{5}\)]

B = \(\dfrac{47}{41}\)[ 4 . \(\dfrac{5}{4}\)]

B = \(\dfrac{47}{41}.5\)

B = \(\dfrac{235}{41}\)

Chúc bn hc tốt!!!

a) \(\frac{1}{3}+\frac{3}{4}-\frac{5}{6}\)

\(=\frac{1}{3}+\frac{3}{4}+\frac{-5}{6}\)

\(=\frac{1.4+3.3+\left(-5\right).2}{12}\)

\(=\frac{4+9+\left(-10\right)}{12}\)

\(=\frac{3}{12}\)

\(=\frac{1}{4}\)

b) \(\frac{-2}{3}+\frac{6}{5}\div\frac{2}{3}-\frac{2}{15}\text{ }\)

\(=\frac{-2}{3}+\frac{6}{5}\times\frac{3}{2}-\frac{2}{15}\)

\(=\frac{-2}{3}+\frac{18}{10}-\frac{2}{15}\)

\(=\frac{-2}{3}+\frac{9}{5}+\frac{-2}{15}\)

\(=\frac{\left(-2\right).5+9.3+\left(-2\right)}{15}\)

\(=\frac{\left(-10\right)+27+\left(-2\right)}{15}\)

\(=\frac{15}{15}\)

\(=1\)

c) \(\frac{-3}{7}+\frac{5}{13}+\frac{-4}{7}\text{ }\)

\(=\left(\frac{-3}{7}+\frac{-4}{7}\right)+\frac{5}{13}\)

\(=\frac{-7}{7}+\frac{5}{13}\)

\(=-1+\frac{5}{13}\)

\(=\frac{-13}{13}+\frac{5}{13}\)

\(=\frac{-8}{13}\)

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) \(=-\dfrac{80}{9}\)

d) \(=-\dfrac{1}{6}\)

a.=\(\dfrac{157}{8}:\dfrac{7}{12}-\dfrac{61}{4}:\dfrac{7}{12}=\dfrac{471}{14}-\dfrac{183}{7}=\dfrac{15}{2}\)

b.=\(\dfrac{2}{15}-\dfrac{2}{3}+\dfrac{1}{5}=-\dfrac{1}{3}\)

c.\(\left(\dfrac{10}{3}+2.5\right):\left(\dfrac{19}{6}-\dfrac{21}{5}\right)-\dfrac{11}{31}=\dfrac{35}{6}:\left(-\dfrac{31}{30}\right)-\dfrac{11}{31}=-\dfrac{175}{31}-\dfrac{11}{31}=-6\)

d.\(\left[6+\dfrac{1}{8}-\dfrac{1}{2}\right]:\dfrac{3}{12}=\dfrac{45}{8}:\dfrac{3}{12}=\dfrac{45}{2}\)