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\(a.\dfrac{1}{3}+\dfrac{4}{5}-\left(\dfrac{4}{5}+\dfrac{5}{8}\right)\)
\(=\dfrac{1}{3}+\dfrac{4}{5}-\dfrac{4}{5}-\dfrac{5}{8}\)
\(=\dfrac{1}{3}+\left(\dfrac{4}{5}-\dfrac{4}{5}\right)-\dfrac{5}{8}\)
\(=\dfrac{1}{3}-\dfrac{5}{8}\)
=\(-\dfrac{7}{24}\)
\(b.8\dfrac{4}{9}-\left(4\dfrac{2}{7}+5\dfrac{4}{9}\right)\)
\(=8\dfrac{4}{9}-4\dfrac{2}{7}-5\dfrac{4}{9}\)
\(=\left(8\dfrac{4}{9}-5\dfrac{4}{9}\right)-4\dfrac{2}{7}\)
\(=\left(8+\dfrac{4}{9}-5-\dfrac{4}{9}\right)-4-\dfrac{2}{7}\)
\(=\left[8-5+\left(\dfrac{4}{9}-\dfrac{4}{9}\right)\right]-4-\dfrac{2}{7}\)
\(=3-4-\dfrac{2}{7}\)
\(=-1-\dfrac{2}{7}\)
\(=-\dfrac{9}{7}\)
\(c.\left(-\dfrac{5}{7}\right).\dfrac{2}{11}+\left(-\dfrac{5}{7}\right).\dfrac{9}{11}+1\dfrac{5}{7}\)
\(=\left(-\dfrac{5}{7}\right).\dfrac{2}{11}+\left(-\dfrac{5}{7}\right).\dfrac{9}{11}+1+\left(-\dfrac{5}{7}\right).\left(-1\right)\)
\(=\left(-\dfrac{5}{7}\right).\left[\dfrac{2}{11}+\dfrac{9}{11}+\left(-1\right)\right]+1\)
\(=\left(-\dfrac{5}{7}\right).0+1=1\)
a) \(6\dfrac{5}{7}-\left(1\dfrac{3}{4}+2\dfrac{5}{7}\right)\)
\(=6\dfrac{5}{7}-1\dfrac{3}{4}-2\dfrac{5}{7}\)
\(=\left(6\dfrac{5}{7}-2\dfrac{5}{7}\right)-1\dfrac{3}{4}\)
\(=4-1\dfrac{3}{4}\)
\(=3\dfrac{3}{4}\)
b) \(7\dfrac{5}{11}-\left(2\dfrac{3}{7}+3\dfrac{5}{11}\right)\)
\(=7\dfrac{5}{11}-2\dfrac{3}{7}-3\dfrac{5}{11}\)
\(=\left(7\dfrac{5}{11}-3\dfrac{5}{11}\right)-2\dfrac{3}{7}\)
\(=4-2\dfrac{3}{7}\)
\(=2\dfrac{3}{7}\)
\(-2\dfrac{1}{4}.\)\(\left(3\dfrac{5}{12}-1\dfrac{2}{9}\right)\)
=\(\dfrac{-9}{4}\).\(\left(\dfrac{41}{12}-\dfrac{11}{9}\right)\)
=\(\dfrac{-9}{4}.\dfrac{41}{12}-\dfrac{-9}{4}.\dfrac{11}{9}\)
=\(\dfrac{-123}{16}-\dfrac{-11}{4}\)
=\(\dfrac{-123}{16}-\dfrac{-44}{16}\)
=\(\dfrac{-79}{16}\)
\(\left(-25\%+0,75+\dfrac{7}{12}\right)\div\left(-2\dfrac{1}{8}\right)\)
=\(\left(\dfrac{-1}{4}+\dfrac{3}{4}+\dfrac{7}{12}\right)\div\left(\dfrac{-17}{8}\right)\)
=\(\left(\dfrac{-3}{12}+\dfrac{9}{12}+\dfrac{7}{12}\right).\dfrac{-8}{17}\)
=\(\dfrac{13}{12}.\dfrac{-8}{17}=\dfrac{-26}{51}\)
B= (7 \(\dfrac{4}{9}\) + 3\(\dfrac{8}{13}\)) - 5\(\dfrac{4}{9}\)
B=(\(\dfrac{67}{9}\)+ \(\dfrac{47}{13}\)) - \(\dfrac{49}{9}\)
B=\(\dfrac{116}{9}\)-\(\dfrac{47}{13}\)
B=\(\dfrac{116-47}{117}\)
B=\(\dfrac{23}{39}\)
tui cũng không biết đúng hông nữa
Đề bài : Áp dụng tính chất các phép tính và quy tắc dấu ngoặc để tính giá trị các biểu thức
a)
C1: \(2\dfrac{1}{4}+1\dfrac{1}{6}\)
\(=\dfrac{9}{4}+\dfrac{7}{6}\)
\(=\dfrac{27}{12}+\dfrac{14}{12}\)
\(=\dfrac{41}{12}=3\dfrac{5}{12}\)
C2: \(2\dfrac{1}{4}+1\dfrac{1}{6}\)
\(=2\dfrac{3}{12}+1\dfrac{2}{12}\)
\(=3\dfrac{5}{12}\)
Đáp án và hướng dẫn giải bài 109:
a) Cách 1
cách 2
b) Cách 1
cách 2
c, cách 1
cách 2
\(\dfrac{7}{12}+\dfrac{4}{9}+\dfrac{1}{3}+\dfrac{5}{12}+2=\dfrac{7+5}{12}+\dfrac{4+3}{9}+2=1+\dfrac{7}{9}+2=3+\dfrac{7}{9}=\dfrac{34}{9}\)
\(\dfrac{6}{5}.\dfrac{3}{8}.\dfrac{5}{8}.\dfrac{6}{5}-\dfrac{4}{5}=\dfrac{6.3.5.6}{5.8.8.5}-\dfrac{4}{5}=\dfrac{27}{80}-\dfrac{4}{5}=-\dfrac{37}{80}\)
a: \(=\dfrac{-3}{7}+\dfrac{-9}{35}-\dfrac{2}{5}\)
\(=\dfrac{-15-9-14}{35}=\dfrac{-38}{35}\)
b: \(=\left(\dfrac{15}{24}-\dfrac{7}{12}\right)\cdot\dfrac{-12}{7}\)
\(=\dfrac{15-14}{24}\cdot\dfrac{-12}{7}=\dfrac{1}{24}\cdot\dfrac{-12}{7}=\dfrac{-1}{14}\)
c: \(=\dfrac{7}{5}\cdot\dfrac{15}{19}\cdot\dfrac{-8}{15}+\dfrac{7}{15}\)
\(=\dfrac{-56}{95}+\dfrac{7}{15}\)
\(=\dfrac{-7}{57}\)
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
Mik làm nhé:
cách 1:
a, \(2\dfrac{1}{4}+1\dfrac{1}{6}\)
= \(\dfrac{9}{4}+\dfrac{7}{6}\)
= \(\dfrac{27}{12}+\dfrac{14}{12}\)
= \(\dfrac{27+14}{12}\)
= \(\dfrac{41}{12}=3\dfrac{5}{12}.\)
b, \(7\dfrac{1}{8}-5\dfrac{3}{4}\)
= \(\dfrac{57}{8}-\dfrac{23}{4}\)
=\(\dfrac{57}{8}-\dfrac{46}{8}\)
= \(\dfrac{57-46}{8}\)
= \(\dfrac{11}{8}=1\dfrac{3}{8}.\)
c, \(4-2\dfrac{6}{7}\)
= \(4-\dfrac{20}{7}\)
= \(\dfrac{28}{7}-\dfrac{20}{7}\)
= \(\dfrac{28-20}{7}\)
= \(\dfrac{8}{7}=1\dfrac{1}{7}.\)
CHÚC BN HỌC GIỎI!!! :) :) :)
Đừng quên bình luận nếu bài tớ sai nhé!!!
5, \(\dfrac{1}{7}.\dfrac{2}{5}+\dfrac{1}{7}.\dfrac{1}{5}+\dfrac{1}{7}.\dfrac{4}{5}\)
= \(\dfrac{1}{7}.\left(\dfrac{2}{5}+\dfrac{1}{5}+\dfrac{4}{5}\right)\)
= \(\dfrac{1}{7}\).1
=\(\dfrac{1}{7}\)
5) \(\dfrac{1}{7}.\dfrac{2}{5}+\dfrac{1}{7}.\dfrac{1}{5}+\dfrac{1}{7}.\dfrac{4}{5}\) = \(\dfrac{1}{7}.\left(\dfrac{2}{5}+\dfrac{1}{5}+\dfrac{4}{5}\right)\)
= \(\dfrac{1}{7}.\dfrac{7}{5}=\dfrac{1}{5}\)
\(\dfrac{2}{-5}-\dfrac{4}{-7}\)
\(=-\dfrac{2}{5}+\dfrac{4}{7}\)
\(=\dfrac{-2\cdot7+4\cdot5}{35}=\dfrac{6}{35}\)
\(\dfrac{2}{-5}-\dfrac{4}{7}=\dfrac{-14}{35}+\dfrac{20}{35}=\dfrac{6}{35}\)