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A = -4/5x(1/2+1/3+1/4)= -4/5x1 = -4/5
B = 6/19 x ( 3/4+4/3+-1/2)= 6/19x 19 = 6
C = 2002/2003x(3/4+5/6-19/12)=2003/2002x0=0
a, \(\frac{3}{8}+\frac{11}{13}-\frac{9}{13}\)
=\(\frac{3}{8}+\frac{2}{13}\)
=\(\frac{55}{104}.\)
b, \(\frac{2}{7}.\left(\frac{5}{9}+\frac{4}{9}\right)+\frac{2}{7}\)
=\(\frac{2}{7}.\frac{9}{9}+\frac{2}{7}\)
=\(\frac{2}{7}+\frac{2}{7}\)
=\(\frac{4}{7}\)
c, \(\frac{3}{11}.\left(\frac{3}{5}-\frac{5}{3}\right)-\frac{3}{10}.\left(\frac{1}{3}-\frac{2}{5}\right)\)
=\(\frac{3}{11}.-\frac{16}{15}-\frac{3}{10}.-\frac{1}{15}\)
=\(-\frac{16}{55}--\frac{1}{50}\)
=\(-\frac{149}{550}.\)
d, \(\frac{-3}{4}.\frac{11}{23}+\frac{3}{23}.\frac{31}{17}-\frac{3}{17}.\frac{19}{23}\)
=\(-\frac{33}{92}+\frac{93}{391}-\frac{57}{391}\)
=\(-\frac{417}{1564}\)
e, \(\frac{3}{17}.\frac{11}{23}+\frac{3}{23}.\frac{31}{17}-\frac{3}{17}.\frac{19}{23}\)
=\(\frac{33}{391}+\frac{93}{391}--\frac{254}{391}\)
=\(\frac{380}{391}.\)
g, \(\frac{3}{7}.\frac{-5}{12}+\frac{11}{17}:\frac{5}{-12}\)
=\(-\frac{5}{28}+-\frac{132}{85}\)
= \(-1.731512605.\)
k cho mình nha làm mỏi tay quá ,.....................kết bạn với mình nha.......................
a ,A = \(a.\frac{1}{3}+a.\frac{1}{4}-a.\frac{1}{6}\)
\(=a.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{6}\right)\)
\(=\frac{-3}{5}.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{6}\right)\\ =\frac{-3}{5}.\frac{5}{12}\)
\(=\frac{-1}{4}\)
b, B = \(b.\frac{5}{6}+b.\frac{3}{4}-b.\frac{1}{2}\)
\(=b.\left(\frac{5}{6}+\frac{1}{4}-\frac{1}{2}\right)\)
\(=\frac{12}{13}.\left(\frac{5}{6}+\frac{1}{4}-\frac{1}{2}\right)\)
\(=\frac{12}{13}.\frac{7}{12}\)
\(=\frac{7}{13}\)
a) Thay \(a=\frac{-3}{5}\)vào biểu thức A ta có :
\(A=\frac{-3}{5}.\frac{1}{3}+\frac{-3}{5}.\frac{1}{4}-\frac{-3}{5}.\frac{1}{6}\)
\(A=\frac{-3}{5}.\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{6}\right)\)
\(A=\frac{-3}{5}.\frac{5}{12}\)
\(A=\frac{-1}{4}\)
Vậy giá trị của biểu thức A tại \(a=\frac{-3}{5}\)là \(\frac{-1}{4}\)
b) Thay \(b=\frac{12}{13}\)vào biểu thức B ta có :
\(B=\frac{12}{13}.\frac{5}{6}+\frac{12}{13}.\frac{3}{4}-\frac{12}{13}.\frac{1}{2}\)
\(B=\frac{12}{13}.\left(\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)\)
\(B=\frac{12}{13}.\frac{13}{12}\)
\(B=1\)
Vậy giá trị của biểu thức B tại \(b=\frac{12}{13}\)là 1
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\(\text{Câu 1 :}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{12.13}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{12}-\frac{1}{13}\)
\(=\frac{1}{1}-\frac{1}{13}\)
\(=\frac{12}{13}\)
\(\text{Câu 2 :}\)
\(\frac{5}{1.3}+\frac{5}{3.5}+\frac{5}{5.7}+...+\frac{5}{99.101}\)
\(=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\left(\frac{1}{1}-\frac{1}{101}\right)\)
\(=\frac{5}{2}.\frac{100}{101}\)
\(=\frac{250}{101}\)
\(1)A=a\frac{1}{3}+a\frac{1}{4}-a\frac{1}{6}=a\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{6}\right)=a\frac{5}{12}\)
Thay \(a=-\frac{3}{5}\) vào A,ta đc:
\(A=-\frac{3}{5}.\frac{5}{12}=-\frac{1}{4}\)
\(2)B=b\frac{5}{6}+b\frac{3}{4}-b\frac{1}{2}=b\left(\frac{5}{6}+\frac{3}{4}-\frac{1}{2}\right)=b\frac{13}{12}\)
Thay \(b=\frac{12}{13}\) vào B, ta đc: \(B=b\frac{13}{12}=\frac{12}{13}.\frac{13}{12}=1\)