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\(\dfrac{38}{11}+\left(\dfrac{16}{13}+\dfrac{6}{11}\right)\\ =\dfrac{38}{11}+\dfrac{16}{13}+\dfrac{6}{11}\\ =\left(\dfrac{38}{11}+\dfrac{6}{11}\right)+\dfrac{16}{13}\\ =\dfrac{44}{11}+\dfrac{16}{13}\\ =4+\dfrac{16}{13}\\ =\dfrac{52}{13}+\dfrac{16}{13}\\ =\dfrac{68}{13}\\ \dfrac{3}{4}:\dfrac{3}{5}-\dfrac{1}{5}\\ =\dfrac{3}{4}\times\dfrac{5}{3}-\dfrac{1}{5}\\ =\dfrac{5}{4}-\dfrac{1}{5}\\ =\dfrac{25-4}{20}\\ =\dfrac{21}{20}\)
Phép tính 1:
\(\dfrac{38}{11}+\left(\dfrac{16}{13}+\dfrac{6}{11}\right)\)
\(=\left(\dfrac{38}{11}+\dfrac{6}{11}\right)+\dfrac{16}{13}\)
\(=4+\dfrac{16}{13}=\dfrac{4\cdot13+16}{13}\)(Dấu "." trong phép tính là dấu nhân)
\(=\dfrac{68}{13}\)
Phép tính 2:
\(\dfrac{3}{4}:\dfrac{3}{5}-\dfrac{1}{5}\)
\(=\dfrac{3}{4}\cdot\dfrac{5}{3}-\dfrac{1}{5}\)
\(=\dfrac{\left(3\cdot5\right):3\cdot5}{\left(4\cdot3\right):3\cdot5}-\dfrac{1\cdot4}{5\cdot4}\)
\(=\dfrac{25}{20}-\dfrac{4}{20}\)
\(=\dfrac{21}{20}\)
\(a,1\dfrac{4}{7}.3\dfrac{4}{11}.3\dfrac{11}{15}.5\dfrac{5}{8}\)
\(=\dfrac{11}{7}.\dfrac{27}{11}.\dfrac{56}{15}.\dfrac{45}{8}\)
\(=\dfrac{11.27.56.45}{7.11.15.8}\)
\(=\dfrac{1.3.7.3}{1.1.1.1}\)
\(=63\)
\(b,\dfrac{3}{4}.1\dfrac{1}{2}+\dfrac{3}{4}.\dfrac{1}{2}\)
\(=\dfrac{3}{4}.\left(1\dfrac{1}{2}+\dfrac{1}{2}\right)\)
\(=\dfrac{3}{4}.2\)
\(=\dfrac{3}{2}\)
là sao cái này mình chưa học nhưng mà mẹ mình tứ bắt mình học í
\(a.\frac{19}{5}\cdot\frac{4}{7}+\frac{3}{7}\cdot\frac{19}{5}-\frac{4}{5}\)
\(=\frac{19}{5}\cdot\left(\frac{4}{7}+\frac{3}{7}\right)-\frac{4}{5}\)
\(=\frac{19}{5}\cdot1-\frac{4}{5}\)
\(=\frac{19}{5}-\frac{4}{5}=\frac{15}{5}=3\)
\(b.2\frac{2}{7}\cdot5\frac{2}{5}+\frac{16}{7}\cdot1\frac{3}{5}+\frac{1}{2}\)
\(=\frac{16}{7}\cdot\frac{27}{5}+\frac{16}{7}\cdot\frac{8}{5}+\frac{1}{2}\)
\(=\frac{16}{7}\cdot\left(\frac{27}{5}+\frac{8}{5}\right)+\frac{1}{2}\)
\(=\frac{16}{7}\cdot7+\frac{1}{2}\)
\(=16+\frac{1}{2}=\frac{33}{2}\)
\(c.\frac{3}{7}\cdot3\frac{3}{4}-\frac{3}{7}\cdot\frac{5}{4}-\frac{1}{4}\)
\(=\frac{3}{7}\cdot\frac{15}{4}-\frac{3}{7}\cdot\frac{5}{4}-\frac{1}{4}\)
\(=\frac{3}{7}\cdot\left(\frac{15}{4}-\frac{5}{4}\right)-\frac{1}{4}\)
\(=\frac{3}{7}\cdot\frac{5}{2}-\frac{1}{4}\)
\(=\frac{15}{14}-\frac{1}{4}=\frac{23}{28}\)
Chú ý: \(\cdot:\times\)
\(\frac{5}{8}+\frac{2}{9}-\frac{2}{5}+\frac{3}{8}+\frac{4}{9}+\frac{1}{3}-\frac{3}{5}\)
\(=\frac{5}{8}+\frac{2}{9}-\frac{2}{5}+\frac{3}{8}+\frac{4}{9}+\frac{3}{9}-\frac{3}{5}\)
\(=\left(\frac{5}{8}+\frac{3}{8}\right)+\left(\frac{2}{9}+\frac{4}{9}+\frac{3}{9}\right)-\left(\frac{2}{5}+\frac{3}{5}\right)\)
\(=\frac{8}{8}+\frac{9}{9}-\frac{5}{5}\)
\(=1+1-1\)
\(=2-1\)
\(=1\)
A=(6 : 35−116 x 67) : (415 x 1011+5211)A=(6 : 35−116 x 67) : (415 x 1011+5211)
A=(6 : 35−76 x 67) : (215 x 1011+5711)A=(6 : 35−76 x 67) : (215 x 1011+5711)
A=(10−1) : (4211+5711)A=(10−1) : (4211+5711)
A=9 : 9A=9 : 9
A=1
Tổng trên có: (2002 -1):1+1 =2002 (số hạng)
Ta có: 1+ (2 -3 -4 +5)+ (6 -7 -8 +9)+...+(1998 -1999 -2000 -2001)+ 2002 (có 500 nhóm, thừa 2 số)
= 1 + 0 + 0 + ... + 0 + 2002
= 2003
Chúc bạn học tốt.
\(P=\left(5-\frac{3}{4}+\frac{1}{5}\right)-\left(6+\frac{7}{4}-\frac{8}{5}\right)-\left(2-\frac{5}{4}+\frac{16}{5}\right)\)
\(P=5-\frac{3}{4}+\frac{1}{5}-6-\frac{7}{4}+\frac{8}{5}-2+\frac{5}{4}-\frac{16}{5}\)
\(P=\left(5-6-2\right)-\left(\frac{3}{4}+\frac{7}{4}-\frac{5}{4}\right)+\left(\frac{1}{5}+\frac{8}{5}-\frac{16}{5}\right)\)
\(P=-2-\frac{5}{4}-7=-\frac{41}{4}\)
P/s: Tớ không nhớ là lớp 5 đã học đến số âm :v Hoặc là cậu để sai hoặc là tớ làm sai ;-;
bạn ơi 5-6-2 mà bằng -2 được à