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a) \(\frac{2}{5}+\frac{2}{3}+\frac{3}{5}\)
= \(\left(\frac{2}{5}+\frac{3}{5}\right)+\frac{2}{3}\)
= \(\frac{2}{3}+1\)= \(\frac{5}{3}\)
b) \(\frac{6}{7}-\frac{5}{14}-\frac{3}{7}\)
= \(\frac{3}{7}-\frac{5}{14}\)
= \(\frac{6-5}{14}\) \(=\frac{1}{14}\)
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a, x + 2/5 = 1/2
x = 1/2 + 2/5
x = 9/10 b,34/5 + x = 15/8
x = 34/5 - 15/8
x =197/40 c, x - 45/6 = 21/6 + 5/6
x - 45/6 = 26/6
x = 26/6 + 45/6
x =71/6 d,14/5 - x = 11/5 - 3/5
14/5 - x = 8/5
x = 14,5 - 8/5
x =6/5
a, x + 2/5 = 1/2
x = 1/2 + 2/5
x = 9/10
b,34/5 + x = 15/8
x = 34/5 - 15/8
x =197/40
c, x - 45/6 = 21/6 + 5/6
x - 45/6 = 26/6
x = 26/6 + 45/6
x =71/6
d,14/5 - x = 11/5 - 3/5
14/5 - x = 8/5
x = 14,5 - 8/5
x =6/5
- Làm lại.
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Bài 1: Tính
a) =1\(\frac{49}{60}\)
b) =2\(\frac{13}{30}\)
Bài 2: Tìm x
a) =4\(\frac{1}{21}\)
Riêng câu b) thì mk nghĩ là bạn viết lộn vì mk thấy cái chỗ xx3/4 là mk ko hiểu rồi
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\(a,\frac{2}{5}:\frac{6}{5}+\frac{3}{4}.\frac{12}{5}\)
=\(\frac{2}{5}.\frac{5}{6}+\frac{9}{5}\)
=\(\frac{1}{3}+\frac{9}{5}\)
=\(\frac{32}{5}\)
b) \(\frac{17}{5}.\frac{3}{5}.\frac{17}{5}.\frac{2}{5}\)
=\(\frac{17}{5}.\left(\frac{3}{5}.\frac{2}{5}\right)\)
=\(\frac{17}{5}.\frac{6}{25}\)
=\(\frac{102}{125}\)
c)\(\frac{1}{4}+\frac{3}{4}.\frac{2}{3}-\frac{1}{2}\)
=\(\frac{1}{4}+\frac{1}{2}-\frac{1}{2}\)
=\(\frac{1}{4}\)
d)\(\frac{6}{7}+\frac{5}{8}:5\)
=\(\frac{6}{7}+\frac{5}{8}.\frac{1}{5}\)
=\(\frac{6}{7}+\frac{1}{8}\)
=\(\frac{48}{56}+\frac{7}{56}\)
=\(\frac{55}{56}\)
tk nhé bn!
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a) 2/5+2/3+3/5=
= ( 2/5 + 3/5 ) + 2/3
= 1 + 2/3
= 5/3
b) 6/7-5/14-3/7=
= ( 6/7 - 3/7 ) - 5/14
= 3/7 - 5/14
= 6/14 - 5/14
= 1/14
c)[2/5+3/5]x 5/3
= 1 x 5/3
= 5/3
d)1:5/6-1
= 1 x 6/5 - 1
= 6/5 - 1
= 6/5 - 5/5
= 1/5
~ HOk tốt ~
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\(M=\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+....+\frac{5}{46.51}\)
\(M=\frac{6-1}{1.6}+\frac{11-6}{6.11}+\frac{16-11}{11.16}+...+\frac{51-46}{46.51}\)
\(M=1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+....+\frac{1}{46}-\frac{1}{51}\)
\(M=1-\frac{1}{51}=\frac{50}{51}\)
\(N=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+....+\frac{1}{199\cdot201}\)
\(N=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{199}-\frac{1}{201}\right)\)
\(N=\frac{1}{2}\cdot\left(1-\frac{1}{201}\right)\)
\(N=\frac{1}{2}\cdot\frac{200}{201}=\frac{100}{201}\)
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