a)4 + 4/7 b)11/3 + 1
nhớ giải đầy đủ nha! :3
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\(\dfrac{4}{3}:\dfrac{2}{5}=\dfrac{4}{3}\cdot\dfrac{5}{2}=\dfrac{20}{6}=\dfrac{10}{3}\)
\(\dfrac{1}{8}:7=\dfrac{1}{8\cdot7}=\dfrac{1}{56}\)
\(3:\dfrac{4}{5}=3\cdot\dfrac{5}{4}=\dfrac{15}{4}\)
\(\dfrac{4}{5}:3=\dfrac{4}{5\cdot3}=\dfrac{4}{15}\)
a.\(y:\dfrac{4}{5}=\dfrac{11}{8}\)
\(y=\dfrac{11}{8}\times\dfrac{4}{5}\)
\(y=\dfrac{11}{10}\)
b.\(\dfrac{11}{3}-y=\dfrac{1}{9}\)
\(y=\dfrac{11}{3}-\dfrac{1}{9}\)
\(y=\dfrac{32}{9}\)
c.\(\dfrac{1}{7}\times x=\dfrac{8}{5}\)
\(x=\dfrac{8}{5}:\dfrac{1}{7}\)
\(x=\dfrac{56}{5}\)
1.
= 10 000 - 29 x 11
= 10 000 - 319
= 9681
2.
= ( 3/7 + 4/7 ) + ( 4/9 + 5/9 )
= 1 + 1
= 2
\(A=\dfrac{4}{1\cdot5}+\dfrac{4}{5\cdot9}+...+\dfrac{4}{2001\cdot2005}\)
\(A=1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-...+\dfrac{1}{2001}-\dfrac{1}{2005}\)
\(A=1-\dfrac{1}{2005}=\dfrac{2004}{2005}\)
\(B=\dfrac{3}{10\cdot12}+\dfrac{3}{12\cdot14}+...+\dfrac{3}{998\cdot1000}\)
\(\dfrac{2}{3}B=\dfrac{2}{10\cdot12}+...+\dfrac{2}{998\cdot1000}\)
\(\dfrac{2}{3}B=\dfrac{1}{10}-\dfrac{1}{12}+\dfrac{1}{12}-...+\dfrac{1}{998}-\dfrac{1}{1000}\)
\(\dfrac{2}{3}B=\dfrac{1}{10}-\dfrac{1}{1000}=\dfrac{99}{1000}\)
\(B=\dfrac{99}{1000}:\dfrac{2}{3}=\dfrac{297}{2000}\)
\(A=\dfrac{4}{1.5}+\dfrac{4}{5.9}+...+\dfrac{4}{2001.2005}\)
\(\Rightarrow A=4\left(\dfrac{1}{1.5}+\dfrac{1}{5.9}+...+\dfrac{1}{2001.2005}\right)\)
\(\Rightarrow A=4.\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+...+\dfrac{1}{2001}-\dfrac{1}{2005}\right)\)
\(\Rightarrow A=1-\dfrac{1}{2005}\)
\(\Rightarrow A=\dfrac{2004}{2005}\)
\(D=\frac{\frac{88}{132}-\frac{33}{132}+\frac{60}{132}}{\frac{55}{132}+\frac{132}{132}-\frac{84}{132}}\)
\(D=\frac{\frac{115}{132}}{\frac{103}{132}}\)
\(D=\frac{115}{103}\)
a)\(\dfrac{4}{1}+\dfrac{4}{7}=\dfrac{28}{7}+\dfrac{4}{7}=\dfrac{32}{7}\)
b)\(\dfrac{11}{3}+\dfrac{1}{1}=\dfrac{11}{3}+\dfrac{3}{3}=\dfrac{14}{3}\)
4 + 4/7 = 32/7
11/3 + 1 =14/3