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a=78/35
b=22/12
c=1/1
d=40202090/4040090
e=1,24025667172...
f=871,82
ko biết đúng ko [0_0'] hihi
\(\dfrac{\left(\dfrac{5}{30}+\dfrac{3}{30}+\dfrac{2}{30}\right):\left(\dfrac{5}{30}+\dfrac{3}{30}-\dfrac{2}{30}\right)}{\left(\dfrac{30}{60}-\dfrac{20}{60}+\dfrac{15}{60}-\dfrac{12}{60}\right):\left(\dfrac{3}{12}-\dfrac{2}{12}\right)}=\dfrac{\dfrac{1}{3}:\dfrac{1}{5}}{\dfrac{13}{60}:\dfrac{1}{12}}=\dfrac{\dfrac{1}{3}\times5}{\dfrac{13}{60}\times12}=\dfrac{\dfrac{5}{3}}{\dfrac{13}{5}}=\dfrac{25}{39}\)
Đặt biểu thực là A
2A = \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+...+\dfrac{2}{21.23}\)
= \(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{21}-\dfrac{1}{23}=1-\dfrac{1}{23}=\dfrac{22}{23}\)
=> A = \(\dfrac{11}{23}\)
\(\dfrac{2}{5}\times15\dfrac{1}{3}-\dfrac{2}{5}\times10\dfrac{1}{3}\)
\(=\dfrac{2}{5}\times\left(15\dfrac{1}{3}-10\dfrac{1}{3}\right)\)
\(=\dfrac{2}{5}\times5\)
\(=2\)
____________________
\(12\dfrac{5}{11}-\left(3\dfrac{1}{4}+2\dfrac{5}{11}\right)\)
\(=12\dfrac{5}{11}-3\dfrac{1}{4}-2\dfrac{5}{11}\)
\(=\left(12\dfrac{5}{11}-2\dfrac{5}{11}\right)-3\dfrac{1}{4}\)
\(=10-3\dfrac{1}{4}\)
\(=\dfrac{27}{4}\)
_______________
\(\dfrac{34}{31}-\dfrac{19}{28}-\dfrac{3}{31}\)
\(=\left(\dfrac{34}{31}-\dfrac{3}{31}\right)-\dfrac{19}{28}\)
\(=\dfrac{31}{31}-\dfrac{19}{28}\)
\(=1-\dfrac{19}{28}\)
\(=\dfrac{9}{28}\)
Lời giải:
a.
$x:3\frac{1}{15}-\frac{3}{4}=2\frac{1}{4}$
$x:\frac{46}{15}-\frac{3}{4}=\frac{9}{4}$
$x: \frac{46}{15}=\frac{9}{4}+\frac{3}{4}=3$
$x=3\times \frac{46}{15}=\frac{46}{5}$
b. $x\times 3\frac{2}{3}-1\frac{2}{3}=2\frac{1}{3}$
$x\times \frac{11}{3}=1\frac{2}{3}+2\frac{1}{3}=4$
$x=4: \frac{11}{3}=\frac{12}{11}$