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a)\(8-\frac{3}{5}=\frac{40}{5}-\frac{3}{5}=\frac{37}{5}\)
b)\(\frac{24}{15}\times\frac{56}{72}\times\frac{10}{14}=\frac{13440}{15120}=\frac{8}{9}\)
c)\(\frac{32}{48}\times\frac{35}{21}\times\frac{63}{14}\times\frac{9}{30}=\frac{2}{3}\times\frac{5}{3}\times\frac{9}{2}\times\frac{3}{10}=\frac{270}{180}=\frac{3}{2}\)
d)\(7-\frac{72}{48}\times\frac{56}{60}-3=7-\frac{3}{2}\times\frac{14}{15}-3\)
\(=7-\frac{7}{5}-3\)
\(=\frac{35}{5}-\frac{7}{5}-\frac{15}{5}\)
\(=\frac{13}{5}\)
#H
(Sai thì sửa)
a: \(A=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^7\)
=>\(2\cdot A=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^6\)
=>\(2A-A=1-\left(\dfrac{1}{2}\right)^7=1-\dfrac{1}{128}=\dfrac{127}{128}\)
=>\(A=\dfrac{127}{128}\)
b: \(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{10\cdot11}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{10}-\dfrac{1}{11}\)
\(=1-\dfrac{1}{11}=\dfrac{10}{11}\)
\(\text{A= 73 - (35 + 1𝑎) : 23}\)
thay \(a=45\) vào biểu thức ta có:
\(A=73-\left(35+45\right):23\)
\(=73-\frac{80}{23}=\frac{1599}{23}\)
b)\(73-\left(35+1a\right):23=1715\)
\(\left(35+1a\right):23=-1642\)
\(35+a=-37766\)
\(a=-37801\)
\(\dfrac{x}{5}\) = \(\dfrac{48}{60}\)
\(x\) = \(\dfrac{48}{60}\) \(\times\) 5
\(x\) = 4
\(\dfrac{x}{7}\) = \(\dfrac{16}{56}\)
\(x\) = \(\dfrac{16}{56}\) \(\times\) 7
\(x\) = 2
\(\dfrac{8}{x}\) = \(\dfrac{72}{108}\)
\(x\) = 8 : \(\dfrac{72}{108}\)
\(x\) = 12
\(x\) - \(\dfrac{3}{12}\) = \(\dfrac{35}{84}\)
\(x\) = \(\dfrac{35}{84}\) + \(\dfrac{3}{12}\)
\(x\) = \(\dfrac{2}{3}\)
a) 1+2+3+4+5+6+7+8+9
= (1+9)+(2+8)+(3+7)+(4+6)+5
= 10+10+10+10=5
= 45
b) 19+28+37+46+81+72+63+54
= (19+81)+(28+72)+(46+54)+(63+37)
= 100+100+100+100
= 400
a) (350+1050):14=1400:14
=100
bn giải hộ mk phần a,b,c với