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\(\dfrac{x-2}{5}=\dfrac{1-x}{6}\\ =>\left(x-2\right)\cdot6=\left(1-x\right)\cdot5\\ =>6x-12=5-5x\\ =>6x+5x=5+12\\ =>11x=17\\ x=\dfrac{17}{11}\)
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e) \(\left(\dfrac{-1}{3}\right)\cdot\dfrac{5}{7}=\dfrac{\left(-1\right)\cdot5}{3\cdot7}=\dfrac{-5}{21}\)
f) \(\dfrac{2}{7}:\dfrac{3}{4}=\dfrac{2}{7}\cdot\dfrac{4}{3}=\dfrac{8}{21}\)
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A = 1/(5.6) + 1/(6.7) + ... + 1/(24.25)
= 1/5 - 1/6 + 1/6 - 1/7 + ... + 1/24 - 1/25
= 1/5 - 1/25
= 4/25
B = 2/(1.3) + 2/(3.5) + 2/(5.7) + ... + 2/(99.101)
= 1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 + ... + 1/99 - 1/101
= 1 - 1/101
= 100/101
`a) A = 1/(5.6) + 1/(6.7)+...+1/(24.25)`
`= 1/5 - 1/6 + 1/6 - 1/7 +...+1/24-1/25`
`= 1/5-1/25`
`= 5/25 - 1/25`
`= 4/25`
Vậy:`A = 4/25`
`b) B = 2/(1.3)+2/(3.5)+...+2/(99.101)`
`= 1- 1/3 + 1/3 - .... +1/99-1/101`
`= 1 - 1/101`
`= 100/101`
Vậy: `B = 100/101`
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\(=4\dfrac{5}{57}-3\dfrac{4}{51}+8\dfrac{13}{29}-3\dfrac{5}{57}+6\dfrac{16}{29}\)
\(=\left(4\dfrac{5}{57}-3\dfrac{5}{57}\right)+\left(8\dfrac{13}{29}+6\dfrac{16}{29}\right)-\dfrac{157}{51}\)
\(=16-\dfrac{157}{51}\)
\(=\dfrac{659}{51}\)
\(\left(4\dfrac{5}{57}-3\dfrac{4}{51}+8\dfrac{13}{29}\right)-\left(3\dfrac{5}{57}-6\dfrac{16}{29}\right)\)
\(=\left(\dfrac{233}{57}-\dfrac{157}{51}+\dfrac{245}{29}\right)-\left(\dfrac{176}{57}-\dfrac{190}{29}\right)\)
\(=\left(\dfrac{233}{57}+\dfrac{-157}{51}+\dfrac{245}{29}\right)-\left(\dfrac{176}{57}+\dfrac{-190}{29}\right)\)
\(=\dfrac{233}{57}+\dfrac{-157}{51}+\dfrac{245}{29}-\dfrac{176}{57}-\dfrac{-190}{29}\)
\(=\left(\dfrac{233}{57}-\dfrac{176}{57}\right)+\left(\dfrac{245}{29}-\dfrac{-190}{29}\right)+\left(\dfrac{-157}{51}\right)\)
\(1+15+\left(\dfrac{-157}{51}\right)\)
\(=16+\left(\dfrac{-157}{51}\right)\)
\(=\dfrac{659}{51}\)
a: Diện tích đào ao thả cá chiếm:
100%-35%-15,5%=49,5%(diện tích miếng đất)
b: Diện tích trồng cây ăn quả là: \(20000\cdot35\%=7000\left(m^2\right)\)
Diện tích trồng rau là \(20000\cdot15,5\%=3100\left(m^2\right)\)
Diện tích đào ao thả cá là:
20000-7000-3100=9900(m2)