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59 : 54 = 1,0935
59 : 27 = 2,185
59 : 18 = 3,2777
59 : 9 = 6,5555
59 : 6 = 9,833
bn tính dư cho mik đc ko, tính dư ko phải số thập phân nhé, dù gì cũng cảm ơn bạn ĐÀO TRẦN TUẤN ANH
a. 7/9 - 16/9 = -9/9 = -1
b. 2/-15 + 7/10 = 17/30
c. (4 2/3 - 4 3/4) : -5/12 - 4/5
= (14/3 - 19/4) : (-5/12) - 4/5
= -1/12 : (-5/12) - 4/5
= 1/5 - 4/5
= -3/5
\(A=\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+...+\frac{3}{59.61}\)
\(A=\frac{3}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{59.61}\right)\)
\(A=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(A=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(A=\frac{84}{305}\)
4/ 5 * 7 + 4 / 7 * 9 + .. + 4 / 59 * 61
= 2/5 - 2/7 + 2/7 - 2/9 + ... + 2/59 - 2/61
= 2/5 - 2/61
= 112/305
Đặt
\(S=\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{59.61}\)
\(S=2.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(S=2.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(S=2.\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(S=2.\frac{56}{305}=\frac{112}{305}\)
Câu 2:
a: x-158=32
=>x=158+32
=>x=190
b: \(x\cdot24=264\)
=>\(x=\dfrac{264}{24}\)
=>x=11
c: \(6x+9=3^7:3^4\)
=>\(6x+9=3^3\)
=>6x+9=27
=>6x=18
=>x=18/6=3
Câu 1:
a: \(86\cdot19+14\cdot19\)
\(=19\left(86+14\right)\)
\(=19\cdot100=1900\)
b: \(4\cdot\left(-5\right)^2-104\cdot\left(-5\right)^2\)
\(=4\cdot25-104\cdot25\)
\(=25\left(4-104\right)=-100\cdot25=-2500\)
c: \(7\cdot\left(-2\right)\cdot8\left(-5\right)\)
\(=7\cdot2\cdot8\cdot5\)
\(=56\cdot10=560\)
d: \(59-\left[59+\left(-76\right)\right]\)
\(=59-59+76\)
=76
B = \(\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+....+\frac{3}{59\cdot61}\)
=> B = \(\frac{1}{2}\cdot\left(\frac{3}{5}-\frac{3}{7}\right)+\frac{1}{2}\cdot\left(\frac{3}{7}-\frac{3}{9}\right)+...+\frac{1}{2}\cdot\left(\frac{3}{59}-\frac{3}{61}\right)\)
=>B= \(\frac{1}{2}\cdot\left(\frac{3}{5}-\frac{3}{7}+\frac{3}{7}-\frac{3}{9}+....+\frac{3}{59}-\frac{3}{61}\right)\)
=> B = \(\frac{1}{2}\cdot\left(\frac{3}{5}-\frac{3}{61}\right)\)= \(\frac{1}{2}\cdot\frac{168}{305}\)= \(\frac{84}{305}\) Vậy B= \(\frac{84}{305}\)
\(\frac{59}{7}+\frac{59}{9}=\frac{944}{63}\)
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