\(\dfrac{125}{93}+\dfrac{125}{43}\)
K
Khách
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30 tháng 11 2017
\(A=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{3^3}\right)....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(A=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)....\left(\dfrac{1}{125}-\dfrac{1}{5^3}\right).....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(A=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)....0......\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(A=0\)
QT
1
3 tháng 1 2021
Ta có: \(\left|97\dfrac{2}{3}-125\dfrac{3}{5}\right|+97\dfrac{2}{5}-125\dfrac{1}{3}\)
\(=\left|97+\dfrac{2}{3}-125-\dfrac{3}{5}\right|+97+\dfrac{2}{5}-125-\dfrac{1}{3}\)
\(=\left|-28+\dfrac{1}{15}\right|-28+\dfrac{1}{15}\)
\(=\left|\dfrac{1}{15}-28\right|-28+\dfrac{1}{15}\)
\(=28-\dfrac{1}{15}-28+\dfrac{1}{15}\)
\(=0\)
QT
1
2 tháng 1 2021
Ta có: \(\left|97\dfrac{2}{3}-123\dfrac{3}{5}+97\dfrac{2}{5}-125\dfrac{1}{3}\right|\)
\(=\left|97\left(\dfrac{2}{3}+\dfrac{2}{5}\right)-125\cdot\left(\dfrac{3}{5}+\dfrac{1}{3}\right)\right|\)
\(=\left|194\cdot\dfrac{8}{15}-125\cdot\dfrac{14}{15}\right|\)
\(=\left|\dfrac{-66}{5}\right|=\dfrac{66}{5}\)
GL
24 tháng 8 2023
a) \(\dfrac{2}{7}+\dfrac{4}{7}=\dfrac{2+4}{7}=\dfrac{6}{7}\)
b) \(\dfrac{23}{13}+\dfrac{8}{13}=\dfrac{23+8}{13}=\dfrac{31}{13}\)
c) \(\dfrac{27}{125}+\dfrac{16}{125}=\dfrac{27+16}{125}=\dfrac{43}{125}\)
NV
Nguyễn Việt Dũng
CTVVIP
24 tháng 8 2023
a)\(\dfrac{2}{7}\) + \(\dfrac{4}{7}\) = \(\dfrac{6}{7}\)
b)\(\dfrac{23}{13}\) + \(\dfrac{8}{13}\) = \(\dfrac{31}{13}\)
c)\(\dfrac{27}{125}\) + \(\dfrac{16}{125}\) = \(\dfrac{43}{125}\)
AT
30 tháng 6 2017
\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(=\dfrac{1}{4}+\dfrac{3}{4}=\dfrac{4}{4}=1\)
13 tháng 9 2017
\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3}{4}\)
\(=\dfrac{1}{4}+\dfrac{3}{4}\)
\(=1\)
13 tháng 9 2017
\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}=\dfrac{1}{4}+\dfrac{3}{4}=1\)
NN
1
CX
23 tháng 7 2021
Cho biểu thức ban đầu là A
Đặt 3 = a ; \(\sqrt{9+\dfrac{125}{27}}\)= b
⇔A = \(\sqrt[3]{a+b} . \sqrt[3]{b-a}\)
⇔A= \(\sqrt[3]{(a+b)(b-a)}\)
⇔A= \(\sqrt[3]{b^2-a^2}\)
⇔A= \(\sqrt[3]{9+\dfrac{125}{27}-9}\)
⇔A= \(\sqrt[3]{\dfrac{125}{27}}\)
⇔A = \(\dfrac{5}{3}\) ( ĐPCM)
TK
23 tháng 7 2021
\(\sqrt[3]{3+\sqrt{9+\dfrac{125}{27}}}\sqrt[3]{-3+\sqrt{9+\dfrac{125}{27}}}\)
=\(\sqrt[3]{-\left(3+\sqrt{9+\dfrac{125}{27}}\right)\left(3-\sqrt{9+\dfrac{125}{27}}\right)}\)
=\(\sqrt[3]{-\left[9-\left(9+\dfrac{125}{27}\right)\right]}\)
=\(\sqrt[3]{\dfrac{125}{27}}\)
=5/3
\(=125\left(\dfrac{1}{93}+\dfrac{1}{43}\right)=125\cdot\dfrac{136}{3999}=\dfrac{17000}{3999}\)