\(\dfrac{7}{9}\) : u = 3
giúp mình với
giúp mình với ạ
\(\dfrac{3}{7}.\dfrac{5}{9}+\dfrac{4}{9}.\dfrac{-3}{7}+\left(-2023^0\right)\)
\(\dfrac{3}{7}.\dfrac{5}{9}+\dfrac{4}{9}.\dfrac{-3}{7}+\left(-2023^0\right)\)
\(=\left(\dfrac{3}{7}.\dfrac{5}{9}\right)+\left(\dfrac{4}{9}.\dfrac{-3}{7}\right)+\left(-1\right)\)
\(=\dfrac{5}{21}+\dfrac{-4}{21}+\left(-1\right)\)
\(=\dfrac{5}{21}+\dfrac{-4}{21}+\dfrac{-21}{21}\)
\(=\dfrac{5-4-21}{21}=\dfrac{-20}{21}\)
\(\dfrac{5}{9}\):\(\dfrac{13}{7}\)+\(\dfrac{5}{9}\):\(\dfrac{13}{9}\)-1\(\dfrac{2}{3}\)
4\(\dfrac{2}{5}\):0,5%-1\(\dfrac{3}{7}\).14%+(-0,5)
Tính giúp mik với!!! Mình cảm ơn
a) \(\dfrac{5}{9}:\left(\dfrac{13}{7}+\dfrac{13}{9}\right)-\dfrac{5}{3}\)(chỗ này mk lười chép lại đề)
=\(\dfrac{5}{9}:\dfrac{208}{63}-\dfrac{5}{3}\)
=\(\dfrac{5}{9}.\dfrac{63}{208}-\dfrac{5}{3}\)
=\(\dfrac{5.63}{9.208}-\dfrac{5}{3}\)
=\(\dfrac{5.7}{1.208}-\dfrac{5}{3}\)
=\(\dfrac{36}{208}-\dfrac{5}{3}\)
=\(\dfrac{108}{624}-\dfrac{1040}{624}\)
=\(\dfrac{-932}{624}\)
=\(\dfrac{233}{156}\)
còn câu b mk chưa học nên mk chịu
Giải:
5/9:13/7+5/9:13/9 -1 2/3
=5/9.7/13+5/9.9/13-5/3
=5/9.(7/13+9/13)-5/3
=5/9.16/13-5/3
=80/117-5/3
=-115/117
4 2/5 : 0,5% -1 3/7 .14% +(-0,5)
=22/5:1/200-10/7.7/50 +(-1/2)
=880-1/5-1/2
=8793/10
\(\dfrac{3}{15}.\dfrac{5}{9}:\dfrac{-18}{17}.\dfrac{14}{17}\)
\(\dfrac{12}{7}.\dfrac{7}{4}+\dfrac{35}{11}:\dfrac{245}{121}\)
giúp mình với
a: \(=\dfrac{1}{3}\cdot\dfrac{1}{3}\cdot\dfrac{-17}{18}\cdot\dfrac{14}{17}=\dfrac{-14}{18\cdot9}=\dfrac{-14}{162}=\dfrac{-7}{81}\)
b: \(=\dfrac{12}{4}+\dfrac{35}{11}\cdot\dfrac{121}{245}=3+\dfrac{11}{7}=\dfrac{32}{7}\)
\(\dfrac{3}{15}.\dfrac{5}{9}:\dfrac{-18}{17}.\dfrac{14}{17}\)
\(\dfrac{12}{7}.\dfrac{7}{4}+\dfrac{35}{11}:\dfrac{245}{121}\)
giúp mình với
a: \(=\dfrac{15}{135}\cdot\dfrac{-17}{18}\cdot\dfrac{14}{17}=\dfrac{1}{9}\cdot\dfrac{-7}{9}=\dfrac{-7}{81}\)
b: \(=3+\dfrac{35}{11}\cdot\dfrac{121}{245}=3+\dfrac{11}{7}=\dfrac{32}{7}\)
1. Tính nhanh:
A, \(\dfrac{7}{9}\) x \(\dfrac{25}{28}\) x \(\dfrac{9}{7}\)
B, \(\dfrac{4}{7}\) x \(\dfrac{17}{18}\) x \(\dfrac{7}{4}\) x \(\dfrac{18}{17}\)
Giúp mình với ạ. Mình sẽ tick
a/\(\left(\dfrac{7}{9}\times\dfrac{9}{7}\right)\times\dfrac{25}{28}\)
\(=1\times\dfrac{25}{28}\)
\(=\dfrac{25}{28}\)
b/\(\dfrac{4}{7}\times\dfrac{17}{18}\times\dfrac{7}{4}\times\dfrac{18}{17}\)
\(=\left(\dfrac{4}{7}\times\dfrac{7}{4}\right)\times\left(\dfrac{17}{18}\times\dfrac{18}{17}\right)\)
\(=1\times1\)
\(=1\)
Giúp mình với:
a) A=\(\dfrac{1}{3}\)-\(\dfrac{3}{5}\)+\(\dfrac{5}{7}\)-\(\dfrac{7}{9}\)+\(\dfrac{9}{11}\)-\(\dfrac{11}{13}\)+\(\dfrac{13}{15}\)+\(\dfrac{11}{13}-\dfrac{9}{11}+\dfrac{7}{9}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{1}{3}\)
b) B= \(\dfrac{1}{9.10}-\dfrac{1}{8.9}-\dfrac{1}{7.8}-...-\dfrac{1}{2.3}-\dfrac{1}{1.2}\)
a)A=\(\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{-3}{5}+\dfrac{3}{5}\right)+\left(\dfrac{5}{7}-\dfrac{5}{7}\right)+\left(\dfrac{-7}{9}+\dfrac{7}{9}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)+\left(\dfrac{-11}{13}+\dfrac{11}{13}\right)+\dfrac{13}{15}\)
A=0+0+0+...+0+\(\dfrac{13}{15}\)
A=\(\dfrac{13}{15}\)
b) Ta có: \(-\dfrac{1}{9\cdot10}-\dfrac{1}{8\cdot9}-\dfrac{1}{7\cdot8}-...-\dfrac{1}{2\cdot3}-\dfrac{1}{1\cdot2}\)
\(=-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\)
\(=-\left(1-\dfrac{1}{10}\right)=\dfrac{-9}{10}\)
so sánh các cặp số hữu tỉ sau
a)\(\dfrac{3}{-7}\)và \(\dfrac{-5}{9}\)
b)-0,625 và \(\dfrac{-19}{50}\)
c)\(-2\dfrac{5}{9}\)và \(-\left(\dfrac{-23}{-9}\right)\)
giúp mình với, mik tick cho
Lời giải:
a. $\frac{3}{-7}=\frac{-27}{63}$
$\frac{-5}{9}=\frac{-35}{63}$
Do $\frac{27}{63}< \frac{35}{63}$ nên $\frac{-27}{63}> \frac{-35}{63}$
$\Rightarrow \frac{3}{-7}> \frac{-5}{9}$
---------
b.
$-0,625=\frac{-625}{1000}=\frac{-5}{8}=\frac{-125}{200}$
$\frac{-19}{50}=\frac{-76}{200}> \frac{-125}{200}$
$\Rightarrow -0,625> \frac{-19}{50}$
c.
$-2\frac{5}{9}=-(2+\frac{5}{9})=\frac{-23}{9}=-(\frac{-23}{-9})$
a, \(5^6\) : \(5^5\) + \(\left(\dfrac{4}{9}\right)^0\) b,\(\left(\dfrac{3}{7}\right)^{21}\) : \(\left(1-\dfrac{40}{49}\right)^3\) c, 3.\(\left(\dfrac{2}{3}\right)^3\) -\(\left(\dfrac{-52}{3}\right)^0\) +\(\dfrac{4}{9}\)
Mọi người giúp mình với
a) \(5^6:5^5+\left(\dfrac{4}{9}\right)^0=5^{6-5}+1=5+1=6\)
b) \(\left(\dfrac{3}{7}\right)^{21}:\left(1-\dfrac{40}{49}\right)^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^6\)
\(=\left(\dfrac{3}{7}\right)^{21-6}=\left(\dfrac{3}{7}\right)^{15}\)
c) \(\left(\dfrac{2}{3}\right)^3-\left(\dfrac{-52}{3}\right)^0+\dfrac{4}{9}\)
\(=\dfrac{8}{27}-1+\dfrac{4}{9}\)
\(=\dfrac{8-27+12}{27}=-\dfrac{7}{27}\)
\(a)5^6:5^5+\left(\dfrac{4}{9}\right)^0=5^1+1=6\)
\(b,\left(\dfrac{3}{7}\right)^{21}:\left(1-\dfrac{40}{49}\right)^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{49-40}{49}\right)^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^3=\left(\dfrac{3}{7}\right)^{21}:[\left(\dfrac{3}{7}\right)^2]^3\)
\(=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^6=\left(\dfrac{3}{7}\right)^{21-6}\)
\(=\left(\dfrac{3}{7}\right)^{15}\)
\(c,3.\left(\dfrac{2}{3}\right)^3-\left(\dfrac{-52}{3}\right)^0+\dfrac{4}{9}\)
\(=3.\dfrac{8}{27}-1+\dfrac{4}{9}\)
\(=\dfrac{8}{9}-1+\dfrac{4}{9}\)
\(=\dfrac{8-9+4}{9}=\dfrac{1}{3}\)
Giúp mình với ạ, giải chi tiết nhé !
Mình xin cảm ơn !
a/ \(\sqrt{25}\) - 3\(\sqrt{\dfrac{4}{9}}\)
b/ (2 - \(\dfrac{5}{3}\)) : (\(\dfrac{2}{7}\) + \(\dfrac{5}{21}\) - 1)
c/ 12,7 - 17,2 + 199,9 - 22,8 - 149,9
d/ (\(\dfrac{-1}{2}\))4 + |\(\dfrac{-2}{3}\)| - 20070
e/ 4(\(\dfrac{-1}{2}\))3 + |\(\dfrac{1}{2}\)| : 5
g/ 3 - (\(\dfrac{-6}{7}\))0 + \(\sqrt{9}\) : 2
h/ \(\dfrac{27}{23}\) + \(\dfrac{5}{21}\) - \(\dfrac{4}{23}\) + \(\dfrac{6}{21}\) + \(\dfrac{1}{2}\)
a, \(\sqrt{25}-3\sqrt{\dfrac{4}{9}}=5-3.\dfrac{2}{3}=3\)
b, \(\left(2-\dfrac{5}{3}\right):\left(\dfrac{2}{7}+\dfrac{5}{21}-1\right)\)
\(=\dfrac{1}{3}:\dfrac{6+5-21}{21}\)
\(=-\dfrac{1}{3}.\dfrac{21}{10}\)
\(=-\dfrac{7}{10}\)
1) Tính:
\(b,\sqrt{2}.\sqrt{7+3\sqrt{5}}-\dfrac{4}{\sqrt{5-1}}\)
\(c,\sqrt{27}-6\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\)
\(d,\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}\)
Giúp mình với, mình cần gấp
\(b,\sqrt{2}.\sqrt{7+3\sqrt{5}}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{14+6\sqrt{5}}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{\sqrt{5^2}+2.3\sqrt{5}+3^2}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{\left(\sqrt{5}+3\right)^2}-\dfrac{4}{\sqrt{5}-1}\\ =\left|\sqrt{5}+3\right|-\dfrac{4}{\sqrt{5}-1}\\ =\dfrac{\left(\sqrt{5}+3\right)\left(\sqrt{5}-1\right)-4}{\sqrt{5}-1}\\ =\dfrac{2+2\sqrt{5}-4}{\sqrt{5}-1}\\ =\dfrac{-2+2\sqrt{5}}{\sqrt{5}-1}\\ =\dfrac{2\left(-1+\sqrt{5}\right)}{\sqrt{5}-1}\\ =2\)
\(c,\sqrt{27}-6\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\\ =3\sqrt{3}-\dfrac{6}{\sqrt{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\)
\(=\dfrac{3\sqrt{3}.\sqrt{3}-6+\sqrt{3}-3}{\sqrt{3}}\\ =\dfrac{9-6+\sqrt{3}-3}{\sqrt{3}}\\ =\dfrac{\sqrt{3}}{\sqrt{3}}\\ =1\)
\(d,\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}\\ =\dfrac{\left(9-2\sqrt{3}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{\left(3\sqrt{6}-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}\\ =\dfrac{27\sqrt{6}+18\sqrt{2}-18\sqrt{2}-4\sqrt{6}}{\left(3\sqrt{6}\right)^2-\left(2\sqrt{2}\right)^2}\\ =\dfrac{23\sqrt{6}}{54-8}\\ =\dfrac{23\sqrt{6}}{46}\\ =\dfrac{\sqrt{6}}{2}\)
Câu b á bạn, chỗ \(\dfrac{4}{\sqrt{5-1}}\) là đề như vậy hay là \(\dfrac{4}{\sqrt{5}-1}\) vậy?