5/7 x 6/11 + 5/11 x 1/7 - 5/7 x 14/11
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\(\dfrac{5}{7}\cdot\dfrac{6}{11}+\dfrac{5}{11}\cdot\dfrac{1}{7}-\dfrac{5}{7}\cdot\dfrac{14}{11}\)
\(=\dfrac{5}{7}\cdot\dfrac{6}{11}+\dfrac{5}{7}\cdot\dfrac{1}{11}-\dfrac{5}{7}\cdot\dfrac{14}{11}\)
\(=\dfrac{5}{7}\cdot\left(\dfrac{6}{11}+\dfrac{1}{11}-\dfrac{14}{11}\right)\)
\(=\dfrac{5}{7}\cdot\dfrac{-7}{11}=\dfrac{-5}{11}\)
`(x-1)^3=1/8`
<=> `(x-1)^3=(1/2)^3`
<=> `x-1=1/2`
<=> `x=1/2` `+ 1`
<=> `x=3/2`
`#3107.101107`
\(\dfrac{27^2\cdot2^3\cdot5^4}{15^2\cdot6^9}\)
\(=\dfrac{\left(3^3\right)^2\cdot2^3\cdot5^4}{3^2\cdot5^2\cdot2^9\cdot3^9}\)
\(=\dfrac{3^6\cdot2^3\cdot5^4}{3^{11}\cdot5^2\cdot2^9}\)
\(=\dfrac{5^2}{3^5\cdot2^6}\)
\(=\dfrac{25}{15552}\)
Đây nha:
\(\left(x-\dfrac{1}{8}\right)^3=-\dfrac{8}{125}\)
\(\left(x-\dfrac{1}{8}\right)^3=\left(\dfrac{-2}{5}\right)^3\)
\(\Rightarrow x-\dfrac{1}{8}=\dfrac{-2}{5}\)
\(x=\dfrac{-2}{5}+\dfrac{1}{8}=\dfrac{-16}{40}+\dfrac{5}{40}\)
\(x=\dfrac{-11}{40}\)
\(#WendyDang\)
\(\left(x-\dfrac{1}{3}\right)^3=\dfrac{-8}{125}\)
\(\left(x-\dfrac{1}{3}\right)^3=\left(\dfrac{-2}{5}\right)^3\)
\(\Rightarrow x-\dfrac{1}{3}=\dfrac{-2}{5}\)
\(x=\dfrac{-2}{5}+\dfrac{1}{3}=\dfrac{-6}{15}+\dfrac{5}{15}\)
\(x=\dfrac{-1}{15}\)
\(\dfrac{4^5\cdot9^4}{8^3\cdot27^3}=\dfrac{\left(2^2\right)^5\cdot\left(3^2\right)^4}{\left(2^3\right)^3\cdot\left(3^3\right)^3}=\dfrac{2^{10}\cdot3^8}{2^9\cdot3^9}=\dfrac{2}{3}\)
\(\dfrac{4^{20}\cdot3^{35}}{2^{37}\cdot27^{12}}=\dfrac{\left(2^2\right)^{20}\cdot3^{35}}{2^{37}\cdot\left(3^3\right)^{12}}=\dfrac{2^{40}\cdot3^{35}}{2^{37}\cdot3^{36}}=\dfrac{2^3}{3}\)
\(\dfrac{5^4\cdot20^4}{25^5\cdot4^5}=\dfrac{5^4\cdot5^4\cdot4^4}{5^5\cdot5^5\cdot4^5}=\dfrac{1}{5^2\cdot4}=\dfrac{1}{100}\)
\(\dfrac{2^{15}\cdot9^4}{6^6\cdot8^3}=\dfrac{2^{15}\cdot\left(3^2\right)^4}{2^6\cdot3^6\cdot\left(2^3\right)^3}=\dfrac{2^{15}\cdot3^8}{2^6\cdot3^6\cdot2^9}=3^2\)
A = 0,(36) - 0,(71)
= 36/99 - 71/99
= -35/99
B = [3,(12) - 2,(5)] : 0,(14)
= (103/33 - 203/99) : 14/99
= 106/99 : 14/99
= 53/7
\(A=0,\left(36\right)-0,\left(71\right)\)
\(A=\dfrac{36}{99}-\dfrac{71}{99}=-\dfrac{35}{99}=-0,\left(35\right)\)
\(B=\left[3,\left(12\right)-2,\left(5\right)\right]:0,\left(14\right)\)
\(B=\left(\dfrac{3\cdot99+12}{99}-\dfrac{2\cdot9+5}{9}\right):\dfrac{14}{99}\)
\(B=\left(\dfrac{309}{99}-\dfrac{23}{9}\right):\dfrac{14}{99}\)
\(B=\dfrac{56}{99}:\dfrac{14}{99}\)
\(B=4\)
`#3107.101107`
\(\dfrac{5}{7}\times\dfrac{6}{11}+\dfrac{5}{11}\times\dfrac{1}{7}-\dfrac{5}{7}\times\dfrac{14}{11}\\ =\dfrac{5}{7}\times\dfrac{6}{11}+\dfrac{5}{7}\times\dfrac{1}{11}-\dfrac{5}{7}\times\dfrac{14}{11}\\ =\dfrac{5}{7}\times\left(\dfrac{6}{11}+\dfrac{1}{11}-\dfrac{14}{11}\right)\\ =\dfrac{5}{7}\times\left(-\dfrac{7}{11}\right)\\ =-\dfrac{5}{11}\)