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\(\frac{3}{5.11}+\frac{5}{11.21}+\frac{7}{21.35}+\frac{9}{35.53}=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{11}+\frac{1}{11}-\frac{1}{21}+\frac{1}{21}-\frac{1}{35}+\frac{1}{35}-\frac{1}{53}\right)=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{53}\right)=\frac{1}{10}-\frac{1}{106}
3/5 . 11/7 + 3/5 . (- 4/7) + 2/5
= 3/5 . [11/7 + (- 4/7)] + 2/5
= 3/5 . 1 + 2/5
= 3/5 + 2/5
= 1
\(\dfrac{3^{11}.11+3^{11}.21}{3^9.2^5}\)
= \(\dfrac{3^{11}.\left(11+21\right)}{3^932}\)
= \(\dfrac{3^{11}32}{3^9.32}\)
= 32
= 9
a \(\dfrac{-2}{5}\)
b -1
c \(\dfrac{2}{5}\)
d \(\dfrac{-18}{11}\)
7x-3.4=196
7x-3=49
x-3=2
x= -1
2x+3+2x=36
2x.9=36
2x=4
x=2
(52x.5x+2):25=1255
53x+2=517
x=5
\(P=\frac{3^{11}\cdot11+3^{11}\cdot21}{3^9\cdot2^5}=\frac{3^{11}\cdot\left(11+21\right)}{3^9\cdot32}=\frac{3^{11}\cdot32}{3^9\cdot32}=3^2=9\)
a.
\(\frac{11^4\times6-11^5}{11^4-11^5}=\frac{11^4\times\left(6-11\right)}{11^4\times\left(1-11\right)}=\frac{-5}{-10}=\frac{1}{2}\)
b.
\(\frac{9^8\times3-3^{18}}{9^8\times5+9^8\times7}=\frac{9^8\times3-\left(3^2\right)^9}{9^8\times\left(5+7\right)}=\frac{9^8\times3-9^9}{9^8\times12}=\frac{9^8\times\left(3-9\right)}{9^8\times12}=-\frac{6}{12}=-\frac{1}{2}\)
c.
\(\frac{10^5-10^5\times3}{10^5\times11}=\frac{10^5\times\left(1-3\right)}{10^5\times11}=-\frac{2}{11}\)
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