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a) \(\dfrac{8,5-8,2}{16}=\dfrac{0,3}{16}=\dfrac{0,3\cdot10}{16\cdot10}=\dfrac{3}{160}\)
b) \(\dfrac{17\cdot5-17}{3-20}=\dfrac{17\cdot\left(5-1\right)}{-17}=\dfrac{1\cdot4}{-1}=-4\)
a)
\(\dfrac{8\cdot5-8\cdot2}{16}=\dfrac{8\left(5-2\right)}{16}=\dfrac{3}{2}\)
b)
\(\dfrac{17\cdot5-17}{3-20}=\dfrac{17\left(5-1\right)}{-17}=\dfrac{4}{-1}=-4\)
q=1/3; u1=2/3
\(S_{100}=\dfrac{\dfrac{2}{3}\cdot\left(\dfrac{1}{3^{100}}-1\right)}{\dfrac{1}{3}-1}=-\dfrac{1}{3^{100}}+1=\dfrac{-1+3^{100}}{3^{100}}\)
a. \(\dfrac{8,5-8,2}{16}=\dfrac{0,3}{16}=\dfrac{3}{160}\)
b. \(\dfrac{2\cdot14}{7\cdot8}=\dfrac{1\cdot2}{1\cdot4}=\dfrac{2}{4}=\dfrac{1}{2}\)
c. \(\dfrac{11\cdot4-11}{2-13}=\dfrac{11\left(4-1\right)}{-11}=\dfrac{1\cdot3}{-1}=-3\)
d. \(\dfrac{49+7\cdot49}{49}=\dfrac{49\cdot\left(1+7\right)}{49}=\dfrac{8}{1}=8\)
1) \(\frac{3^{10}+6^2}{5\cdot3^8+20}=\frac{3^{10}+3^2\cdot2^2}{5\cdot3^8+5\cdot2^2}=\frac{3^2\left(3^8+2^2\right)}{5\left(3^8+2^2\right)}=\frac{9}{5}\)
2) \(\frac{28^{15}\cdot3^{17}}{84^{16}}=\frac{28^{15}\cdot3^{17}}{28^{16}\cdot3^{16}}=\frac{3}{28}\)
\(=\dfrac{3^4\left(5-9\right)}{3^4\left(13+1\right)}=\dfrac{-4}{14}=\dfrac{-2}{7}\)
Lời giải:
$A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2023}}$
$2A=2+1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2022}}$
$2A-A=2-\frac{1}{2^{2023}}$
$A=2-\frac{1}{2^{2023}}$
6-(y+15)+17
=6-y-15+17
=6-15+17-y
=8-y
vậy kết quả là 8-y
k cho mik zới
\(A=\frac{2^{15}.3^{12}-3^{11}.2^{17}}{2^{15}.3^{11}+3^{11}.2^{17}}\)
\(A=\frac{2^{15}.3^{11}.\left(3-2^2\right)}{2^{15}.3^{11}.\left(1+2^2\right)}\)
\(A=\frac{3-2^2}{1+2^2}\)
\(A=\frac{-1}{5}\)
-158/17
=-158/17