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Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\)
\(\Leftrightarrow\)\(3A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\)
\(\Leftrightarrow\)\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^8}\right)\)
\(\Leftrightarrow\)\(2A=1-\frac{1}{3^8}\)
\(\Leftrightarrow\)\(2A=\frac{3^8-1}{3^8}\)
\(\Leftrightarrow\)\(A=\frac{3^8-1}{3^8}:2\)
\(\Leftrightarrow\)\(A=\frac{3^8-1}{3^8}.\frac{1}{2}\)
\(\Leftrightarrow\)\(A=\frac{3^8-1}{2.3^8}\)
b) Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+.....+\frac{1}{3^8}\)
\(\Rightarrow\)\(3A=1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^7}\)
\(\Rightarrow\)\(3A-A=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^7}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+....+\frac{1}{3^8}\right)\)
\(\Rightarrow\)\(2A=1-\frac{1}{3^8}\)
\(\Rightarrow\)\(A=\frac{1-\frac{1}{3^8}}{2}\)
a, ( -18 + 25 ) - ( 125 - 18 + 25 )
= -18+25-125+18-25
= (-18+18) +(25-25) -125
= 0+0-125
= -125
b, 45 . ( 18 - 15 ) -18 . ( 45 + 45 )
= 45.18 - 45.15 -18.45 +18.45
= 45(18-15-18+18)
= 45.3=135
a ) = -18 + 25 - 125 + 18 - 25
= (-18 + 18 ) - ( 25 -25 ) + 125
= 0 - 0 + 125
=125
a,
85.(35-27)-35.(85-27)
=85.35-85.27-35.85-35.27
=85.35-35.85-85.27.35.27
=85.27-35.27
=(85-35).27
=100.27/2
=2700/2
=1350
a: 2015 x 102
= 2015 x ( 100 + 2 )
=2015 x 100 + 2015 x 2
=201500 + 4030
=205530
b: 998 x 45
=(1000 - 2 ) x 45
=1000 x 45 - 2 x 45
=45000 - 90
=44910
a 2015 x 102 = 2015 x (100 + 2) = 2015 x100 + 2015 x 2 = 201500 + 4030= 205530
b 998 x 45 = (1000 - 2) x 45= 45 x 1000 - 45 x 2= 45000- 90 = 45910
\(\frac{45.86+45.15-45}{50.37-50.5+50.58}\)
\(=\frac{45.\left(86+15-1\right)}{50.\left(37-5+58\right)}\)
\(=\frac{45.100}{50.90}\)
\(=1\)
học tốt
A = \(\frac{45x\left(86+15-1\right)}{50x\left(37-5+58\right)}\)
A=\(\frac{45x100}{50x90}\)= \(\frac{2}{2}\)= 1
(Cái phần cuối thì bạn rút họn 45 với 90, 50 với 100)