(5mu 15.18+5mu15.7):5mu17
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\(5^{100}-5^{99}-5^{98}=5^{98}\left(5^2-5-1\right)=5^{98}\cdot19=5^{97}\cdot5\cdot19=5^{97}\cdot95⋮95\left(đpcm\right)\)
5100 - 599 - 598
= 598 (52-5-1)
=598 . 19
=597 .5 . 19
=597 .95
vì 95 chia hết cho 95
=> 5100 - 599 - 598 chia hết cho 95
\(P=1+5+5^2+..........+5^{2016}\)
\(\Leftrightarrow5P=5+5^2+5^3+............+5^{2017}\)
\(\Leftrightarrow5P-P=\left(5+5^2+..........+5^{2017}\right)-\left(1+5+..........+5^{2016}\right)\)
\(\Leftrightarrow4P=5^{2017}-1\)
\(\Leftrightarrow P=\dfrac{5^{2017}-1}{4}\)
Mà \(Q=\dfrac{5^{2017}}{4}\)
\(\Leftrightarrow Q-P=\dfrac{5^{2017}}{4}-\dfrac{5^{2017}-1}{4}=\dfrac{5^{2017}-5^{2017}-1}{4}=-\dfrac{1}{4}\)
\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{15}-\dfrac{1}{18}\right)\)
\(=\dfrac{1}{3}\left(\dfrac{1}{3}-\dfrac{1}{18}\right)=\dfrac{1}{3}\cdot\dfrac{5}{18}=\dfrac{5}{54}\)
\(\frac{6}{15.18}+\frac{6}{18.21}+\frac{6}{21.24}+...+\frac{6}{87.90}\)
\(\rightarrow\frac{6}{3}.\left(\frac{3}{15.18}+\frac{3}{18.21}+...+\frac{3}{87.90}\right)\)
\(\rightarrow2.\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{87}-\frac{1}{90}\right)\)
\(\rightarrow2.\left(\frac{1}{15}-\frac{1}{90}\right)\)
\(\rightarrow\frac{1}{9}\)
A = 6/3 . ( 1/15.18 + 1/18.21 + 1/21/24 + . . . + 1/87.90 )
A = 6/3 . ( 1/15 - 1/18 + 1/18 - 1/21 + 1/21 - 1/24 + . . . + 1/87 - 1/90 )
A = 2 . ( 1/15 - 1/90 )
A = 2. 5/90
A = 10/90 = 1/9
\(\frac{6}{15.18}+\frac{6}{18.21}+\frac{6}{21.24}+...+\frac{6}{84.87}+\frac{6}{87.90}\)
\(=\frac{6}{3}\left(\frac{3}{15.18}+\frac{3}{18.21}+\frac{3}{21.24}+...+\frac{3}{84.87}+\frac{3}{87.90}\right)\)
\(=2\left(\frac{1}{15}-\frac{1}{18}+\frac{1}{18}-\frac{1}{21}+...+\frac{1}{84}-\frac{1}{87}+\frac{1}{87}-\frac{1}{90}\right)\)
\(=2\left(\frac{1}{15}-\frac{1}{90}\right)=2\left(\frac{6-1}{90}\right)=2\times\frac{1}{18}=\frac{1}{9}\)
( 515 . 18 + 515 . 7 ) : 517
= [515 . ( 18 + 15 ) ] : 517
= ( 5 15 . 33 ) : 517
= ( 517 : 515 ) . 33
= 52 .33
= 25 . 33
= 825
khong biet lam