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Mình làm bài a thôi nhé:
a)7^11 x 7^13 x 7^17= 7^11+13+17=7^41=>7^2 x 7^39=>49 x 7^39
=>49 x7^39 chia hết cho 49
k mình nhé
Ta có :
\(\frac{1}{3^2}< \frac{1}{2.3}\)
\(\frac{1}{4^2}< \frac{1}{3.4}\)
............
\(\Rightarrow\frac{1}{100^2}< \frac{1}{99.100}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{100^2}< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{49}{100}\)
Mà \(\frac{49}{100}< \frac{1}{2}\)
\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{2}\)
\(\RightarrowĐPCM\)
\(a)\) Ta có :
\(\frac{1}{100}A=\frac{100^{2009}+1}{100^{2009}+100}=\frac{100^{2009}+100}{100^{2009}+100}-\frac{99}{100^{2009}+100}=1-\frac{99}{100^{2009}+100}\)
\(\frac{1}{100}B=\frac{100^{2010}+1}{100^{2010}+100}=\frac{100^{2010}+100}{100^{2010}+100}-\frac{99}{100^{2010}+100}=1-\frac{99}{100^{2010}+100}\)
Vì \(\frac{99}{100^{2009}+100}>\frac{99}{100^{2010}+100}\) nên \(1-\frac{99}{100^{2009}+100}< 1-\frac{99}{100^{2010}+100}\)
Do đó :
\(\frac{1}{100}A< \frac{1}{100}B\)\(\Rightarrow\)\(A< B\)
Vậy \(A< B\)
Chúc bạn học tốt ~
a) Giải
Đặt \(M=\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{98}{99}\)
\(\Rightarrow A< A.M\)
hay \(A< \left(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\right).\left(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{98}{99}\right)\)
\(\Rightarrow A< \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.\dfrac{5}{6}.\dfrac{6}{7}...\dfrac{98}{99}.\dfrac{99}{100}\)
\(\Leftrightarrow A< \dfrac{1.2.3.4.5.6...98.99}{2.3.4.5.6.7...99.100}\)
\(\Rightarrow A< \dfrac{1}{100}< \dfrac{1}{10}\)
Vậy \(A< \dfrac{1}{10}\)
B = ( 1 + 2 ) + ( 2^2 + 2^3 ) + ( 2^4 + 2^5 ) + ... + ( 2^99 + 2^100 )
B = 1( 1 + 2 ) + 2^2( 1 + 2 ) + 2^4( 1 + 2 ) + ... + 2^99( 1 + 2)
B = 1 . 3 + 2^2 . 3 + 2^4 . 3 + ... + 2^99 . 3
B = 3( 1 + 2^2 + 2^4 + ... + 2^99 )
=> B chia hết cho 3
B = ( 1 + 2 + 2^2 ) + ( 2^3 + 2^4 + 2^5 ) + ... + ( 2^98 + 2^99 + 2^100 )
B = 1( 1 + 2 + 2^2 ) + 2^3( 1 + 2 + 2^2 ) + ... + 2^98( 1 + 2 + 2^2 )
B = 1.7 + 2^3.7 + ... + 2^98.7
B = 7( 1 + 2^3 + ... + 2^98 )
=> B chia hết cho 7
B = ( 1 + 2 + 2^2 + 2^3 ) + ( 2^4 + 2^5 + 2^6 + 2^7 ) + ... ( 2^97 + 2^98 + 2^99 + 2^100 )
B = 1( 1 + 2 + 2^2 + 2^3 ) + 2^4( 1 + 2 + 2^2 + 2^3 ) + ... + 2^97( 1 + 2 + 2^2 + 2^3 )
B = 1.15 + 2^4.15 + ... + 2^98.15
B = 15( 1 + 2^4 + ... + 2^98 )
=> B chia hết cho 15
Mà 15 = 3 . 5
=> B chia hết cho 5
a)\(B=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{99}+2^{100}\right)\)
\(B=\left(1+2\right)+2^2\left(1+2\right)+...+2^{99}\left(1+2\right)\)
\(B=3+2^2.3+...+2^{99}.3\)
\(B=3\left(1+2^2+...+2^{99}\right)⋮3\left(đpcm\right)\)
b)\(B=\left(1+2+2^2\right)+...+\left(2^{98}+2^{99}+2^{100}\right)\)
\(B=\left(1+2+2^2\right)+...+2^{98}\left(1+2+2^2\right)\)
\(B=7+....+2^{98}.7\)
\(B=7\left(1+...+2^{98}\right)⋮7\left(đpcm\right)\)
c)\(B=\left(1+2+2^2+2^3\right)+...+\left(2^{97}+2^{98}+2^{99}+2^{100}\right)\)
\(B=\left(1+2+2^2+2^3\right)+...+2^{97}\left(1+2+2^2+2^3\right)\)
\(B=15+...+2^{97}.15\)
\(B=15\left(1+...+2^{97}\right)⋮5\left(đpcm\right)\)