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\(125^7-25^{10}+5^{19}\)
\(=\left(5^3\right)^7-\left(5^2\right)^{10}+5^{19}\)
\(=5^{21}-5^{20}+5^{19}\)
\(=5^{19}.\left(5^2-5+1\right)\)
\(=5^{19}.21\)
\(=5^{18}.5.21\)
\(=5^{18}.105\)
Ta có: \(105⋮105\)
\(\Rightarrow5^{18}.105⋮105\)
\(\Rightarrow125^7-25^{10}+5^{19}⋮105\)
đpcm
\(125^7-25^{10}+5^{19}\)
\(=\left(5^3\right)^7-\left(5^2\right)^{10}+5^{19}\)
\(=5^{21}-5^{20}+5^{19}\)
\(=5^{19}.\left(5^2-5+1\right)\)
\(=5^{19}.21\)
\(=5^{18}.5.21=5^{18}.105⋮105\)
Vậy ......
Câu 1:
\(=\dfrac{5}{4}\left(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{4n-1}-\dfrac{1}{4n+3}\right)\)
\(=\dfrac{5}{4}\left(\dfrac{1}{3}-\dfrac{1}{4n+3}\right)\)
\(=\dfrac{5}{4}\cdot\dfrac{4n+3-3}{3\left(4n+3\right)}=\dfrac{5}{4}\cdot\dfrac{4n}{3\left(4n+3\right)}=\dfrac{5n}{3\left(4n+3\right)}\)
Câu 2:
\(=\dfrac{3}{5}\left(\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{5n-1}-\dfrac{1}{5n+4}\right)\)
\(=\dfrac{3}{5}\left(\dfrac{1}{9}-\dfrac{1}{5n+4}\right)\)
\(=\dfrac{3}{5}\cdot\dfrac{5n+4-9}{9\left(5n+4\right)}=\dfrac{3}{5}\cdot\dfrac{5\left(n-1\right)}{9\left(5n+4\right)}=\dfrac{n-1}{3\left(5n+4\right)}< \dfrac{1}{15}\)
Gọi số đó là ab
ta có ab = a.10 + b = 3a + 7b + b
vì 7b chia hết cho 7 => để 3a + 7a + b chia hêt cho 7
=> 3a + b chia hêt cho 7
=> 3a + b + 14b chia hêt cho 7
=> 3a + 15b chia hêt cho 7
=> 3 ( a + 5b ) chia hêt cho 7
mà 3 ko chia hêt cho 7 => a + 5b chia hêt cho 7 ( đpcm )
Gọi số đó là ab (không phải là a.b đâu, đành phải chuyển dấu nhân thành dấu x)
\(ab=a\times10+b=7a+3a+b⋮7\)
\(\Leftrightarrow3a+b⋮7\)
\(\Leftrightarrow3a+b+14b⋮7\)
\(\Leftrightarrow3a+15b⋮7\)
\(\Leftrightarrow3\left(a+5b\right)⋮7\left(1\right)\)
Vì UCLN(3;7) = 1
\(\Rightarrow\left(1\right)\Leftrightarrow a+5b⋮7\)
XONG RỒI ĐÓ BẠN.
2n-3 \(⋮\) n+1
=> 2n+2-5 \(⋮\) n+1
=> 2(n+1)-5 \(⋮\) n+1
Mà 2(n+1) \(⋮\) n+1
=> 5 \(⋮\) n+1
=> n+1 ∈ Ư(5)
mà Ư(5) ∈ {1;-1;5;-5}
⇒ n+1 ∈ {1;-1;5;-5}
TH1: n+1=1 => n=0 ∈ Z
TH2: n+1=-1 => n=-2 ∈ Z
TH3: n+1=5 => n= 4 ∈ Z
TH4: n+1=-5 => n= -6∈ Z
=> n ∈ {0;-2;4;6}
\(B=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+\frac{7}{3^2.4^2}+....+\frac{19}{9^2.10^2}\)
\(B=\frac{3}{1.4}+\frac{5}{4.9}+\frac{7}{9.16}+....+\frac{19}{81.100}\)
\(B=\frac{4-1}{1.4}+\frac{9-4}{4.9}+\frac{16-9}{9.16}+....+\frac{100-81}{81.100}\)
\(B=\frac{4}{1.4}-\frac{1}{1.4}+\frac{9}{4.9}-\frac{4}{4.9}+\frac{16}{9.16}-\frac{9}{9.16}+...+\frac{100}{81.100}-\frac{81}{81.100}\)
\(B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+....+\frac{1}{81}-\frac{1}{100}\)
\(B=1-\frac{1}{100}< 1\)
=> B < 1 (Đpcm)
B = 3/12.22 + 5/22.32 + 7/32.42 + ... + 19/92.102
B = 3/1.4 + 5.4.9 + 7/9.16 + ... + 19/81.100
B = 1 - 1/4 + 1/4 - 1/9 + 1/9 - 1/16 + ... + 1/81 - 1/100
B = 1 - 1/100 < 1 ( đpcm)
