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a)\(\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.....\frac{2499}{2500}\) \(=\frac{2.4}{3^2}.\frac{3.5}{4^2}.\frac{4.6}{5^2}.....\frac{49.51}{50^2}\)\(=\frac{\left(2.3.4.....49\right)\left(4.5.6....51\right)}{\left(3.4.5.....50\right)\left(3.4.5.....50\right)}=\frac{2.51}{50.3}=\frac{1.17}{25}=\frac{17}{25}\)
b)\(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\left(1-\frac{1}{10}\right)......\left(1-\frac{1}{780}\right)\)
\(=\frac{2}{3}.\frac{5}{6}.\frac{9}{10}......\frac{779}{780}\)\(=\frac{4}{6}.\frac{10}{12}.\frac{18}{20}.....\frac{1558}{1560}\)
\(=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.....\frac{38.41}{39.40}\)
\(=\frac{\left(1.2.3.....38\right)\left(4.5.6.....41\right)}{\left(2.3.4.....39\right)\left(3.4.5.....40\right)}=\frac{1.41}{39.3}=\frac{41}{117}\)
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\(\text{ta có: }\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3}\)
\(\frac{1}{3.4}=\frac{1}{3}-\frac{1}{4}\)
\(\frac{1}{4.5}=\frac{1}{4}-\frac{1}{5}\)
...........................
\(\frac{1}{39.40}=\frac{1}{39}-\frac{1}{40}\)
Đồng nhất 2 vế ta có:
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{39}-\frac{1}{40}=\frac{1}{2}-\frac{1}{40}=\frac{19}{40}\)
Ta có :
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{39.40}\)
\(=\)\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{39}-\frac{1}{40}\)
\(=\)\(\frac{1}{2}-\frac{1}{40}\)
\(=\)\(\frac{20}{40}-\frac{1}{40}\)
\(=\)\(\frac{19}{40}\)
Vậy \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{39.40}=\frac{19}{40}\)
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\(S=1^2+2^2+3^2+...+30^2\)
<=>\(S=1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+30\left(31-1\right)\)
<=>\(S=1.2-1+2.3-2+3.4-3+...+30.31-30\)
<=>\(S=\left(1.2+2.3+3.4+...+30.11\right)-\left(1+2+3+...+30\right)\)
Đặt \(A=1.2+2.3+3.4+...+30.11\) và \(B=1+2+3+...+30\)
Thôi đến bước này bạn tự tính đi, dễ rồi, ghi hết thì ... mình mệt =)))
được: \(A=\frac{30.31.32}{3}=10.31.32=9920;B=\frac{30.31}{2}=15.31=465\)
=>S=9920+465=10385 không phải số chính phương
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a) Để â nhận giá trị nguyên
\(\Rightarrow8n-9⋮2n+5\)
\(\Rightarrow8n+20-29⋮2n+5\)
\(\Rightarrow4.\left(2n+5\right)-29⋮2n+5\)
mà \(4.\left(2n+5\right)⋮2n+5\)
\(\Rightarrow-29⋮2n+5\)
\(\Rightarrow2n+5\inƯ\left(-29\right)\)
tự làm nốt nhé, tick nha
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2n+1=2(n-6)+12+1=2(n-6)+13 chia hết cho n-6
=> 13 chia hết cho n-6
=> n-6 thuộc Ư(13)={1;13}
=> n-6=1 hoặc n-6=13
n=7 n=19
=> n thuộc {7;19}