1. tìm tích của A= \(\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\times..\times\frac{899}{900}\)
2. CMR \(\frac{1}{5}+\frac{1}{6}+..+\frac{1}{17}< 2\)
3. tính \(M=\frac{1}{1.2.3}+\frac{1}{2.3.4}+..+\frac{1}{10.11.12}\)
3. \(M=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{10.11.12}\)
\(\Leftrightarrow2M=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{10.11.12}\)
\(\Leftrightarrow2M=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{10.11}-\frac{1}{11.12}\)
\(\Leftrightarrow2M=\frac{1}{1.2}-\frac{1}{11.12}\)
\(\Leftrightarrow2M=\frac{1}{2}-\frac{1}{132}\)
\(\Leftrightarrow2M=\frac{65}{132}\)
\(\Leftrightarrow M=\frac{65}{132}\div2\)
\(\Leftrightarrow M=\frac{65}{264}\)
1\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{899}{900}\)
\(\Leftrightarrow A=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{29.31}{30.30}\)
\(\Leftrightarrow A=\frac{1.3.2.4.3.5...29.31}{2.2.3.3.4.4...30.30}\)
\(\Leftrightarrow A=\frac{\left(1.2.3....29\right)\left(3.4.5...31\right)}{\left(2.3.4...30\right)\left(2.3.4...30\right)}\)
\(\Leftrightarrow A=\frac{1.31}{30.2}\)
\(\Leftrightarrow A=\frac{31}{60}\)
2. Đặt \(A=\frac{1}{5}+\frac{1}{6}+...+\frac{1}{17}\)
\(\Rightarrow A< \frac{1}{5}+\frac{1}{5}+...+\frac{1}{5}+\frac{1}{8}+\frac{1}{8}+...+\frac{1}{8}\)
\(\Rightarrow A< 1+1=2\)
Vậy a < 2 (đpcm)
Bài 1 : Chứng tỏ rằng : \(\frac{14n+3}{21n+5}\) là phân số tối giản với mọi n ϵ Z
Bài 2 Tìm x , biết
30 : \(\left(\frac{1}{4}x+\frac{3}{4}x\right)^2=\frac{5}{6}\)
Bài 3 Tính tích : A= \(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{899}{900}\)
gọi UCLN ( 14n+ 3 ; 21n +5 ) là d
=> 14n+ 3⋮d và 21n +5⋮d
=> 42n + 9⋮d và 42n + 10⋮d
=> 42n + 10 - (42n + 9) ⋮ d
=> 42n + 10 - 42n - 9⋮ d
=> 1⋮ d
=> p/s ...là phân số tối giản
1) Để phân số \(\frac{14n+3}{21n+5}\) là PSTG thì
ƯC(14n+3, 21n+5)={-1,1}
Gọi d là UC của 14n+3 và 21n+5
⇒14n+3⋮d
21n+5⋮d
⇒3(14n+3)⋮d
2(21n+5)⋮d
⇒42n+9⋮d
42n+10⋮d
⇒42n+9-(42n+10)⋮d
⇒42n+9-42n-10⋮d
⇒-1⋮d
⇒d={1, -1)
⇒ƯC(14n+3, 21n+5)={-1,1}
Vậy phân số................
2)\(\text({\frac{1}{4}.x+\frac{3}{4}.x})^{2}\)=\(\frac{5}{6}\)
⇒\(\text((\frac{1}{4}+\frac{3}{4}).x)^2=\frac{5}{6}\)
⇒\(\text{(1x)}^2\)=\(\frac{5}{6}\)
⇒x=....(mình ko tính dc)
Vậy x∈ϕ
3) A=\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{899}{900}\)
=\(\frac{3.8.15...899}{4.9.16...900}\)
=\(\frac{1.3.2.4.3.5...29.31}{2.2.3.3.4.4...30.30}\)
=\(\frac{1.2.3...29}{2.3.4...30}.\frac{3.4.5....31}{2.3.4...30}\)
=\(\frac{1}{30}.\frac{31}{2}\)
=\(\frac{31}{60}\)
A. \(\frac{3}{4}\) x \(\frac{8}{9}\)x \(\frac{15}{16}\)x .... x \(\frac{899}{900}\)
= \(\frac{1.3}{2^2}\) x \(\frac{2.4}{3^3}\)x \(\frac{3.5}{4^2}\)x ... x \(\frac{29.31}{30^2}\)
= \(\left(\frac{1.2.3...29}{2.3.4...30}\right).\left(\frac{3.4.5...31}{2.3.4...30}\right)\)
= \(\frac{1}{30}.\frac{31}{2}\)= \(\frac{31}{60}\)
B.
\(\frac{1}{3}+\frac{3}{8}-\frac{7}{12}=\frac{8}{24}+\frac{9}{24}-\frac{14}{24}=\frac{8+9-14}{24}=\frac{3}{24}=\frac{1}{8}\)
Tính tích:A=\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.......\frac{899}{900}\)\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.........\frac{899}{900}\)
A=\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.........\frac{899}{900}\)
A=\(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}..........\frac{29.31}{30.30}\)
A=\(\frac{1.2.3.......29}{2.3.4.......30}.\frac{3.4.5........31}{2.3.4.......30}\)
A=\(\frac{1}{30}.\frac{2}{31}=\frac{1}{465}\)
bài 1 : tính phân số:
a) \(\frac{5}{7}+\frac{4}{9}=?;\frac{4}{5}-\frac{2}{3}=?;\frac{9}{11}+\frac{3}{8}=?;\frac{16}{25}-\frac{2}{5}=?\)=?
b)\(5+\frac{3}{5}=?;10-\frac{9}{16}=?;\frac{2}{3}-\left(\frac{1}{6}+\frac{1}{8}\right)=?\)
c)\(\frac{5}{7}+\frac{7}{6}=?;\frac{7}{12}+\frac{17}{18}=?;\frac{9}{8}+\frac{15}{32}=?;4+\frac{35}{45}=?\)
d)\(\frac{11}{4}-\frac{15}{16}=?;\frac{5}{6}-\frac{5}{8}=?;\frac{196}{64}-2=?;3-\frac{13}{9}=?\)
e)\(\frac{8}{5}+\frac{7}{6}+\frac{5}{9}-2=?;3-\frac{5}{6}-\frac{4}{9}+\frac{32}{24}=?\)
a)\(\dfrac{5}{7}+\dfrac{4}{9}=\dfrac{45}{63}+\dfrac{28}{63}=\dfrac{73}{63}\) ; \(\dfrac{9}{11}+\dfrac{3}{8}=\dfrac{72}{88}+\dfrac{33}{88}=\dfrac{105}{88}\)
\(\dfrac{4}{5}-\dfrac{2}{3}=\dfrac{12}{15}-\dfrac{10}{15}=\dfrac{2}{15}\); \(\dfrac{16}{25}-\dfrac{2}{5}=\dfrac{16}{25}-\dfrac{10}{25}=\dfrac{6}{25}\)
a)Tính\(\frac{\left(17\frac{8}{19}-16\frac{9}{18}\right)\left(17,5+16\frac{17}{51}-32\frac{15}{22}\right)}{\frac{7}{3.13}+\frac{7}{13.23}+\frac{7}{23.33}}\)
b) Chứng tò rằng:\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{199}-\frac{1}{200}\)\(=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\)
phần a dễ bạn tự làm đi tử thì bạn tính như bình thường còn mẫu thì:7.(\(\frac{1}{3.13}\)+\(\frac{1}{13.23}\)+\(\frac{1}{23.33}\))
\(\frac{7}{10}\).(\(\frac{1}{3}\)-\(\frac{1}{33}\))=\(\frac{7}{33}\)
b)(1+1/3+1/5+..+1/199)-(1/2+1/4+...+1/200)
(1+1/2+1/3+...+1/199+1/200)-(1/2+1/2+1/4+1/4+...+1/200+1/200)
=1+1/2+1/3+...+1/199+1/200-(1+1/2+1/3+...+1/100)
=1/101+1/102+...+1/200
https://olm.vn/hoi-dap/question/60726.html
câu1: P=(1-\(\frac{1}{2}\)).(1-\(\frac{1}{3}\)).(1-\(\frac{1}{4}\))....(1-\(\frac{1}{99}\))
câu 2: chứng tỏ rằng \(\frac{1}{101}\)+\(\frac{1}{102}\)+....+\(\frac{1}{299}\)+\(\frac{1}{300}\)>\(\frac{2}{3}\)
câu 3: tính tích A=\(\frac{3}{4}\).\(\frac{8}{9}\).\(\frac{15}{16}\)...\(\frac{899}{900}\)
giúp mình giải 3 câu này với mấy bạn
Câu 1 :\(P=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).....\left(1-\frac{1}{99}\right)=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{98}{100}=\frac{1}{100}\)
vì bài này dài quá , tui đg vội nên bạn có thể tham khảo câu 2 ở đây : http://olm.vn/hoi-dap/question/55580.html
Chứng tỏ rằng: \(1< \frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+......+\frac{1}{16}+\frac{1}{17}< 2\)2
Tính tích \(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.......\frac{899}{900}\)
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.......\frac{899}{900}=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}......\frac{29.31}{30.30}=\frac{1.2.3.....29}{2.3.4......30}.\frac{3.4.5......31}{2.3.4......30}\)
\(=\frac{1}{30}.\frac{31}{2}=\frac{31}{60}\)
chứng tỏ rằng 1/5 + 1/6 + 1/7 + .....+ 1/17 < 2
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}......\frac{99}{100}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}......\frac{9.11}{10.10}\)
\(=\frac{\left(1.2.3...9\right).\left(3.4.5...11\right)}{\left(2.3.4...10\right).\left(2.3.4...10\right)}\)
\(=\frac{1}{10}.\frac{11}{2}=\frac{11}{20}\)
