Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(P=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+...+\frac{1}{121}+\frac{1}{144}\)
\(\Rightarrow P=\frac{1}{4}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{11^2}+\frac{1}{12^2}\)
Ta có : \(P< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10.11}+\frac{1}{11.12}\)
\(\Rightarrow P< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(\Rightarrow P< \frac{1}{4}+\frac{1}{2}-\frac{1}{12}\)
\(\Rightarrow P< \frac{2}{3}\left(đpcm\right)\)
\(P=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+...+\frac{1}{121}+\frac{1}{144}\)
\(P=\frac{1}{4}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{11^2}+\frac{1}{12^2}\)
Có : \(P< \frac{1}{4}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}+\frac{1}{11.12}\)
\(\Rightarrow P< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}\)
\(\Rightarrow P< \frac{1}{4}=\frac{1}{2}-\frac{1}{12}\)
\(\Rightarrow P< \frac{2}{3}\)( đpcm )
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{120}{121}=\frac{3.8.15...120}{4.9.16...121}\)
\(=\frac{\left(1.3\right).\left(2.4\right).\left(3.5\right)...\left(10.12\right)}{\left(2.2\right).\left(3.3\right).\left(4.4\right)...\left(11.11\right)}\)
\(=\frac{\left(1.2.3...10\right).\left(3.4.5...12\right)}{\left(2.3.4...11\right).\left(2.3.4...11\right)}=\frac{1.12}{11.2}=\frac{6}{11}\)
ta có :
A=\(\left(-\frac{3}{4}\right)\left(-\frac{8}{9}\right)\left(-\frac{15}{16}\right)...\left(-\frac{120}{121}\right)\)(có 10 số hạng)
= \(\frac{3\cdot8\cdot15\cdot...\cdot120}{4\cdot9\cdot16\cdot...\cdot121}=\frac{\left(1.3\right)\left(2\cdot4\right)\left(3\cdot5\right)\cdot...\cdot\left(10\cdot12\right)}{2^2\cdot3^2\cdot4^2\cdot...\cdot11^2}=\frac{\left(1\cdot2\cdot3\cdot...\cdot10\right)\left(3\cdot4\cdot5\cdot...\cdot12\right)}{\left(2\cdot3\cdot4\cdot..\cdot11\right)\left(2\cdot3\cdot4\cdot..\cdot11\right)}\)
=\(\frac{12}{11\cdot2}=\frac{12}{22}\)
A= 13;21;34
B= 37;70;135
C= 64;128;256
D= 22;29;37
E= 53;68;75
F= 127;255;511
G= 49;64;81
H= 324;841;2209
I= chịu
k cho mk nha!
a, A={x thuộc các số nguyên tố |2<hoặc bằng x<hoặc bằng 7}
oặc A={x thuộc R |(x^2-5*x+6)*(x^2-12*x+35)=0}
b,B={x thuộc Z | -3<hoặc bằng x<hoặc bằng 3}
c,C={5*x thuộc Z |-1<hoặc bằng x<hoặc bằng 3}
`= 3/4 . 8/9 . 15/16. ... . 143/144`
`= (1.2.3...12)/(2.3.4....12) . (3.4.....12.13)/(2.3....11.12)`
`= 1/12 . 13/2 = 13/24`
=(1-1/2)(1+1/2)(1-1/3)(1+1/3)*...*(1-1/12)(1+1/12)
=1/2*2/3*...*11/12*3/2*4/3*...*13/12
=1/12*13/2=13/24
Sửa đề : \(S=\left(1-\frac{1}{4}\right).\left(1-\frac{1}{9}\right).\left(1-\frac{1}{16}\right).....\left(1-\frac{1}{144}\right)\)
\(\Rightarrow S=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{143}{144}\)
\(\Rightarrow S=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{11.13}{12.12}\)
\(\Rightarrow S=\frac{1.2.3.....11}{2.3.4.....12}.\frac{3.4.5.....13}{2.3.4.....12}\)
\(\Rightarrow S=\frac{1}{12}.\frac{13}{2}\)
\(\Rightarrow S=\frac{13}{24}\)